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Default Looking to drop 6 volts dc to 4.5 volts dc

I'm looking to add a mod to my pinball machine with a couple of led's and
need to drop 6 volts dc down to 4.5 volts dc. Can anyone suggest which
resistor to use?

Thanks.

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Default Looking to drop 6 volts dc to 4.5 volts dc

On Thu, 28 Sep 2006 18:41:29 +0000, .D.E wrote:

I'm looking to add a mod to my pinball machine with a couple of led's and
need to drop 6 volts dc down to 4.5 volts dc. Can anyone suggest which
resistor to use?


R = (Vsupply - Vled) / Iled

Good Luck!
Rich


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Default Looking to drop 6 volts dc to 4.5 volts dc


".D.E" schreef in bericht
news:tHUSg.64297$5R2.18444@pd7urf3no...
I'm looking to add a mod to my pinball machine with a couple of led's and
need to drop 6 volts dc down to 4.5 volts dc. Can anyone suggest which
resistor to use?

Thanks.

--
_____________________________
Later.
Have a better one.
D.E.
To email me back, remove "forget.the.spam"


No, due to lack of information. Suppose the LEDs do 4.5V when lighting, what
current do they use? You'd realize that LEDs are current driven devices so
the current can easily double while the voltage hardly changes.

petrus bitbyter


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Default Looking to drop 6 volts dc to 4.5 volts dc

In article ,
"petrus bitbyter" wrote:

".D.E" schreef in bericht
news:tHUSg.64297$5R2.18444@pd7urf3no...
I'm looking to add a mod to my pinball machine with a couple of led's and
need to drop 6 volts dc down to 4.5 volts dc. Can anyone suggest which
resistor to use?

Thanks.

--
_____________________________
Later.
Have a better one.
D.E.
To email me back, remove "forget.the.spam"


No, due to lack of information. Suppose the LEDs do 4.5V when lighting, what
current do they use? You'd realize that LEDs are current driven devices so
the current can easily double while the voltage hardly changes.

petrus bitbyter



If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.
Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.

Al
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Default Looking to drop 6 volts dc to 4.5 volts dc


"Al" schreef in bericht
...
In article ,
"petrus bitbyter" wrote:

".D.E" schreef in bericht
news:tHUSg.64297$5R2.18444@pd7urf3no...
I'm looking to add a mod to my pinball machine with a couple of led's
and
need to drop 6 volts dc down to 4.5 volts dc. Can anyone suggest which
resistor to use?

Thanks.

--
_____________________________
Later.
Have a better one.
D.E.
To email me back, remove "forget.the.spam"


No, due to lack of information. Suppose the LEDs do 4.5V when lighting,
what
current do they use? You'd realize that LEDs are current driven devices
so
the current can easily double while the voltage hardly changes.

petrus bitbyter



If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.
Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.

Al


That diodes will work when you have a resistive load. But neither LEDs nor
other diodes behave like resistors. Once mo You need to regulate the
current, not the voltage. If the LED has been specified for 10mA at 4.5V
you'll need a series resistor of (6-4.5)/10=0.15k that's 150 Ohm. Most LEDs
are specified for 20mA or more. FAIK only low power LEDs require less.

petrus bitbyter




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Default Looking to drop 6 volts dc to 4.5 volts dc

On 2006-09-29, Al wrote:

If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.


yeah, if the current doubles the voltage may increase maybe 10%
how's that going to help?

Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.


it seems to me that regular resistors would work better,

--

Bye.
Jasen
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Default Looking to drop 6 volts dc to 4.5 volts dc

In article , jasen
wrote:

On 2006-09-29, Al wrote:

If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.


yeah, if the current doubles the voltage may increase maybe 10%
how's that going to help?

Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.


it seems to me that regular resistors would work better,


If your current doubled for some reason, you have other big problems.

Al
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Default Looking to drop 6 volts dc to 4.5 volts dc

On Sat, 30 Sep 2006 14:12:59 GMT, Al wrote:

In article , jasen
wrote:

On 2006-09-29, Al wrote:

If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.


yeah, if the current doubles the voltage may increase maybe 10%
how's that going to help?

Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.


it seems to me that regular resistors would work better,


If your current doubled for some reason, you have other big problems.


---
I'm not trying to be rude, but the problem would rear its ugly head
if your suggestion was followed. Take a look at what it takes to
double the current through pretty much _any_ diode once it's on the
far side of the Vf knee and you should be able to see that your
suggestion, if followed, would be inviting disaster. The LED(s)
need to be fed from a constant current source. Period. End of
story.


--
John Fields
Professional Circuit Designer
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Default Looking to drop 6 volts dc to 4.5 volts dc

In article ,
John Fields wrote:

On Sat, 30 Sep 2006 14:12:59 GMT, Al wrote:

In article , jasen
wrote:

On 2006-09-29, Al wrote:

If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.

yeah, if the current doubles the voltage may increase maybe 10%
how's that going to help?

Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.

it seems to me that regular resistors would work better,


If your current doubled for some reason, you have other big problems.


---
I'm not trying to be rude, but the problem would rear its ugly head
if your suggestion was followed. Take a look at what it takes to
double the current through pretty much _any_ diode once it's on the
far side of the Vf knee and you should be able to see that your
suggestion, if followed, would be inviting disaster. The LED(s)
need to be fed from a constant current source. Period. End of
story.


A series resistor is not a constant current source.

And, this method is not for manufacturing millions of items. It's for
solving a special home problem. Like I said, I do it and it works for
me. No, I don't bake the circuit nor do I freeze it. And as I mentioned
above, diodes with soft knees should be used.

Al
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Default Looking to drop 6 volts dc to 4.5 volts dc

On Sat, 30 Sep 2006 19:45:31 GMT, Al wrote:

In article ,
John Fields wrote:

On Sat, 30 Sep 2006 14:12:59 GMT, Al wrote:

In article , jasen
wrote:

On 2006-09-29, Al wrote:

If your LED is specified at 10mA, then get two diodes whose forward
voltage is 0.75V at 10ma. Put them in series with your LED. If the
supply voltage goes up and the current goes up, the forward voltage of
the diodes will go up also thereby dropping more voltage across them.

yeah, if the current doubles the voltage may increase maybe 10%
how's that going to help?

Thus the dynamic resistance of the dropping diodes serves as a voltage
regulator. Really good diodes with a low dynamic resistance won't do.

it seems to me that regular resistors would work better,

If your current doubled for some reason, you have other big problems.


---
I'm not trying to be rude, but the problem would rear its ugly head
if your suggestion was followed. Take a look at what it takes to
double the current through pretty much _any_ diode once it's on the
far side of the Vf knee and you should be able to see that your
suggestion, if followed, would be inviting disaster. The LED(s)
need to be fed from a constant current source. Period. End of
story.


A series resistor is not a constant current source.


---
No but it's a hell of a lot closer than forward biased diodes.
---

And, this method is not for manufacturing millions of items. It's for
solving a special home problem. Like I said, I do it and it works for
me. No, I don't bake the circuit nor do I freeze it. And as I mentioned
above, diodes with soft knees should be used.


---
Ridiculous. Take a diode and straighten out its knee and what have
you got? a resistor. The point is that the softest knee you can
find won't help you regulate (or even limit) the current into the
LEDs if the source voltage varies.



--
John Fields
Professional Circuit Designer


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Default Looking to drop 6 volts dc to 4.5 volts dc

In article ,
John Fields wrote:


big snip

---

And, this method is not for manufacturing millions of items. It's for
solving a special home problem. Like I said, I do it and it works for
me. No, I don't bake the circuit nor do I freeze it. And as I mentioned
above, diodes with soft knees should be used.


---
Ridiculous. Take a diode and straighten out its knee and what have
you got? a resistor. The point is that the softest knee you can
find won't help you regulate (or even limit) the current into the
LEDs if the source voltage varies.



The poster said his source voltage was 6 volts. Since he was using it to
drive LEDs, I presumed, perhaps falsely, that it was a steady DC. He
should have specified a range, such as 6Vdc +/- 0.5V.

