can i use a 12 volt battery on a 9 volt device ?
My friends dad is building a Tardis for his grandson.
We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? tafankuverymuch |
can i use a 12 volt battery on a 9 volt device ?
"hhgggff" wrote in message om... My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? tafankuverymuch You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. |
can i use a 12 volt battery on a 9 volt device ?
On Sat, 11 Jul 2009 14:33:07 GMT, "hhgggff" wrote:
My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? tafankuverymuch Go to your local radioshack and buy a 9v voltage regulator. 12 volts in 9 volts out. Get a 5 or 3 amp fuse so the player will only draw the amps it needs. |
can i use a 12 volt battery on a 9 volt device ?
you can only jump a few more lightyears with 12 volt.
"hhgggff" wrote in message om... My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? tafankuverymuch |
can i use a 12 volt battery on a 9 volt device ?
"richard" wrote in message ... On Sat, 11 Jul 2009 14:33:07 GMT, "hhgggff" wrote: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? tafankuverymuch Go to your local radioshack and buy a 9v voltage regulator. 12 volts in 9 volts out. Get a 5 or 3 amp fuse so the player will only draw the amps it needs. Richard You are implying the fuse will limit the current, which of course it will not. OK the regulator may have some current limiting in it, but a fuse will not be much good -- Bill Naylor www.electronworks.co.uk Electronic Kits for Education and Fun |
can i use a 12 volt battery on a 9 volt device ?
"hhgggff" wrote in
om: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? The tape motor will run at the wrong speed for one. What you need is a voltage regulator. A 7809 should do the trick. The 7800 series convert DC voltages. Granted a transformer does it, but these handle *much* wider voltage inputs and are tiny in comparison. A 7812 hooked to the +12v connection with the other connection hooked to a 200 microfarad capacitor should give you a rock solid +9v out. You'll need a piece of breadboard about 1" square to mount these on. http://en.wikipedia.org/wiki/78xx -- (setq (chuck nil) car(chuck) ) |
can i use a 12 volt battery on a 9 volt device ?
"chuckcar" wrote in message ... "hhgggff" wrote in om: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? The tape motor will run at the wrong speed for one. What you need is a voltage regulator. A 7809 should do the trick. The 7800 series convert DC voltages. Granted a transformer does it, but these handle *much* wider voltage inputs and are tiny in comparison. A 7812 hooked to the +12v connection with the other connection hooked to a 200 microfarad capacitor should give you a rock solid +9v out. You'll need a piece of breadboard about 1" square to mount these on. You should read up on a subject you know little about before giving advice to others. Even in older cassette recorders the motors had centrifugal speed governors and modern ones have an IC speed controller (otherwise they'd run slower and slower as the battery was used). Your mention of transformers is misleading, without a "chopper" circuit to convert DC into AC a transformer will burn out. The OP is best advised to ask people who are equipped to give accurate advice, such as the folk on News:sci.electronics.basics . |
can i use a 12 volt battery on a 9 volt device ?
"ian field" wrote in
: "chuckcar" wrote in message ... "hhgggff" wrote in om: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? The tape motor will run at the wrong speed for one. What you need is a voltage regulator. A 7809 should do the trick. The 7800 series convert DC voltages. Granted a transformer does it, but these handle *much* wider voltage inputs and are tiny in comparison. A 7812 hooked to the +12v connection with the other connection hooked to a 200 microfarad capacitor should give you a rock solid +9v out. You'll need a piece of breadboard about 1" square to mount these on. You should read up on a subject you know little about before giving advice to others. Considering the fact that my post is *far* more detailed than yours, you're hardly a person to make such a judgement = along with your errors detaied below. Even in older cassette recorders the motors had centrifugal speed governors and modern ones have an IC speed controller (otherwise they'd run slower and slower as the battery was used). And none of this would burn out with 1 1/3 times the voltage input? dubious. Your mention of transformers is misleading, without a "chopper" circuit to convert DC into AC a transformer will burn out. He's *using* a DC power source - a car battery. So a rectifier circuit or an analogue is completely unncessary The OP is best advised to ask people who are equipped to give accurate advice, such as the folk on News:sci.electronics.basics . One *minor* correction of my post however - that *should* have been a 7809 not a 7812. This I picked up immediately when I read your reply. A pretty basic error you didn't even notice. a 7812 converts *to* 12v whereas a 7809 converts to 5v. Hence their names. -- (setq (chuck nil) car(chuck) ) |
can i use a 12 volt battery on a 9 volt device ?
