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Constant Current Source Design

Discussion in 'Electronic Basics' started by Confused Soul, Jan 24, 2005.

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  1. I need to design a constant current source of 0.1 mA. The load varies
    between 1Mohm and 1 Kohm. Using a constant voltage source, is it
    possible to design a constant current source?
    Any help in this is greatly appreciated.

    --Confused Soul
     
  2. Let's see....

    At 100uA,

    1k requires 1k*100u = 0.1V
    1M requires 1M*100u = 100V

    So, you require a voltage compliance of three decades and a voltage of 100V
    across your 1M load. And, when you don't want the 100V across the load, because
    the load is only 1k, the transistors themselves will have to stand off that much
    voltage and still provide a reasonably close 100uA?

    Is that about it?

    Jon
     
  3. Jamie

    Jamie Guest

    look for current Regulating diodes.

    a couple of Fets and biasing resistors will
    do it which is really all the CRD are.

    http://www.linearsystems.com/datasheets/J500.pdf

    those there don't come in a .1ma size but will give
    you and idea of how stable they are.

    if the range is not tight enough then i suggest using
    an OP-Amp as a comparator as the limiter circuit.
    basicly you couple the - and + inputs with a resistor
    which feeds also your output.
    when the inputs swing apart due to loading, the Op-Amp
    out put will bias a transistor that is supplying the source.
    something to look into.
     
  4. 100 V power supply seems impractical in my case. So instead of 100 uA,
    I could go for 10 uA or even 1 uA. This would reduce the voltage
    required to 10 V or 1 V.
    only 1k, the >>transistors themselves will have to stand off that much
    voltage and still provide a >>reasonably close 100uA?
    -- I couldnt figure out the circuit you are talking about !!! Could you
    please elaborate on this?
     
  5. Andrew Holme

    Andrew Holme Guest

    Here's a constant(ish) current source using a PNP transistor:

    http://www.ece.osu.edu/ee327/Figures/current_source_pnp_schem.gif

    Current in load resistor RL = (Vcc * R1 / (R1+R2) - 0.7) / Re
     
  6. What do you really need? If you reduce to 0uA you need nothing at all.

    petrus bitbyter
     
  7. Certainly. Ohm's law tells you how much voltage is required to push a
    given current through a resistance.
    The concept is to also have the current pass through a second resistor
    and a variable resistance device like a mosfet or junction transistor.

    Some additional circuitry must subtract the voltage drop across the
    second resistor from a reference voltage and amplify that difference,
    using that result to control the variable resistance device so that it
    burns up all the extra voltage not needed to push the same current
    through the first resistor. If the amplifier has high gain, the
    current will vary very little as the load resistance changes.

    How fancy this circuit gets depends on how accurate and temperature
    stable the current must be and how fast the load resistance can
    change.

    Some examples:
    See precision current source on page 7 of:
    http://www.national.com/an/AN/AN-20.pdf
    See bilateral current source on page 10 of:
    http://www.national.com/an/AN/AN-31.pdf
    Note that the current is proportional to Vin (that you will have to
    provide).

    You might also browse Google for [current source] or [constant
    current]
    or [current regulator]
     
  8. Simple constant current device using tvo PNP transistors and two resistors:

    ___
    .____/ |
    | \___|
    |
    .-.
    | |
    | |
    '-' o Constant current
    | |
    | |
    | |
    | |/
    o-----------|
    | |>
    | |
    | |
    \| |
    |-----------o
    <| |
    | |
    | .-.
    | | |
    | | |
    | '-'
    | |
    | |
    === ===
    GND GND
    (created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

    The current is about 0.6 divided by value to the lower resistor.

    Hope this helps
     
  9. Chris

    Chris Guest

    I guess there are two answers to your question. First, it's relatively
    easy to make a fairly stiff current source out of a voltage source with
    a series resistor. For the example you specified, though, you require
    a very high voltage (view in fixed font or M$ Notepad):

    Stiff Current Source From Voltage Source
    ` + ___
    ` o----|___|-------.
    ` 60 megohms |
    ` |
    ` o-----.
    ` | |
    ` R(L) .-. |
    ` 6000V 1K-1meg | |<---'
    ` | |
    ` '-'
    ` |
    ` I(L) |
    ` - <------ |
    ` o----------------'

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    As you can see, a load resistance of zero ohms will produce a current
    of 0.1mA. If you have a load resistance of 1 meg, the current will
    droop to about 0.098mA. That's pretty good, within a couple of
    percent. If you want better, you have to bump up the voltage source
    and series resistor more. Not very practical in this case.

    That's the homework-type answer, specific answer to specific question.
    If you actually have to make one, you probably want something a little
    more practical (and a little safer). In that case, you might try
    something like this:

    100uA HV Current Source

    ` + +
    ` o-----o------------------o------o-----.
    ` | | |
    ` | | |
    ` | R(L) .-. |
    ` | | |<---'
    ` | | |
    ` |20K 1 Watt '-'
    ` .-. I(L) |
    ` | | <----- |
    ` | | .-------o------'
    ` '-' | -
    ` 120VDC | |/
    ` o--------| TIP47
    ` | |>
    ` | |
    ` 1N4740 | .-.
    ` V(z)=10V /-/ 93K | |
    ` ^ | |
    ` | '-'
    ` | |
    ` - | |
    ` o----o----------'

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    Use an isolated lab 120VDC supply, and safety first -- if you don't
    know what you're doing, ask someone who does.

    Here's how it works: The 20K resistor in series with the zener applies
    10VDC to the base of the TIP47. As a first cut, you would expect that
    would mean 9.3V at the emitter of the TIP47. That makes the emitter
    current 100uA (9.3V / 93K, use an 82K and an 11K in series). Since the
    TIP47 has a minimum h(FE) of 30, you can be guaranteed that at least
    97% of that emitter current is coming from the collector, and 3% or
    less from the base. That will give you around 100uA to the load
    resistor, and it should be a stiffer current source than the first
    example over the range of 1K to 1Meg. You might want to replace the
    93K resistor with a 91K resistor in series with a 10K pot to tweak it
    in to exactly 100uA.
    Again, be sure to play safe around high voltage.

    Good luck
    Chris
     
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