Need help building a split voltage supply

Can someone help me design a simple dual voltage power supply.

I have some IC's that require 5V and others that require 3V (actually a battery operated device that I am interfacing to). I want to build a power supply that can power both devices.

I was going to use a 7805 for the 5V supply. How should I drop the voltage for the 3V device. A simple schematic would also help if possible.

Thank you all.

JoJo

No - just normal rectifier diodes. A forward-biased silicon diode will have a voltage drop of about 0.7 volts across it.

To allow some fine tuning of the voltage, a forward-biased Schottky diode has a drop of about 0.3 volts.

In both cases, the drop will vary slightly with current and with the current rating of the diode.

Perhaps, or perhaps not - but it is something to be aware of.

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The most simple way is to put three silicon diodes in series with the 5V if the tolerance on the 3V allows. Do some calculation (or measurments) to check that within the temperature range you are operating the changes in the diodes threshold voltage are acceptable. Gene

normal diodes are fine, probably better than zeners, because they are cheaper. 1N4001 is a good choice for low power applications. You can get them most places electronics parts are sold.

Its just the way diodes are. They drop a small voltage when passing current. The current is an exponential function of the voltage, so for most purposes, you can assume that they drop around 0.7V no matter what the current. 10x the current will change this by about 60mV.

The 0.7V is referred to as the barrier potential of the diode.

The amount of voltage that a diode drops for a given current will change, depending on temperature. However, its not normally a concern for this kind of circuit. The change is around -2.1mV/C, meaning that the voltage the thing will drop decreases by 2.1mV for each increase of 1 degree C. However, that refers to the temperature of the junction, not the ambient temperature, and so is also related to how much current the thing is passing (self-heating), and the ability of the case and leads to transmit heat to the surroundings.

So, I guess you can ignore this if you want.

regards,

-- Bob Monsen

One possibility is to use this 5V to 3.3V regulator:

If you have trouble finding this specialized part, a widely available adjustable regulator is

which can be programmed to give 3V (or any value from 1.2V to 37V) with two resistors. The schematic is included on the data sheet.

power

Thanks for the info.

I had some 1N4002 diodes laying around and tried them. I got 1V drop using 4 in parallel. Is that really my best way, I would need 8 (I think) to get me down to 3V?

Would 4001 make a difference?

Thanks

in

Please respond on the bottom of the post... its the convention here.

Try putting 3 in series. That would work better. Also, use a resistor for your test, maybe 1k.

The number is a code indicating the max reverse voltage you should put across these, so 1N4002 and 1N4001 are identical unless you are putting 75V reverse across them, in which case the 1N4001 would break down, whereas the 1N4002 would prevent current flow (maybe, these are just what the mfgr guarantees. I'm guessing they probably come out of the same manufacturing line)

Bob Monsen

I tried 3 4001 in series and a 1K both as a load across the 5V and in series with the diodes.

3 diodes in series - 1K as load = 4.36V (same reading using 4 diodes too) 3 diodes in series - 1K in series =4.75V

Here is some information that might be helpful, some of which you might already know.

(please view in courier font)

On real schematics, the symbol for a diode is a filled in black arrow with a line on the point. The line on the diagram corresponds to the line on the actual diode. This is an ascii equivalent:

->|-

Current flows in the direction of the arrow, from left to right, when you put voltage across it.

This is the ascii symbol for a resistor:

--/\/\/---

Series means end to end, whereas parallel means all ends tied together.

Series: 1k

5V ---->|---->|---->|----/\/\/--- GND

Parallel:

+--->|----+ | | 1k 5V-+--->|----+--/\/\/--- GND | | +--->|----+

If you put them in series, then the voltage across the resistor shown will be about 2.9V. Replacing the resistor with your circuit that requires 3V is what the original responder was saying.

If you put them in parallel, then the voltage across the resistor will be about 4.3V. This is probably not what you want, cause it doesn't drop the voltage to near 3 volts for the remainder of the series elements.

If you don't have a series resistor, like this

Series:

5V ---->|---->|---->|--- GND

or

Parallel:

+--->|----+ | | 5V-+--->|----+---- GND | | +--->|----+

then you are shorting out your 5V supply, and probably pulling the supply down to 4.75 by trying to suck far too much current through it. Don't do this, you will burn out your power supply and fry your diodes. You should always have a resistor, or something that provides resistance between the 5V and GND terminals.

One other thing, a Zener diode is a special diode, which is used reversed from the usual orientation (the line points to the positive side.) It works by preventing current from flowing until a voltage threshold is reached, and then allowing allowing a virtual short above that. If you put it in series with a resistor, and put more voltage across it than the voltage rating of the zener, the voltage across the zener will be approximately the voltage limit, and the voltage across the resistor will be whatever is left over, that is, V+ - V- - Vzener.

Hope this helps!

Regards, Bob Monsen

Thanks for clarifying, worked like a charm :-)

This depends on how much current you need on the 3 V line, and how precise the voltage needs to be.

For low-power devices normally powered from a battery (whose voltage changes with age) a simple resistor may be sufficient. Measure the current and then calculate the resitor required to drop 2 V by Ohms law (R = 2 V / current).

Resistor +5 V in o-----/\/\/\-----o +3 V out

If the voltage required needs to be better regulated, use a resistor of somewhat lower value than calculated above and a 3 V Zener diode:

Resistor +5 V in o-----/\/\/\-----o +3 V out | | ----- / /\ Zener Diode / \ 3V ---- | | --- Gnd

the 7805 is a voltage regulator also don't forget to calculate the wattage of your resistor which is " current" X " voltage" you could also use two 7805's..the 1st one is for straight 5 volts the 2nd one is for the 3 volts..(put your series resistor from the supply to the input of the 7805)..also don't forget to put a large cap ...1000 mfd on the input leg to gnd..and a smaller cap..100 mfd on the output pin to gnd.. depending on the case type for your 7805..whether it is a power type usually 1 amp rating or a signal style case usually less than 100 ma..you may have to heat sink the case.. also note that your input to the 5v 7805 must be approx two volts higher ..perhaps 7 volts to get the two volt differential needed for the circuit to function..so use the higher input voltage to calculate the resistance drop..and wattage value I believe the 7805 is also thermally protected in case you overload it..the output will drop to a very low value until you remove the excessive load.. hope this helps your design..