On Sat, 11 Jul 2009 21:50:55 +0000 (UTC), chuckcar
wrote: "ian field" wrote in : "chuckcar" wrote in message ... "hhgggff" wrote in om: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? The tape motor will run at the wrong speed for one. What you need is a voltage regulator. A 7809 should do the trick. The 7800 series convert DC voltages. Granted a transformer does it, but these handle *much* wider voltage inputs and are tiny in comparison. A 7812 hooked to the +12v connection with the other connection hooked to a 200 microfarad capacitor should give you a rock solid +9v out. You'll need a piece of breadboard about 1" square to mount these on. You should read up on a subject you know little about before giving advice to others. Considering the fact that my post is *far* more detailed than yours, you're hardly a person to make such a judgement = along with your errors detaied below. --- The fact that your post was more "detailed" than Ian's doesn't mean that his was wrong. --- Even in older cassette recorders the motors had centrifugal speed governors and modern ones have an IC speed controller (otherwise they'd run slower and slower as the battery was used). And none of this would burn out with 1 1/3 times the voltage input? dubious. --- Just conjecture without knowing more about the cassette recorder. --- Your mention of transformers is misleading, without a "chopper" circuit to convert DC into AC a transformer will burn out. He's *using* a DC power source - a car battery. So a rectifier circuit or an analogue is completely unncessary --- But, since his application runs on DC, your reference to a transformer was misleading since a transformer takes an AC input and supplies an AC output which must then be rectified, filtered, and possibly regulated before it can be used by the DC input device. If, by "transformer", you meant an AC to DC converter, then your terminology was wrong. --- The OP is best advised to ask people who are equipped to give accurate advice, such as the folk on News:sci.electronics.basics . One *minor* correction of my post however - that *should* have been a 7809 not a 7812. This I picked up immediately when I read your reply. A pretty basic error you didn't even notice. a 7812 converts *to* 12v whereas a 7809 converts to 5v. Hence their names. --- ??? A 7809 converts to 9V, not 5V. In addition, not knowing the current requirements of the recorder, your comment to build it on one square inch of breadboard (perfboard?) may well be inconsistent with the heat-sinking required for the regulator. And why the 200µF capacitor? Why not 10? Why not 1000? JF |
can i use a 12 volt battery on a 9 volt device ?
snip
A 7809 converts to 9V, not 5V. In addition, not knowing the current requirements of the recorder, your comment to build it on one square inch of breadboard (perfboard?) may well be inconsistent with the heat-sinking required for the regulator. And why the 200µF capacitor? Why not 10? Why not 1000? JF Forget the regulator. Connect 4 diodes in series with the power supply and that will get you into the ballpark. Just make sure the diodes can handle the max current the thing will draw. |
can i use a 12 volt battery on a 9 volt device ?
On Sun, 12 Jul 2009 11:58:58 -0700, "Zootal"
wrote: snip A 7809 converts to 9V, not 5V. In addition, not knowing the current requirements of the recorder, your comment to build it on one square inch of breadboard (perfboard?) may well be inconsistent with the heat-sinking required for the regulator. And why the 200µF capacitor? Why not 10? Why not 1000? JF Forget the regulator. Connect 4 diodes in series with the power supply and that will get you into the ballpark. --- But, (although it may not matter if the recorder's motor speed is maintained constant by the recorder itself) the output voltage will fall as the battery voltage does. JF |
Thanks
I will look into a voltage regulator but will probably just get an old car
stereo as it's easier tafankuverymuch |
can i use a 12 volt battery on a 9 volt device ?
"hhgggff" wrote in message om... My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? tafankuverymuch If the old player isnt that important, just wire it up and see what happens ;).. Worst case is that your throwing away something that probably shoulda been discarded years ago anyhow.. If it just plays too fast then you can make a slower recording ;P.. http://www.national.com/mpf/LM/LM2575.html throw in a diode, inductor, couple of caps, and a variable resistor and you have a up to 1amp source :/.. Course, it might just be cheaper and easier to find some crappy used phone charger or something that has a output close to 9v ;).. |
can i use a 12 volt battery on a 9 volt device ?