Is it a true DC as derived from a dry cell? Is it pulsating DC derived
from either a half-wave or a full wave diode bridge? Does he have an LC
filter on the output of the bridge? Or is it just a big rectifier across
the bridge? Is there a linear regulator or a switcher involved? As
someone else in this thread had suggested, it might be the output of a
6.3V filament transformer that is rectified.

All of these factors, and probably many others, would have to be
considered.

The brightness of an LED is a function of the current through it.
Typically it specified to have a certain light output level at a
specified current. You may increase or decrease the current as you will.
The lifetime of the LED will depend on the current as will its light
output. Even the specified current is just a normalization of the
readings from a large sample. Your specfic LED may need more or less
current for the specified light output.

So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.

The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0

Two diodes in series would give me the approximate 1.4V drop close to
what I would need.

Al
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Default Looking to drop 6 volts dc to 4.5 volts dc

On Sun, 01 Oct 2006 15:31:55 GMT, Al wrote:

In article ,
John Fields wrote:


big snip

---

And, this method is not for manufacturing millions of items. It's for
solving a special home problem. Like I said, I do it and it works for
me. No, I don't bake the circuit nor do I freeze it. And as I mentioned
above, diodes with soft knees should be used.


---
Ridiculous. Take a diode and straighten out its knee and what have
you got? a resistor. The point is that the softest knee you can
find won't help you regulate (or even limit) the current into the
LEDs if the source voltage varies.



The poster said his source voltage was 6 volts. Since he was using it to
drive LEDs, I presumed, perhaps falsely, that it was a steady DC. He
should have specified a range, such as 6Vdc +/- 0.5V.

Is it a true DC as derived from a dry cell? Is it pulsating DC derived
from either a half-wave or a full wave diode bridge? Does he have an LC
filter on the output of the bridge? Or is it just a big rectifier across
the bridge? Is there a linear regulator or a switcher involved? As
someone else in this thread had suggested, it might be the output of a
6.3V filament transformer that is rectified.

All of these factors, and probably many others, would have to be
considered.

The brightness of an LED is a function of the current through it.
Typically it specified to have a certain light output level at a
specified current. You may increase or decrease the current as you will.
The lifetime of the LED will depend on the current as will its light
output. Even the specified current is just a normalization of the
readings from a large sample. Your specfic LED may need more or less
current for the specified light output.

So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.

The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0

Two diodes in series would give me the approximate 1.4V drop close to
what I would need.



---
The problem with thinking that it's the voltage which is what must
be controlled is that it isn't. What must be controlled/limited is
the current.

LEDs are specified to operate at a certain current, and when that
current is pumped through them then the voltage dropped across the
LED will vary according to the range given in the data sheet.

For example, take an LED rated at 20 mA with a minimum voltage of 2V
and a maximum voltage of 3V across it.

That means that with 20mA through the diode the voltage dropped
across it can vary from 2 to 3V. It _doesn't_ mean that if you put
a voltage source across the LED and crank it up to 3V everything
will be OK. More than likely you'll toast the LED.

With that in mind, the proper way to limit the current through an
LED is to subtract the _maximum_ LED Vf from the supply and then to
divide that by the LED current.

For example, let's say you have a white LED rated for 20mA with a Vf
somewhere between 3.5V and 4.5V and you have a 12VDC supply.

The current limiting resistor would be:


Vcc - Vf (max) 12V - 4.5V
Rs = ---------------- = ------------ = 375 ohms
If 0.02A

The closest standard 5% resistors are 360 and 390 ohms, but to err
on the side of caution the 390 ohm resistor should be chosen.


--
John Fields
Professional Circuit Designer
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Al wrote:


The poster said his source voltage was 6 volts. Since he was using it to
drive LEDs, I presumed, perhaps falsely, that it was a steady DC. He
should have specified a range, such as 6Vdc +/- 0.5V.

