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1.5V bypass

Discussion in 'Electronic Basics' started by [email protected], Jan 8, 2007.

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  1. Guest

    I have a 4.5V power supply, and I need make a 1.5 V bypass.
    How can I do this ?

    Regards.
     
  2. I am not familiar with your use of the word bypass. Do you mean a second
    output at 1.5 volts? If so, a simple shunt regulator using two silicon
    rectifiers in series might do it.
     
  3. JeffM

    JeffM Guest

    Yup. Using more words to describe the concept
    would have been useful.
    You've used words clumbsily as well.
    You think that a guy who has this much trouble
    clearly describing what he needs can decypher "shunt"?

    The diodes are in series WITH EACH OTHER
    but are in PARALLEL with the load.

    There is also a resistor between the original supply and the diodes.
    With no further details about what the OP is doing,
    the resistor's value can not be stated.

    View in a monospaced font like Courier.
    (Click "Fixed font" at the top of the Google Groups page.)

    ---/\/\--+--- OUT
    ¦
    _V_
    ¦
    _V_
    ¦
    ---------+---
     
  4. PeteS

    PeteS Guest

    I'll admit the biggest impediment to the transfer of knowledge around
    here is the inability to write clearly and concisely.

    Cheers

    PeteS
     
  5. Yup. Using more words to describe the concept
    would have been useful.
    You've used words clumbsily as well.
    You think that a guy who has this much trouble
    clearly describing what he needs can decypher "shunt"?

    Clumbsily? Googling for "shunt regulator" yields a ton of information that
    I have little interest in regurgitating here.

    The diodes are in series WITH EACH OTHER
    but are in PARALLEL with the load.

    There is also a resistor between the original supply and the diodes.
    With no further details about what the OP is doing,
    the resistor's value can not be stated.

    Oh, you are indeed a sage among mere strugglers!

    Piss off and PLONK!
     
  6. Chris

    Chris Guest

    If your 4.5V is being supplied from three 1.5V batteries, you already
    have the 1.5V available. All you have to do is make the connection on
    the other side of the battery closest to the "minus" end of the output,
    and you're done. Total cost: zero.

    If your setup isn't like that, and you just need a steady 1.5V, and
    aren't drawing any load current (or a very small amount) from your 1.5V
    source, you can just use a resistive voltage divider, like this (view
    in fixed font or M$ Notepad):

    | .--------.
    | | |
    | | .-.
    | | | |2K
    | | | |
    | +|4.5V '-'
    | --- | 1.5V
    | - o-----o
    | | |
    | | .-.
    | | | |1K
    | | | |
    | | '-'
    | | |
    | '--------'
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    If you need to draw some current (even 1mA) from the 1.5V source,
    though, that will mess up the voltage divider and lower your voltage.
    By using lower value resistors (again, in the 2:1 ratio), you can
    increase the current you have available a little, at an enormous
    increase in the power you'll need to waste. Total cost: $0.02 to $0.20
    USD.

    If you can live with something close (say, 1.3V to 1.5V), you can just
    use two diodes and a series resistor to get the lower voltage, like
    this:

    |
    | .-------.
    | | |
    | | .-.
    | | | |33 ohm 1/2 watt
    | | | |
    | | '-'
    | +|4.5V | 1.5V
    | --- o------o
    | - |
    | | V 1N4001
    | | -
    | | |
    | | V 1N4001
    | | -
    | | |
    | '-------'
    |
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    This will give you a fairly steady 1.4V or so with a load current
    between zero and 90mA. It's probably the simplest way to get a lower
    battery voltage, but it's very wasteful of power. You will draw 90mA
    from your 4.5V source no matter what your 1.5V load. Of course, if you
    need less current, you can bump up the 33 ohm resistor to a higher
    value. If all you'll need is a maximum of 30mA. you can make the
    resistor a 100 ohm 1/4 watt resistor and save 2/3 of the current.
    Total cost: $0.20 to $0.75 USD.

    But again, this is very wasteful of the 4.5V power. A more complex and
    somewhat more expensive way to do this is to use the LM317 linear
    voltage regulator IC, along with a small capacitor and two resistors.
    This setup will only waste 6mA, and will provide a steady, precise 1.5V
    for a load current of zero to 1/4 amp, and over 1 amp if you use a heat
    sink on the IC. Total cost: $1.00 to $4.00 USD.

    To determine which is best, you could provide more information. Is the
    4.5V from a battery, or a series of three batteries? What kind of load
    current do you need from the 1.5V "bypass"? If you don't know, could
    you at least describe the load or what you're doing?

    Cheers
    Chris
     
  7. ehsjr

    ehsjr Guest



    "clumbsily" ???

    pot/kettle
     
  8. default

    default Guest

    This guy is posting through Google Groups. If one is used to
    AOL,Earthlink, etc., there's a tendency to treat Usenet like some AOL
    bulletin board - or chat room - or myspace.

    Google, for their part, does nothing to educate or dissuade one from
    that conclusion - no mention of what Usenet is, its history, or
    etiquette. They even encourage you to put in a personal profile -

    Half the time, the original poster can't remember what group he posted
    to, or was asking a question he had only a passing interest in -
    attention span limitations, or is too clueless to understand the
    answer, or some combination thereof.
     
  9. Rich Grise

    Rich Grise Guest

    Yes, it makes the sentence clumbersome. ;-)

    Cheers!
    Rich
     
  10. Guest

    Chris escreveu:
    Thanks a lot, Chris. I'm using your circuit below with total success !

    | .--------.
    | | |
    | | .-.
    | | | |2K
    | | | |
    | +|4.5V '-'
    | --- | 1.5V
    | - o-----o
    | | |
    | | .-.
    | | | |1K
    | | | |
    | | '-'
    | | |
    | '--------'


    Regards !
     
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