John Fields wrote in
: On Sat, 11 Jul 2009 21:50:55 +0000 (UTC), chuckcar wrote: "ian field" wrote in : "chuckcar" wrote in message ... "hhgggff" wrote in om: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? The tape motor will run at the wrong speed for one. What you need is a voltage regulator. A 7809 should do the trick. The 7800 series convert DC voltages. Granted a transformer does it, but these handle *much* wider voltage inputs and are tiny in comparison. A 7812 hooked to the +12v connection with the other connection hooked to a 200 microfarad capacitor should give you a rock solid +9v out. You'll need a piece of breadboard about 1" square to mount these on. You should read up on a subject you know little about before giving advice to others. Considering the fact that my post is *far* more detailed than yours, you're hardly a person to make such a judgement = along with your errors detaied below. --- The fact that your post was more "detailed" than Ian's doesn't mean that his was wrong. --- Even in older cassette recorders the motors had centrifugal speed governors and modern ones have an IC speed controller (otherwise they'd run slower and slower as the battery was used). And none of this would burn out with 1 1/3 times the voltage input? dubious. --- Just conjecture without knowing more about the cassette recorder. --- Your mention of transformers is misleading, without a "chopper" circuit to convert DC into AC a transformer will burn out. He's *using* a DC power source - a car battery. So a rectifier circuit or an analogue is completely unncessary --- But, since his application runs on DC, your reference to a transformer was misleading since a transformer takes an AC input and supplies an AC output which must then be rectified, filtered, and possibly regulated before it can be used by the DC input device. If, by "transformer", you meant an AC to DC converter, then your terminology was wrong. --- The OP is best advised to ask people who are equipped to give accurate advice, such as the folk on News:sci.electronics.basics . One *minor* correction of my post however - that *should* have been a 7809 not a 7812. This I picked up immediately when I read your reply. A pretty basic error you didn't even notice. a 7812 converts *to* 12v whereas a 7809 converts to 5v. Hence their names. --- ??? A 7809 converts to 9V, not 5V. Message-ID: In addition, not knowing the current requirements of the recorder, your comment to build it on one square inch of breadboard (perfboard?) may well be inconsistent with the heat-sinking required for the regulator. And why the 200µF capacitor? Why not 10? Too small to absorb the variations Why not 1000? Way too big for the voltage. -- (setq (chuck nil) car(chuck) ) |
can i use a 12 volt battery on a 9 volt device ?
And why the 200µF capacitor?
Why not 10? Too small to absorb the variations Why not 1000? Way too big for the voltage. -- With a 78xx, you don't need 200µF on the output. 10 is more then adequate - that is what the regulator is for. Be sure to put a 100nF or so cap in parallel with it. |
can i use a 12 volt battery on a 9 volt device ?
On Mon, 13 Jul 2009 02:34:52 +0000 (UTC), chuckcar
wrote: John Fields wrote in : On Sat, 11 Jul 2009 21:50:55 +0000 (UTC), chuckcar wrote: "ian field" wrote in : "chuckcar" wrote in message ... "hhgggff" wrote in om: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? The tape motor will run at the wrong speed for one. What you need is a voltage regulator. A 7809 should do the trick. The 7800 series convert DC voltages. Granted a transformer does it, but these handle *much* wider voltage inputs and are tiny in comparison. A 7812 hooked to the +12v connection with the other connection hooked to a 200 microfarad capacitor should give you a rock solid +9v out. You'll need a piece of breadboard about 1" square to mount these on. You should read up on a subject you know little about before giving advice to others. Considering the fact that my post is *far* more detailed than yours, you're hardly a person to make such a judgement = along with your errors detaied below. --- The fact that your post was more "detailed" than Ian's doesn't mean that his was wrong. --- Even in older cassette recorders the motors had centrifugal speed governors and modern ones have an IC speed controller (otherwise they'd run slower and slower as the battery was used). And none of this would burn out with 1 1/3 times the voltage input? dubious. --- Just conjecture without knowing more about the cassette recorder. --- Your mention of transformers is misleading, without a "chopper" circuit to convert DC into AC a transformer will burn out. He's *using* a DC power source - a car battery. So a rectifier circuit or an analogue is completely unncessary --- But, since his application runs on DC, your reference to a transformer was misleading since a transformer takes an AC input and supplies an AC output which must then be rectified, filtered, and possibly regulated before it can be used by the DC input device. If, by "transformer", you meant an AC to DC converter, then your terminology was wrong. --- The OP is best advised to ask people who are equipped to give accurate advice, such as the folk on News:sci.electronics.basics . One *minor* correction of my post however - that *should* have been a 7809 not a 7812. This I picked up immediately when I read your reply. A pretty basic error you didn't even notice. a 7812 converts *to* 12v whereas a 7809 converts to 5v. Hence their names. --- ??? A 7809 converts to 9V, not 5V. Message-ID: In addition, not knowing the current requirements of the recorder, your comment to build it on one square inch of breadboard (perfboard?) may well be inconsistent with the heat-sinking required for the regulator. And why the 200µF capacitor? Why not 10? Too small to absorb the variations --- What variations? --- Why not 1000? Way too big for the voltage. --- Why? JF |
can i use a 12 volt battery on a 9 volt device ?