Is it a true DC as derived from a dry cell? Is it pulsating DC derived
from either a half-wave or a full wave diode bridge? Does he have an LC
filter on the output of the bridge? Or is it just a big rectifier across
the bridge? Is there a linear regulator or a switcher involved? As
someone else in this thread had suggested, it might be the output of a
6.3V filament transformer that is rectified.

All of these factors, and probably many others, would have to be
considered.

The brightness of an LED is a function of the current through it.
Typically it specified to have a certain light output level at a
specified current. You may increase or decrease the current as you will.
The lifetime of the LED will depend on the current as will its light
output. Even the specified current is just a normalization of the
readings from a large sample. Your specfic LED may need more or less
current for the specified light output.

So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.

The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0

Two diodes in series would give me the approximate 1.4V drop close to
what I would need.


That doesn't sound like it would work. Now you have an extra 0.1 V on
the LED. The current could be easily double (or more) what is wanted.
"Close to the right voltage", in an LED or any diode, just doesn't cut
it in terms of limiting the current.

And if the source voltage is NOT a constant, exact 6V, it's even worse.

Resistors are the standard, simple easy way to limit current through an
LED.

Mark

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Default Looking to drop 6 volts dc to 4.5 volts dc

On 1 Oct 2006 10:04:49 -0700, "redbelly" wrote:


Al wrote:


The poster said his source voltage was 6 volts. Since he was using it to
drive LEDs, I presumed, perhaps falsely, that it was a steady DC. He
should have specified a range, such as 6Vdc +/- 0.5V.

Is it a true DC as derived from a dry cell? Is it pulsating DC derived
from either a half-wave or a full wave diode bridge? Does he have an LC
filter on the output of the bridge? Or is it just a big rectifier across
the bridge? Is there a linear regulator or a switcher involved? As
someone else in this thread had suggested, it might be the output of a
6.3V filament transformer that is rectified.

All of these factors, and probably many others, would have to be
considered.

The brightness of an LED is a function of the current through it.
Typically it specified to have a certain light output level at a
specified current. You may increase or decrease the current as you will.
The lifetime of the LED will depend on the current as will its light
output. Even the specified current is just a normalization of the
readings from a large sample. Your specfic LED may need more or less
current for the specified light output.

So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.

The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0

Two diodes in series would give me the approximate 1.4V drop close to
what I would need.


That doesn't sound like it would work. Now you have an extra 0.1 V on
the LED. The current could be easily double (or more) what is wanted.
"Close to the right voltage", in an LED or any diode, just doesn't cut
it in terms of limiting the current.

And if the source voltage is NOT a constant, exact 6V, it's even worse.

Resistors are the standard, simple easy way to limit current through an
LED.

Mark


Has anyone mentioned to just use a current source? Even 0.5V headroom
would be adequate.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
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Default Looking to drop 6 volts dc to 4.5 volts dc

On 2006-10-01, Al wrote:

So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.


Typically leds are specified with a voltage range for their limit current.
get the curren right and the voltage will be somewhere in that range, it
depends on the device and the environment.

If that particulasr led is 4.4V with 10mA flowing through it it could with
4.5V the current could be 30mA or more.

The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0


And when they get warm, it drops, this can lead to thermal runaway.

ordinary resistors consistently out-perform ordinary diodes as a LED
current source.

Bye.
Jasen


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Default Looking to drop 6 volts dc to 4.5 volts dc

In article , jasen
wrote:

On 2006-10-01, Al wrote:

So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.


Typically leds are specified with a voltage range for their limit current.
get the curren right and the voltage will be somewhere in that range, it
depends on the device and the environment.

If that particulasr led is 4.4V with 10mA flowing through it it could with
4.5V the current could be 30mA or more.

The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0


And when they get warm, it drops, this can lead to thermal runaway.

ordinary resistors consistently out-perform ordinary diodes as a LED
current source.

Bye.
Jasen


Sigh! Yes, that's true. But I routinely use one or two diodes, depending
upon current, to drop my 6Vdc battery down to a nominal 5Vdc that my
circuits use. It works. What can I say? It's a cheap and dirty way of
doing it for home projects.

