Looking to drop 6 volts dc to 4.5 volts dc

R = (Vsupply - Vled) / Iled

Good Luck! Rich

".D.E" schreef in bericht news:tHUSg.64297$5R2.18444@pd7urf3no...

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

Pinball machines have light bulbs in them that work off 6.3V like valves do. It's all been standardised. If you are replacing your 6.3V light bulbs with LEDs then you do not deserve the machine in the first place.

It was also a really bad idea to post your request because all this stuff gets logged and the American Society for Repatriation of Pinball Machines WILL cross the border.

DNA

Try it with 2 regular silicon diodes in there. That will drop around a volt and a half.

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

Next to go will be the solenoids and relays and ratchet and pawl counters! No more clickety-clack!This abomination must be STOPPED! Alas poor Yorick, I knew him well.

and

do.

with

"Al" schreef in bericht news: snipped-for-privacy@news.verizon.net...

That diodes will work when you have a resistive load. But neither LEDs nor other diodes behave like resistors. Once more: 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

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

it seems to me that regular resistors would work better,

--

Bye.
   Jasen

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

Al

Alas, poor Hamlet may have known him "well", but he never said it. The emphasis was on the word "knew" - an acknowledgment that death is indiscriminant and mysterious.

For the pin-ball machine, try this:

"For in that sleep of death what dreams may come when we have shuffled off this mortal coil..."

-- Joe Legris

Ha! Evidently a new word - use at your own risk:

"indiscriminant" - (noun) a mathematical expression that has nothing to do with an actual solution.

-- Joe Legris

--
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

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

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

--
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.

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            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

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.

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