"John Fields" wrote in message ... On Mon, 13 Jul 2009 02:34:52 +0000 (UTC), chuckcar wrote: John Fields wrote in m: On Sat, 11 Jul 2009 21:50:55 +0000 (UTC), chuckcar wrote: "ian field" wrote in : "chuckcar" wrote in message ... "hhgggff" wrote in om: My friends dad is building a Tardis for his grandson. We have rigged up the blue flashing light using a 12 volt car battery, we now need to sort out the sounds. I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Will it burn down the TARDIS if I use that instead ? The tape motor will run at the wrong speed for one. What you need is a voltage regulator. A 7809 should do the trick. The 7800 series convert DC voltages. Granted a transformer does it, but these handle *much* wider voltage inputs and are tiny in comparison. A 7812 hooked to the +12v connection with the other connection hooked to a 200 microfarad capacitor should give you a rock solid +9v out. You'll need a piece of breadboard about 1" square to mount these on. You should read up on a subject you know little about before giving advice to others. Considering the fact that my post is *far* more detailed than yours, you're hardly a person to make such a judgement = along with your errors detaied below. --- The fact that your post was more "detailed" than Ian's doesn't mean that his was wrong. --- Even in older cassette recorders the motors had centrifugal speed governors and modern ones have an IC speed controller (otherwise they'd run slower and slower as the battery was used). And none of this would burn out with 1 1/3 times the voltage input? dubious. --- Just conjecture without knowing more about the cassette recorder. --- Your mention of transformers is misleading, without a "chopper" circuit to convert DC into AC a transformer will burn out. He's *using* a DC power source - a car battery. So a rectifier circuit or an analogue is completely unncessary --- But, since his application runs on DC, your reference to a transformer was misleading since a transformer takes an AC input and supplies an AC output which must then be rectified, filtered, and possibly regulated before it can be used by the DC input device. If, by "transformer", you meant an AC to DC converter, then your terminology was wrong. --- The OP is best advised to ask people who are equipped to give accurate advice, such as the folk on News:sci.electronics.basics . One *minor* correction of my post however - that *should* have been a 7809 not a 7812. This I picked up immediately when I read your reply. A pretty basic error you didn't even notice. a 7812 converts *to* 12v whereas a 7809 converts to 5v. Hence their names. --- ??? A 7809 converts to 9V, not 5V. Message-ID: In addition, not knowing the current requirements of the recorder, your comment to build it on one square inch of breadboard (perfboard?) may well be inconsistent with the heat-sinking required for the regulator. And why the 200µF capacitor? Why not 10? Too small to absorb the variations --- What variations? --- Why not 1000? Way too big for the voltage. --- Why? A 1000µF cap on the output of a 78xx or 79xx regulator will work just fine. It's just not necessary, but I don't think the power up surge of charging the cap would hurt the regulator any, and once it's charged the circuit should work fine. If you are making a half million gizmos, you would want to use the smallest cap you can get away with. Use a cap that cost a dollar less, and you just saved a half million dollars. |
UPDATE : can i use a 12 volt battery on a 9 volt device ?
I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Well, In the end I just plugged it in and it works,correct speed. tafankuverymuch |
UPDATE : can i use a 12 volt battery on a 9 volt device ?
"hhgggff" wrote in message om... I was thinking of using an old cars cassette tape, but I have an old 9 volt portable cassette handy. Well, In the end I just plugged it in and it works,correct speed. tafankuverymuch If you're running a 9V cassette player from a car battery, don't run it at full volume - the audio output might be the weak link. |
can i use a 12 volt battery on a 9 volt device ?
You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. R |
can i use a 12 volt battery on a 9 volt device ?
Roger Dewhurst wrote:
You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. |
can i use a 12 volt battery on a 9 volt device ?