For example, I have a circuit which nominally draws 50ma during
operation. Upon the needs of the project, a 200mA pulse (100mS) drives a
latching relay. If I used a 20 ohm resistor to drop the voltage to 5Vdc
during norminal operation, it would really screw up the circuit during
the pulse as the series resistor would drop the voltage by 4V!

So, tell me, how many components would you use to drop your 6Vdc to 5Vdc?

Al
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Default Looking to drop 6 volts dc to 4.5 volts dc

Al wrote:
In article , jasen
wrote:


On 2006-10-01, Al wrote:


So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.


Typically leds are specified with a voltage range for their limit current.
get the curren right and the voltage will be somewhere in that range, it
depends on the device and the environment.

If that particulasr led is 4.4V with 10mA flowing through it it could with
4.5V the current could be 30mA or more.


The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0


And when they get warm, it drops, this can lead to thermal runaway.

ordinary resistors consistently out-perform ordinary diodes as a LED
current source.

Bye.
Jasen



Sigh! Yes, that's true. But I routinely use one or two diodes, depending
upon current, to drop my 6Vdc battery down to a nominal 5Vdc that my
circuits use. It works. What can I say? It's a cheap and dirty way of
doing it for home projects.

For example, I have a circuit which nominally draws 50ma during
operation. Upon the needs of the project, a 200mA pulse (100mS) drives a
latching relay. If I used a 20 ohm resistor to drop the voltage to 5Vdc
during norminal operation, it would really screw up the circuit during
the pulse as the series resistor would drop the voltage by 4V!

So, tell me, how many components would you use to drop your 6Vdc to 5Vdc?

Al


What someone else would use is irrelevant. You use
whatever you want to meet the circuit requirements.
Here's a possibility for your pulsed 200 mA.
Substitute whatever you're using in place of the switch.

/
+6 ----+-----o o--------+
| |
[20R] [5R]
| |
[50 mA Load] [200 mA Load]
| |
Gnd ---+----------------+

For the LED circuit that was discussed a resistor
is fine. For something that requires regulated 5,
you can use an LDO like
http://www.national.com/ds/LP/LP38690.pdf

For something like a 4.5V toy motor, your diodes
may be fine.

Ed
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Posts: 65
Default Looking to drop 6 volts dc to 4.5 volts dc

In article N7AUg.2324$6S2.1177@trndny02,
ehsjr wrote:

Al wrote:
In article , jasen
wrote:


On 2006-10-01, Al wrote:


So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.

Typically leds are specified with a voltage range for their limit current.
get the curren right and the voltage will be somewhere in that range, it
depends on the device and the environment.

If that particulasr led is 4.4V with 10mA flowing through it it could with
4.5V the current could be 30mA or more.


The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0

And when they get warm, it drops, this can lead to thermal runaway.

ordinary resistors consistently out-perform ordinary diodes as a LED
current source.

Bye.
Jasen



Sigh! Yes, that's true. But I routinely use one or two diodes, depending
upon current, to drop my 6Vdc battery down to a nominal 5Vdc that my
circuits use. It works. What can I say? It's a cheap and dirty way of
doing it for home projects.

For example, I have a circuit which nominally draws 50ma during
operation. Upon the needs of the project, a 200mA pulse (100mS) drives a
latching relay. If I used a 20 ohm resistor to drop the voltage to 5Vdc
during norminal operation, it would really screw up the circuit during
the pulse as the series resistor would drop the voltage by 4V!

So, tell me, how many components would you use to drop your 6Vdc to 5Vdc?

Al


What someone else would use is irrelevant. You use
whatever you want to meet the circuit requirements.
Here's a possibility for your pulsed 200 mA.
Substitute whatever you're using in place of the switch.

/
+6 ----+-----o o--------+
| |
[20R] [5R]
| |
[50 mA Load] [200 mA Load]
| |
Gnd ---+----------------+

For the LED circuit that was discussed a resistor
is fine. For something that requires regulated 5,
you can use an LDO like
http://www.national.com/ds/LP/LP38690.pdf

For something like a 4.5V toy motor, your diodes
may be fine.

Ed


You are correct!

