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

"hhgggff" wrote in message
news
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.

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.

you can only jump a few more lightyears with 12 volt.

"hhgggff" wrote in message
news
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

"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

"hhgggff" wrote in
news
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) )

"chuckcar" wrote in message
...
"hhgggff" wrote in
news
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 .

"ian field" wrote in
:


"chuckcar" wrote in message
...
"hhgggff" wrote in
news
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) )

On Sat, 11 Jul 2009 21:50:55 +0000 (UTC), chuckcar
wrote:

"ian field" wrote in
:


"chuckcar" wrote in message
...
"hhgggff" wrote in
news
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

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.

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

I will look into a voltage regulator but will probably just get an old car
stereo as it's easier

tafankuverymuch

"hhgggff" wrote in message
news
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 ..

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
news
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) )

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.

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
news
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

"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
news
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.

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

"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.

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

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 $.

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

"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.

"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.

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

"Ian Jackson" wrote in message
news
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.

"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.

In message , ian field
writes

"Ian Jackson" wrote in message
news
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

"Ian Jackson" wrote in message
...
In message , ian field
writes

"Ian Jackson" wrote in message
news
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.

In message , Zootal
writes

"Ian Jackson" wrote in message
...
In message , ian field
writes

"Ian Jackson" wrote in message
news 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

"Ian Jackson" wrote in message
...
In message , Zootal
writes

"Ian Jackson" wrote in message
...
In message , ian field
writes

"Ian Jackson" wrote in message
news 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.

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
news 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

"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
news 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.

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
news 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

"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
news 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.

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

"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.

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

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

"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.