In message , rf
writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. Exactly one 7809 for less than a buck. No other circutry required. Perfect regulation. |
can i use a 12 volt battery on a 9 volt device ?
In message , ian field
writes "Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. |
can i use a 12 volt battery on a 9 volt device ?
"rf" wrote in message ... "Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. Exactly one 7809 for less than a buck. No other circutry required. Perfect regulation. Not *quite* no extra circuitry, they require decoupling capacitors on the input and output otherwise they can break into oscillation. If a 3-terminal regulator feeds a circuit with a large supply decoupling electrolytic (possibility of stored charge feeding backwards through the regulator when the input voltage is switched off) its advisable to strap a diode between the input and output terminals, otherwise the regulator can be damaged - not conducting in the normal condition of input voltage being higher than the output but conducts if the output tries to go higher than the input. |
can i use a 12 volt battery on a 9 volt device ?
In message , ian field
writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian It won't even be noticeable. The capacitors won't charge up that high to start with, and they don't "dump" into the circuit, they just quickly discharge down to the lower voltage that is present at the output of the last diode - how fast this happens depends on the size of the caps and the load. I wouldn't even call it a surge. A resistor from the last diode to ground will prevent them from charging more then a few tenths of a volt and is a good idea. And the capacitor doesn't need to be that big. I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible to failure. Use 2 or 3 amp diodes, they are cheap and readily available. They are bigger and will run cooler and won't fail as easily. Or use two strings of 1n400x in series, that is good enough for an app like this. So - 12v - diode - diode - diode - diode - ~9.2v Output From the output run a 470 ohm half watt resistor through a standard 1/4" LED to ground (I like the LED so you can see when the circuit is on). In parallel to this connect a capacitor, say 100uF. Presto, ~9.2V, adequately regulated, and minimal voltage increase when the load is off. |
can i use a 12 volt battery on a 9 volt device ?
In message , Zootal
writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian It won't even be noticeable. The capacitors won't charge up that high to start with, and they don't "dump" into the circuit, they just quickly discharge down to the lower voltage that is present at the output of the last diode - how fast this happens depends on the size of the caps and the load. I wouldn't even call it a surge. A resistor from the last diode to ground will prevent them from charging more then a few tenths of a volt and is a good idea. Indeed. A bleed of a few mA will prevent the off-load voltage rising to 12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W will do). And the capacitor doesn't need to be that big. Which capacitor do you mean? I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible to failure. Use 2 or 3 amp diodes, they are cheap and readily available. They are bigger and will run cooler and won't fail as easily. 1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will run a bit warm, so maybe a physically larger (higher current) diode might be better. But it depends on how much current the Tardis takes! Or use two strings of 1n400x in series, that is good enough for an app like this. So - 12v - diode - diode - diode - diode - ~9.2v Output That's only one string of diodes in series. Did you mean parallel? If so, no, you shouldn't parallel diodes. As you suggest, use higher current diodes. From the output run a 470 ohm half watt resistor through a standard 1/4" LED to ground (I like the LED so you can see when the circuit is on). In parallel to this connect a capacitor, say 100uF. Presto, ~9.2V, adequately regulated, and minimal voltage increase when the load is off. The cap isn't a bad idea, but 470 ohms will give you around 14mA through the LED (allowing 2V for the LED). Anything between 470 ohms and 1k should be fine. -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , Zootal writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian It won't even be noticeable. The capacitors won't charge up that high to start with, and they don't "dump" into the circuit, they just quickly discharge down to the lower voltage that is present at the output of the last diode - how fast this happens depends on the size of the caps and the load. I wouldn't even call it a surge. A resistor from the last diode to ground will prevent them from charging more then a few tenths of a volt and is a good idea. Indeed. A bleed of a few mA will prevent the off-load voltage rising to 12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W will do). And the capacitor doesn't need to be that big. Which capacitor do you mean? I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible to failure. Use 2 or 3 amp diodes, they are cheap and readily available. They are bigger and will run cooler and won't fail as easily. 1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will run a bit warm, so maybe a physically larger (higher current) diode might be better. But it depends on how much current the Tardis takes! Or use two strings of 1n400x in series, that is good enough for an app like this. So - 12v - diode - diode - diode - diode - ~9.2v Output That's only one string of diodes in series. Did you mean parallel? If so, no, you shouldn't parallel diodes. As you suggest, use higher current diodes. Actually if you have two strings each having the same number of diodes and put the two strings in parallel it doesn't matter, when you have a few or more diodes in series as the variation in Vf for each diode averages out. |
can i use a 12 volt battery on a 9 volt device ?
