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Simplest current regulator

Discussion in 'Electronic Design' started by Phil Endecott, Jan 28, 2008.

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  1. Hi All,

    I have a bright white LED that takes 300 mA with a forward voltage of
    about 3.4 V. I'm powering it from 3 AA cells, and it's clear that a
    series resistor is not good enough as a regulator: choosing a value to
    suit 1.5V per cell, I only get half the current once the per-cell
    voltage has dropped to 1.3V.

    So I'm looking for a current-regulator that will give better regulation
    than this - but it doesn't need to be anywhere near perfect, and
    simplicity is vital. The low dropout requirement is the main challenge:
    most circuits that I can think of need a series current sensing
    resistor, which needs to develop a significant voltage. I'm currently
    experimenting with something like this:

    Vin --------------------+
    | |
    / _|_
    \ \ / ->
    / _V_ ->
    | |
    | |/
    +-------------|
    |\>
    |
    GND --------------------+


    This gives enough regulation - though a bit more would be nice. Its
    main deficiency is that it's sensitive to the gain of the transistor. So:

    - Are there any tricks to reduce the sensitivity to the gain of the
    transistor, that don't involve an additional series voltage drop?

    - Is the FET equivalent of this circuit any less sensitive to device
    variability?

    - Are there any particular transistors that have more repeatable gain in
    this sort of configuration? (I'm currently using a ZTX349A, because it
    was handy.)

    - Any better circuits?


    Cheers, Phil.
     
  2. Guest

    There are a lot of white LED driver integrated circuits around. A
    quick session with Google threw up this

    http://www.kin-track.com/led/ZD850.pdf

    which might do what you want - the current sense resistor drops 0.2V.

    You could do almost as well using an National Semiconductor LM10CLN to
    control a PNP series transistor

    http://www.ortodoxism.ro/datasheets/nationalsemiconductor/DS005652.PDF

    The LM10 includes a 200mV voltage reference. The output stage can't
    handle 300mA so you've got to drive some kind of buffer - a MOSFET
    with a low drop-out voltage would do, but a decent-sized PNP
    transsitor would probably be cheaper.

    The LM334 only requires a 64mV drop across the current sensing
    resistor - the reference voltage is temperature dependent (see the
    graph at the bottom of page 3 of the LM334 data sheet), but probably
    won't move enough to worry you if you use it indoors.

    http://www.worldtorch.com/LDO-fixed-current.php

    http://www.ortodoxism.ro/datasheets/nationalsemiconductor/DS005697.PDF
     
  3. Min input is 4V, which is a bit high. And it's more complex than I was
    hoping for...
    Getting closer!
    Ah, now that looks interesting - 64mV is excellent, and it's a
    4-component circuit (plus an extra R-C which I reckon it would probably
    work without...)

    I think I might even have an LM334 somewhere... rummage... no, it's a
    '336. Oh well.

    Does anyone know why we can't buy discrete zenners (or similar) down to
    65mV ?


    Many thanks.

    Phil.
     
  4. except it turns out that I would need a 0R22 current sense resistor. I
    can only find that value as a huge 3W ceramic-clad wirewound. Do
    resistors of that sort of value exist in "normal" packages? It will
    dissipate 20mW in my circuit.


    Phil.
     
  5. Winfield

    Winfield Guest

    No, you need the RC to keep it from oscillating, with
    the extra loop gain added by the LM3906 pnp transistor.
    In a pinch, you can use a higher-value sense resistor,
    and attenuate its output to 64mV with 2 more resistors.
     
  6. RHRRC

    RHRRC Guest

    Read the datasheet.
    The LM334 will not give you 300mA and requires around a 1V overhead at
    currents as low as 5mA.
     
  7. I have done.
    Have you looked at the schematic that Bill linked to?:

    http://www.worldtorch.com/LDO-fixed-current.php
    Use an additional PNP transistor to increase the current. The 1V
    requirement is not a problem in this configuration.

    Phil.
     
  8. Keith

    Keith Guest

    <snipped>

    You could use a small incandescent bulb in place of your series
    resistor. These have a positive temperature coefficient and
    thus will tend to keep the current constantish. There are
    several available for flashlights/torches which work at a volt
    or so - you'd need to experiment and may need to parallel a
    couple of bulbs to get the necessary current. An incandescent
    lamp run at below its rated voltage will last pretty much for
    ever.

    The regulation isn't good, but is better than a plain resistor
    and very simple. I use this method for regulating a 9.6V LED
    device from a 12V battery on my bike. A side effect is that the
    glow from the bulb provides a useful battery level indication.

    Cheers
     
  9. default

    default Guest

    Bob Pease did that circuit some time ago for Electronic Design
    magazine.

    http://electronicdesign.com/Articles/Index.cfm?AD=1&ArticleID=4703
    http://electronicdesign.com/Files/29/4703/Figure_02.gif

    THIS circuit doesn't have a low tempco. Its output current increases
    at +0.33%/°C. But that's plenty good enough for many cases—like in a
    flashlight! If you need a better tempco than that, we know several
    ways to do it. Still, this will let you regulate the current into a
    red LED down to 2.1 V of supply voltage, or a white one down to 4.2
    V—MUCH better than 7.4 V.

