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Need help building a split voltage supply

Discussion in 'Electronic Basics' started by JoJo, Aug 26, 2004.

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  1. JoJo

    JoJo Guest

    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.

  2. Gene

    Gene Guest

    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.
  3. JoJo

    JoJo Guest

    Thanks for your reply.

    I am a newbie at this so I just would like to clarify.

    Would these be Zener diodes?
    If you can explain why this would cause a voltage drop It would be of
    interest to me.
    If I understand you, there would be a change in voltage with a change in
    temp. If I am operating at room tempeteture is this a REAL factor?
  4. 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.

    -- Bob Monsen
  5. 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.
  6. JoJo

    JoJo Guest

    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?

  7. JoJo

    JoJo Guest

    Thanks for the info.
  8. 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
  9. JoJo

    JoJo Guest

    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
  10. Blake

    Blake Guest

  11. 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

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


    | | 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

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


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



    | |
    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!

    Bob Monsen
  12. JoJo

    JoJo Guest

    Thanks for clarifying, worked like a charm :)
  13. JoJo wrote:

    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).

    +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:

    +5 V in o-----/\/\/\-----o +3 V out
  14. Guest

    On Mon, 30 Aug 2004 14:09:41 +0200, Dr Engelbert Buxbaum

    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 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 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..
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