In message , ian field
writes "Ian Jackson" wrote in message ... In message , Zootal writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message .. . In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian It won't even be noticeable. The capacitors won't charge up that high to start with, and they don't "dump" into the circuit, they just quickly discharge down to the lower voltage that is present at the output of the last diode - how fast this happens depends on the size of the caps and the load. I wouldn't even call it a surge. A resistor from the last diode to ground will prevent them from charging more then a few tenths of a volt and is a good idea. Indeed. A bleed of a few mA will prevent the off-load voltage rising to 12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W will do). And the capacitor doesn't need to be that big. Which capacitor do you mean? I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible to failure. Use 2 or 3 amp diodes, they are cheap and readily available. They are bigger and will run cooler and won't fail as easily. 1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will run a bit warm, so maybe a physically larger (higher current) diode might be better. But it depends on how much current the Tardis takes! Or use two strings of 1n400x in series, that is good enough for an app like this. So - 12v - diode - diode - diode - diode - ~9.2v Output That's only one string of diodes in series. Did you mean parallel? If so, no, you shouldn't parallel diodes. As you suggest, use higher current diodes. Actually if you have two strings each having the same number of diodes and put the two strings in parallel it doesn't matter, when you have a few or more diodes in series as the variation in Vf for each diode averages out. Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You don't see many circuits with paralleled diodes (stringed or otherwise). -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , Zootal writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message . .. In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian It won't even be noticeable. The capacitors won't charge up that high to start with, and they don't "dump" into the circuit, they just quickly discharge down to the lower voltage that is present at the output of the last diode - how fast this happens depends on the size of the caps and the load. I wouldn't even call it a surge. A resistor from the last diode to ground will prevent them from charging more then a few tenths of a volt and is a good idea. Indeed. A bleed of a few mA will prevent the off-load voltage rising to 12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W will do). And the capacitor doesn't need to be that big. Which capacitor do you mean? I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible to failure. Use 2 or 3 amp diodes, they are cheap and readily available. They are bigger and will run cooler and won't fail as easily. 1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will run a bit warm, so maybe a physically larger (higher current) diode might be better. But it depends on how much current the Tardis takes! Or use two strings of 1n400x in series, that is good enough for an app like this. So - 12v - diode - diode - diode - diode - ~9.2v Output That's only one string of diodes in series. Did you mean parallel? If so, no, you shouldn't parallel diodes. As you suggest, use higher current diodes. Actually if you have two strings each having the same number of diodes and put the two strings in parallel it doesn't matter, when you have a few or more diodes in series as the variation in Vf for each diode averages out. Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You don't see many circuits with paralleled diodes (stringed or otherwise). -- Ian Larger power supplies will have diodes in parallel, and they will even put the output trannies in parallel. You can get away with it with diodes but you have to be carefull to use the same type, preferably from the same batch. Otherwise you end up with some saturated and others barely turned on, and you get a high failure rate. |
can i use a 12 volt battery on a 9 volt device ?