I was just using the 200mA load situation just to show how a resistor
would not do the job. And, one should use what works in their particular
situation.

I've always wondered why the industry settled on 5 volts as the standard
for TTL esp. since 6 volt batteries were so common. In the 60's I
actually worked on a logic system which did use +6 and -6 volts for the
logic. I can't remember if +6 was a one and -6 was a zero, or was it
visa versa? For experimentation with this logic type, we actually used 6
volt lead/acid batteries for power. We quickly learned never to short
anything here!

Al
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Posts: 6
Default Looking to drop 6 volts dc to 4.5 volts dc

Al wrote:
In article N7AUg.2324$6S2.1177@trndny02,
ehsjr wrote:

Al wrote:
In article , jasen
wrote:


On 2006-10-01, Al wrote:


So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.
Typically leds are specified with a voltage range for their limit current.
get the curren right and the voltage will be somewhere in that range, it
depends on the device and the environment.

If that particulasr led is 4.4V with 10mA flowing through it it could with
4.5V the current could be 30mA or more.


The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0
And when they get warm, it drops, this can lead to thermal runaway.

ordinary resistors consistently out-perform ordinary diodes as a LED
current source.

Bye.
Jasen

Sigh! Yes, that's true. But I routinely use one or two diodes, depending
upon current, to drop my 6Vdc battery down to a nominal 5Vdc that my
circuits use. It works. What can I say? It's a cheap and dirty way of
doing it for home projects.

For example, I have a circuit which nominally draws 50ma during
operation. Upon the needs of the project, a 200mA pulse (100mS) drives a
latching relay. If I used a 20 ohm resistor to drop the voltage to 5Vdc
during norminal operation, it would really screw up the circuit during
the pulse as the series resistor would drop the voltage by 4V!

So, tell me, how many components would you use to drop your 6Vdc to 5Vdc?

Al

What someone else would use is irrelevant. You use
whatever you want to meet the circuit requirements.
Here's a possibility for your pulsed 200 mA.
Substitute whatever you're using in place of the switch.

/
+6 ----+-----o o--------+
| |
[20R] [5R]
| |
[50 mA Load] [200 mA Load]
| |
Gnd ---+----------------+

For the LED circuit that was discussed a resistor
is fine. For something that requires regulated 5,
you can use an LDO like
http://www.national.com/ds/LP/LP38690.pdf

For something like a 4.5V toy motor, your diodes
may be fine.

Ed


You are correct!

I was just using the 200mA load situation just to show how a resistor
would not do the job. And, one should use what works in their particular
situation.

I've always wondered why the industry settled on 5 volts as the standard
for TTL esp. since 6 volt batteries were so common. In the 60's I
actually worked on a logic system which did use +6 and -6 volts for the
logic. I can't remember if +6 was a one and -6 was a zero, or was it
visa versa? For experimentation with this logic type, we actually used 6
volt lead/acid batteries for power. We quickly learned never to short
anything here!

Al


When I did my electronics training in mid-70's, I can recall a cct that
was part positive logic (1=-5V, 0=+5V) and part negative logic (1=+5V,
0=0V).

Just a little confusing.

Daniel

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Posts: 75
Default Looking to drop 6 volts dc to 4.5 volts dc


..D.E wrote:
I'm looking to add a mod to my pinball machine with a couple of led's and
need to drop 6 volts dc down to 4.5 volts dc. Can anyone suggest which
resistor to use?

Thanks.

--
_____________________________
Later.
Have a better one.
D.E.
To email me back, remove "forget.the.spam"


If I understand your question correctly, most people makes voltage
regulators out of Zener diodes. However, I suppose you could make a
voltage regulator using a couple of LEDs although it probably wouldn't
be as accurate.

For information on how to do this look at instructions for doing it
with a Zener and then simply adapt them for using LEDs instead. For
information on making a regulator using a Zener, type in ZENER VOLTAGE
REGULATOR in Google. Here's a couple of examples:

http://www.allaboutcircuits.com/vol_6/chpt_5/7.html
http://www.ycmou.com/st/Download/ALM...r%20ze ner%22

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