    OH—I almost forgot to say—the LM334 sometimes needs a series RC
    damper. My first guess was 2 µF and 22 across the base-emitter of the
    PNP. Actually, this circuit didn't oscillate or ring badly without an
    added capacitor, but the noise was a bit quieter when I added a 2- or
    10-µF electrolytic.

    If you really want a low tempco, use copper wire (magnet wire) for the
    resistor—that will cancel out nicely. You'll need 6 ft. of #34 gauge,
    or 10 ft. of #32.


    --
     
  10. Guest

    It depends on what you mean by a "normal" package. Farnell lists a
    number of 0.22R resistors in surface mount packages which are
    relatively easy to adapt to through-hole circuit boards.

    They do a Phicomp 0,22R part under order code 940-3248 for about $1,
    a much smaller - 0402 - Tyco part under order code 886-7283 which will
    dissipate 100mW. The Tyco part is about half the price, but you can't
    order fewer than 25.

    There's also a 2W Tyco surface mount part, order code108-6266 which
    less than a dollar and availalbe in a minimum qunatity of five.

    You can also buy small conventional 1R axial lead resistors - four or
    five of them in parallel don't take up too much space and would do the
    job.

    As Default has pointed out, you can also use enamelled copper wire
    (transformer wire) to make a temperature sensitive resistor that will
    track with the LM334 reference voltage though he talks about using
    relatively thick wire. I'd go for 0.1 mm diameter wire myself - I
    could wind that without breaking it - where you'd only need 10cm (four
    inches) of wire to get 0.22R. Farnell doesn't stock 0.1 mm wire but it
    does sell 38 SWG copper wire which isn't quite as thin, but you'd
    still only need 25cm (10 inches). You might want to use a non-
    inductive winding scheme.
     
  11. whit3rd

    whit3rd Guest

    Zener knees get soft below 4V or so. If you want a 65 mV reference,
    you can always use a jFET with gate/source connected
    as a current reference, into a suitable load resistor.

    You can also use a resistor divider to drop the 200 mV
    reference to something more convenient in terms of LED
    current sense resistor dissipation.
     
  12. Can someone show me how I could use a potential divider to get a lower
    reference voltage than a zenner can provide in this sort of current source?

    \
    /
    \ Load
    / |
    | |
    | |/
    +------|
    | |\>
    _|_| |
    | ^ \
    /_\ / R
    | \
    | /
    | |
    --+--------+


    V_zenner = V_be + IR

    If I want R to be small, I need V_zenner to be just a little more than V_be.

    Or I could put a regular diode in series with the zenner, and have
    V_zenner equal the voltage across the sense resistor. Or use a second
    trasistor.

    Of course I can use a zenner with a potential divider in an op-amp
    circuit, but I need help to use it in something simpler.

    Thanks for all the useful suggestions.

    Phil.
     
  13. Guest

    Have you considered using niads or NiMH. Secondary cells have less
    change in voltage over the discharge curve. Note that an alkaline
    really needs to be used from 1.5V to 0.5V to get the full capacity.
    For nicads, this range is 1.25V to 1V. The nicad will probably start a
    bit higher than 1.25V, but will drop very quickly.

    Any LDO can be turned into a current regulator. You make the load be a
    resistor, then put the leds in series with the power supply. Of
    course, this is not every efficient, especially for one LED.
     
  14. I'd really like it to work with both, hence the need for a regulator of
    some sort.

    But if you did want to work only with nicads, it would still be
    difficult to use just a series resistor; the difference between the
    nominal 3.6V from the cells and the nominal 3.4V for the LED is small,
    making it sensitive to variations in both.

    Although the LM334 circuit and other things we've discussed are
    interesting, I'm actually quite happy with my single transistor circuit;
    I have superglued the transistor on to the back of the 20mm aluminimum
    disc that's the LED's heatsink. I'm just concerned about potential
    variability in transistor gain, and wonder if anyone has any tips about
    how to get consistent gain, e.g. are there any particular parts that
    have more repeatable gain from batch to batch. It needs to handle 300
    mA and about 400 mW, with a somewhat-elevated ambient temperature; I'm
    currently using a ZTX549.


    Phil.
     
  15. MooseFET

    MooseFET Guest

    Why not use an LED as the zener? The forward voltage on an LED has
    about the same temp-co as the EB drop of a transistor.
     
  16. whit3rd

    whit3rd Guest

    Well, I'd think an LM10 and pass transistor is fairly
    simple (half a square inch of circuit board), but there's
    another option that hasn't been discussed yet: a current
    mirror.

    Just run a resistor from V+ to base of two transistors,
    both emitters to GND, and connect collector 1 to base,
    and LED from V+ to collector 2.

    The trick is, the transistors have to be on a common heat sink
    and the first has to have much smaller emitter area than the
    second (the collector current is scaled by the emitter area,
    all else being equal). LED current is N*((V+) - Vbe))/R
    and the 'N' represents the area ratio.