In message , Zootal
writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , Zootal writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message .. . In message , ian field writes "Ian Jackson" wrote in message .. . In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian It won't even be noticeable. The capacitors won't charge up that high to start with, and they don't "dump" into the circuit, they just quickly discharge down to the lower voltage that is present at the output of the last diode - how fast this happens depends on the size of the caps and the load. I wouldn't even call it a surge. A resistor from the last diode to ground will prevent them from charging more then a few tenths of a volt and is a good idea. Indeed. A bleed of a few mA will prevent the off-load voltage rising to 12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W will do). And the capacitor doesn't need to be that big. Which capacitor do you mean? I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible to failure. Use 2 or 3 amp diodes, they are cheap and readily available. They are bigger and will run cooler and won't fail as easily. 1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will run a bit warm, so maybe a physically larger (higher current) diode might be better. But it depends on how much current the Tardis takes! Or use two strings of 1n400x in series, that is good enough for an app like this. So - 12v - diode - diode - diode - diode - ~9.2v Output That's only one string of diodes in series. Did you mean parallel? If so, no, you shouldn't parallel diodes. As you suggest, use higher current diodes. Actually if you have two strings each having the same number of diodes and put the two strings in parallel it doesn't matter, when you have a few or more diodes in series as the variation in Vf for each diode averages out. Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You don't see many circuits with paralleled diodes (stringed or otherwise). -- Ian Larger power supplies will have diodes in parallel, and they will even put the output trannies in parallel. You can get away with it with diodes but you have to be carefull to use the same type, preferably from the same batch. Otherwise you end up with some saturated and others barely turned on, and you get a high failure rate. During 50 years 'in electronics', I can't immediately recall seeing any power supplies with diodes in parallel. I'm not saying it's never done but, if it is, there should also be some current-balancing resistance in each path. This could be low-value resistors, or even the resistance of the secondary windings of paralleled transformers which feed each set of diodes. I don't think that having to use diodes from the same batch is a very good design criterion! -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , Zootal writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message ... In message , Zootal writes "Ian Jackson" wrote in message ... In message , ian field writes "Ian Jackson" wrote in message . .. In message , ian field writes "Ian Jackson" wrote in message . .. In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. More than good enough for the job. -- Ian The forward conduction knee curve on diodes isn't *that* sharp, depending on current draw and rating of the diode the drop can be as low as 0.55V and as high as 1.1V. For most 'normal' Si diodes, that isn't really the case. The actual voltage drop does, of course, increase with current, but at 'sensible' currents, you can reckon on around 0.65V per diode. How much current is the Tardis toy going to take? 1A max? 4 or 5 1N4000-series diodes should work fine in this application. I've used this non-elegant 'KISS' technique on several occasions, and haven't found any problems. -- Ian A potential danger with a cassette recorder is the difference in current draw between motor on and motor off. In the condition of low current draw (and low diode drop) supply decoupling electrolytic capacitors can charge to a higher voltage which is then dumped into the circuit when switched to play. True, true. But I reckon that a momentary short burst of a rapidly-decaying additional 3V won't hurt too much. -- Ian It won't even be noticeable. The capacitors won't charge up that high to start with, and they don't "dump" into the circuit, they just quickly discharge down to the lower voltage that is present at the output of the last diode - how fast this happens depends on the size of the caps and the load. I wouldn't even call it a surge. A resistor from the last diode to ground will prevent them from charging more then a few tenths of a volt and is a good idea. Indeed. A bleed of a few mA will prevent the off-load voltage rising to 12V. 9V at 10mA would require 900 ohms (say 1k), 90mW (so even a 1/4W will do). And the capacitor doesn't need to be that big. Which capacitor do you mean? I would *not* use 1n400x diodes. ~1 amp will make them hot and susceptible to failure. Use 2 or 3 amp diodes, they are cheap and readily available. They are bigger and will run cooler and won't fail as easily. 1A through diodes dropping 0.6V means 600mW per diode. 1N4000-type will run a bit warm, so maybe a physically larger (higher current) diode might be better. But it depends on how much current the Tardis takes! Or use two strings of 1n400x in series, that is good enough for an app like this. So - 12v - diode - diode - diode - diode - ~9.2v Output That's only one string of diodes in series. Did you mean parallel? If so, no, you shouldn't parallel diodes. As you suggest, use higher current diodes. Actually if you have two strings each having the same number of diodes and put the two strings in parallel it doesn't matter, when you have a few or more diodes in series as the variation in Vf for each diode averages out. Possibly. Possibly not. Murphy and his Laws move in mysterious ways. You don't see many circuits with paralleled diodes (stringed or otherwise). -- Ian Larger power supplies will have diodes in parallel, and they will even put the output trannies in parallel. You can get away with it with diodes but you have to be carefull to use the same type, preferably from the same batch. Otherwise you end up with some saturated and others barely turned on, and you get a high failure rate. During 50 years 'in electronics', I can't immediately recall seeing any power supplies with diodes in parallel. I'm not saying it's never done but, if it is, there should also be some current-balancing resistance in each path. This could be low-value resistors, or even the resistance of the secondary windings of paralleled transformers which feed each set of diodes. I don't think that having to use diodes from the same batch is a very good design criterion! -- Ian A company I used to work for used paralleled rectifiers in ultrasonic cleaner generators up to 1500W, I suspect reliability might not have been all it could have been but they weren't exactly dropping like flies. |
can i use a 12 volt battery on a 9 volt device ?
