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PWM circuit...

Discussion in 'Electronic Design' started by [email protected], Feb 23, 2005.

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  1. Guest

    Hello,

    I was searching the web for easy-to-build DC motor PWM control
    circuits, and I came across this one:

    http://www.cpemma.co.uk/555pwm.html

    and I am wondering what the 100 uF capacitor C4 is for. It seems a bit
    strange to me that it is wired between the +12V and ground, and not in
    parallel with the motor... then again, I do not have a great deal of
    experience with these types of circuits.

    Thanks in advance,

    Mike Darrett
    www.darrettenterprises.com
     
  2. Decoupling. C4 is trying to help keep the 12v rail smooth, and C3, is
    doing the same for higher frequencies (which larger capacitors are
    ineffective at).
    You don't wan't capacitance across the motor, this would give massive
    'switch on' surges as the PWM turns on, and reduce the peak voltage across
    the motor, reducing the torque at low speed settings. A faster diode than
    the IN4001, is probably preferable to tap the flyback as the switch turns
    off, but given the fairly generous rating of the other parts is probably
    not worthwhile.

    Best Wishes
     
  3. Q1 switches motor current between the 12 volt supply and the catch
    diode, D3 each PWM cycle. That means that the 12 volt supply is
    loaded with a pulse current that bounces it around, especially is
    there is much wire length between the Q1 and motor connection and some
    other storage for the 12 volt supply. For best effect, C4 should be
    connected as close as possible to Q1's emitter, and the positive side
    of the motor and the cathode of D3 should connect to the other end of
    C4. This keeps most of that pulsing current local to C4 and out of
    the 12 volt supply wiring.
     
  4. Guest

    Thanks for the kind replies.

    When substituting an IRF530 for the TIP31, how do the base, collector,
    and emitter for the TIP31 correspond to the gate, source and drain for
    the IRF530? (This will be my first time using a MOSFET in a
    circuit...)

    Mike
     
  5. The gate replaces base, source replaces emitter, drain replaces
    collector. You can also eliminate the base resistor or reduce it to
    something like 47 ohms, since its only purpose in the gate circuit is
    to suppress high frequency switching oscillations, not set DC current
    as it does with the junction transistor.

    I would also replace D3 with a much faster, lower drop diode, like a
    1N5817 (20 volt) or 1N5818 (30 volt) 1 amp schottky diode. It will
    increase the efficiency and lower the noise the switching operation
    creates. The specified 1N4001 acts as a short circuit for some
    microseconds when the transistor first switches on, pulling a big load
    spike of current from the 12 volt supply. The schottky type turns off
    much faster.
     
  6. Guest

    Ah, so R2 can be as low as 47 ohms with the IRF530? The article said
    "... the same value or a higher (up to a 1k) gate resistor R2 can be
    used." so now I'm a bit confused if I should go higher or lower than
    100 ohms...

    Would an NTE587 (UF4001 equivalent) be as good as an 1N5817/5818?

    Thanks for the kind replies,

    Mike Darrett
     
  7. Tim Wescott

    Tim Wescott Guest

    The diode depends on switching speed. Often with motors you _want_ a
    PWM that's slow enough for the motor to see the full effects of the
    voltage change (see
    http://www.wescottdesign.com/articles/Friction/friction.html). If
    you're running the PWM at a kHz or lower the specified diode is fine --
    but a Schottkey wouldn't hurt.
     
  8. Guest

    Lower gives faster (higher efficiency, cooler transistor temperature)
    switching, but more radio waves. Higher resistance gives slower
    switching (quieter, but hotter transistor, lower overall efficiency).
    I can see why they want the switching to be slow, with that 1N4001
    diode struggling to turn off as the transistor turns on.
    Not as good, because it is a junction diode (I think) so will drop
    about twice as much voltage in the forward direction as the schottky
    (will run hotter and lower overall efficiency a little). But it will
    certainly switch off faster than the 1N4001.
     
  9. Sure. I would replace the 555 with the CMOS version (LMC555) and use
    a low turn on voltage mosfet (like an IRLD024
    http://www.irf.com/product-info/datasheets/data/irld024.pdf or IRL3202
    http://www.irf.com/product-info/datasheets/data/irl3202.pdf) in place
    of Q1. And, at low voltage, using a schottky diode becomes more
    important in the overall efficiency. There are very low voltage
    schottkys that have even lower forward drop that the ones I listed
    earlier, like:
    10BQ015 http://www.irf.com/product-info/datasheets/data/10bq15.pdf
    30BQ015 http://www.irf.com/product-info/datasheets/data/30bq015.pdf
    95SQ015 http://www.irf.com/product-info/datasheets/data/95sq015.pdf
     
  10. Guest


    Could I wire the output from the MOSFET to the control coil of a (20
    amp or so) relay? Would the switching frequency need to be slowed down
    a bit for the relay to respond?

    Thanks,

    Mike
     
  11. To quote an AOL commercial, "Slowed way down. Slowed down like holiday
    traffic". Contacts can be the switching mechanism for a PWM output
    but they are not very reliable. All the heat that would normally be
    dumped into the mosfet during turn on and turn off will appear in the
    tiny gas pocket between the contacts, converting it to little blasts
    of plasma, which will quickly either heat the contacts till they melt
    together, or will oxidize them till they no longer conduct. That 20
    amp rating is based on the contacts cooling after each break before
    the next make, and a PWM regulator does not meet that requirement.

    You may cool a large mosfet quite a bit just by adding a complementary
    emitter follower between the 555 and the mosfet as a current booster,
    with something like a 10 ohm gate resistor. If this is for the 6 volt
    version, you might try paralleling all 6 buffers in a 74HC4050 to act
    as a high current gate driver. This will boost the drive
    considerably, compared to the LMC555 output. But I would probably
    still keep 10 ohms right at the gate or put a ferrite bead around the
    source lead.
     
  12. Guest

    Thanks for the warning. You just saved me some research $$$. 8)
    Ok; I think I follow this. (You mean, use the 74HC4050 as a
    "pre-amplifier" for the mosfet gate, right?) But what exactly is a
    "complementary emitter follower"?

    I will also investigate the possibility that the motor was originally
    designed for 6V, as opposed to 12V. That would probably explain the
    temperature rise, if I've been exceeding the design voltage for the
    motor...!

    Thanks again,

    Mike
     
  13. Right. Lots of other noninverting chips with high current capability
    could work, too, including 3 state buffers (with the output strapped
    to be constantly enabled. These come in 4s, 6s and 8s per pack.
    It is a current amplifier made up of a pnp and npn transistor with
    their bases connected together as the input and their emitters
    connected as the output. The PNP collector ties to the negative rail,
    and the NPN collector to the positive rail. Unfortunately, you lose
    ..6 volts in each direction to forward bias the base emitter junctions.
    And with only 6 volts available, you cannot afford to lose any gate
    drive voltage and expect to get minimum temperature out of the fet.
    That would do it. Of course, you could limit the duty cycle of the
    PWM to no more than 50% and thus limit the average motor voltage to 6
    volts, even though the circuit runs on 12 volts. Then you have lots
    of gate voltage and can apply the complementary emitter follower. :)
     
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