    ICs use this kind of trick all the time, but buying transistors in
    onesies it's hard to know what N is going to be.
     
  17. Fred Bloggs

    Fred Bloggs Guest

    Right, my thoughts exactly, but he really needs to go to MOSFETs,
    transistors will waste a lot of current saturating...it seems ridiculous:
    View in a fixed-width font such as Courier.

    ..
    ..
    ..
    .. .---------------+------+------.
    .. | | | |
    .. | R1| | |
    .. | [560] | |
    .. ---BATT | |< |
    .. - +----| Q1 |
    .. | | |\ |
    .. | | | |
    .. | | | |
    .. | | | |
    .. | | | |
    .. | R2 | | |<
    .. +---+--[100]-----------+----| Q2
    .. | | | |\
    .. | R3| | |
    .. | [10k] | |
    .. | | | D1 |
    .. | | R4 | 1N914 |
    .. | +--[910]----|-------|<|---+
    .. | | | |
    .. | | | |
    .. | | | TL431 | LED
    .. | | ---U1 ---
    .. | '----------/ \ \ / ~
    .. | --- ---
    .. | | |
    .. | R5 | |
    .. '------[0.22]---+-------------'
    ..
    ..
     
  18. Guest

    You can buy dual transistors - Farnell list a bunch, of which the
    PBSS4240Y looks good - and mimic the effect of emitter area by
    arranging the Vbe for one transistor to be higher than that of the
    other. At room temperature 86mV of extra bias gives you a 30:1 current
    ratio.

    You'd use a 300R resistor to draw some 12mA from the battery; 10mA of
    that would go into the collector of the Vbe-setting transistor, 1mA
    would go through an 82R resistor to the base of the Vbe-setting
    transistor, and from there most of that 1mA would go on to the other
    terminal of the battery through a 680R resistor across the Vbe of the
    Vbe-setting transistor, while the remaining 1mA would provide the base
    current for the other half of the dual transistor, which would be
    sinking roughly 300mA from the battery via the white LED.

    The junction of the drive transistor would be dissipating of the order
    of 300mW, so it would be warmer than the junction of the Vbe-setting
    setting transistor (which would be dissioating only about 8mW); in a
    dual transistor both junctions should be on the same substrate and
    physically close, so the difference ought to be small.

    If I were building the circuit, I wouldn't start of with a 82R and
    680R to bias the base of the Vbe-sensing transistor, but rather with a
    200R pot in series with 560R, and use the wiper of the pot to drive
    the base; you'd begin with the pot set to give you 0R and 760R, then
    move the wiper down towards 82R and 678R while monitoring the current
    drawn through the LED.

    The circuit would be a bit crude, but it would work.
     
  19. Guest

    I would consider using a 4th battery. You will be barely using the
    capacity of the alkaline cells.

    Your circuit goes up a zener, then down a vbe, so you will get a vbe
    of shift over temperature. You could go up a zener, then buffer it
    with two transistors, i.e. up and down a vbe. But zener voltages are
    too high, so you will end up getting a bandgap reference. However, at
    that point, you can just use a LDO.

    There is a circuit called the "peaking current reference."
    http://users.ece.gatech.edu/~rincon/classes/ece4430/handouts/l36_iref.pdf
    I've used in in situations where the spec calls for no trim. I'm not
    all that impressed with the performance, but it is better than moving
    a whole vbe over temperature.

    Have you considered going to Harbor Freight and getting a LED
    flashlight for $5? Seriously, I always buy stuff off the shelf if it
    suits my needs. You can spend a lot of time reinventing the wheel.
     
  20. Fred Bloggs

    Fred Bloggs Guest

    Is this what you're talking about, a Vbe multiplier with current source
    setting???:
    View in a fixed-width font such as
    Courier.

    ..
    ..
    ..
    .. .------+--------------------.
    .. | | |
    .. | | ---
    .. | [300] \ /
    .. | | ---
    .. | | |
    .. | | |
    .. | | |
    .. | | |/
    .. | +-------+----------|
    .. | | | |>
    .. --- | [82] |
    .. - \| | |
    .. | |-----+ |
    .. | <| | |
    .. | | [680] |
    .. | | | |
    .. ------+-------+------------'
    ..
    .. ZXTDAM832

    2X NPN

    You will end up with a current hogging problem when the high current
    transistor approaches saturation. This will have a double whammy effect
    of 1) lowering the Vbe of the Vbe-setting transistor, and 2) increased
    base drop due to base resistance of the high current transistor. My
    initial estimate will be a LED current roll off to 100mA or less. Then
    the current setting has a substantial dependence on Vbatt, maybe 15-20%,
    like all other reference-less schemes. It's a good idea but will not
    handle low battery.
    The LED is almost certainly one of these:
    http://www.lumileds.com/pdfs/DS25.pdf
    They have very substantial dynamic resistance in the 1-2.5R range,
    making them easy to drive with a voltage source.
     
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