In message , ian field
writes A company I used to work for used paralleled rectifiers in ultrasonic cleaner generators up to 1500W, I suspect reliability might not have been all it could have been but they weren't exactly dropping like flies. That's nearly enough to power a REAL Tardis, let alone a toy one! Was there any current equalisation (obvious, or 'hidden')? -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , ian field writes A company I used to work for used paralleled rectifiers in ultrasonic cleaner generators up to 1500W, I suspect reliability might not have been all it could have been but they weren't exactly dropping like flies. That's nearly enough to power a REAL Tardis, let alone a toy one! Was there any current equalisation (obvious, or 'hidden')? -- Ian Absolutely none - the (pair of) 2 press fit rectifiers were pressed into a small square slab of aluminium that was bolted to an elongated cube of more aluminium that pressed into the middle of a forced air cooled heatsink assembly, each pair of rectifiers was linked together with a triangle of 16SWG TC wire, the 2 ends that came together were crimped into a solder tag whose "eye" was the solder point for the big fat wires from the mains transformer. |
can i use a 12 volt battery on a 9 volt device ?
Larger power supplies will have diodes in parallel, and they will even put the output trannies in parallel. You can get away with it with diodes but you have to be carefull to use the same type, preferably from the same batch. Otherwise you end up with some saturated and others barely turned on, and you get a high failure rate. During 50 years 'in electronics', I can't immediately recall seeing any power supplies with diodes in parallel. I'm not saying it's never done but, if it is, there should also be some current-balancing resistance in each path. This could be low-value resistors, or even the resistance of the secondary windings of paralleled transformers which feed each set of diodes. I don't think that having to use diodes from the same batch is a very good design criterion! -- Ian Having diodes from the same batch is something we do at home when we make our own klugey experimental stuff. Of course they aren't going to do this in production equipment! I only have 25 years against your 50, but I've seen so many things I can't immediately recall seeing a lot of the things I've seen :). At one time I was doing the work on our power supplies, and we had a lot. And of course there were resistors in series with each diode, the purpose of which is to minimize variations in the operating parameters of the diodes themselves. I never bothered to test them to see how equally they shared the load, but the things worked. I've never had to replace one of the diodes themselves. It's possible that the diodes would turn on such that one didn't turn on until another approached saturation. So long as they don't burn out I guess it would work OK. Hmm...Let's use big doides mounted in a huge heat sink because some are going to be near saturation while others are idling. I've seen enough klugey designes that I would not be surprised if that really was how it worked :) |
can i use a 12 volt battery on a 9 volt device ?
In message , Any one
writes Ian Jackson wrote on 29-Jul-09 14:40 : In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. BWHAHAHAHAHAHAHAHA!!!! *cough* *wheeze* *wheeze* -if- junction temperature remains constant. -if- load current remains constant. -if- each selected diode returns precisely 0.6v @ 'load current'. More than good enough for the job. AHAHAHAHAHAHAHA!!! -if- source voltage remains constant -if- there aren't any ICs that could suffer over-voltage damage -if- you're as cheap and gullible as a Republi**** lemme guess -- you're a close relative of teh chucktard? Lemme guess.... You have a degree in Electrical Over-Engineering? -- Ian |
can i use a 12 volt battery on a 9 volt device ?
"Ian Jackson" wrote in message ... In message , Any one writes Ian Jackson wrote on 29-Jul-09 14:40 : In message , rf writes Roger Dewhurst wrote: You can get simple to use regulator chips that drop the voltage down, you only need a couple of components to make a working voltage regulator. Ask on News:sci.electronics.basic - the folk on there are usually very helpful and should explain all you need. Why not just drop the voltage through a few diodes? Very simple. Very cheap. A *few* diodes at a couple of ten cents per each. A single 7809 for less that a $. More like "a *few* diodes at a couple of cents per each". 3V to drop = 5 diodes @ 0.6V per diode. No other circuitry required. BWHAHAHAHAHAHAHAHA!!!! *cough* *wheeze* *wheeze* -if- junction temperature remains constant. -if- load current remains constant. -if- each selected diode returns precisely 0.6v @ 'load current'. More than good enough for the job. AHAHAHAHAHAHAHA!!! -if- source voltage remains constant -if- there aren't any ICs that could suffer over-voltage damage -if- you're as cheap and gullible as a Republi**** lemme guess -- you're a close relative of teh chucktard? Lemme guess.... You have a degree in Electrical Over-Engineering? -- Ian The Nazi's invaded Russia with everything over engineered and nothing winterised. |
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