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I was looking at a friend's design for a simple PWM motor drive and I'm trying to figure out what's the use of some components. Here's the schematic:

pwm motor drive

I get what most parts are for:

  • power supply
  • fuse (not sure if needed when using a battery or a lab power supply)
  • PWM signal that drives the optocoupler
  • BJT acting as a switch (R4 to limit base current and R3 as a pull-down resistor when the PWM signal is LOW)
  • DC motor header with freewheeling diode (not shottky but should be fine)

I don't get why R2 is there... maybe to suppress noise/spikes together with the two capacitors?

Some more questions:

  • are C1 and C2 needed anyway?
  • should I add a capacitor across the motor leads?
  • should I expect voltage spikes even if there's a freewheeling diode?

It's a lot of questions so... please be gentle, I'm a newcomer :)


Edit:

I talked to my friend again and read some stuff. He still thinks R2 is needed and acting with C1 and C2 to suppress motor EMF and noise. I definitely think R2 is not needed and not useful for filtering - an RC snubber should be across motor leads and this is not the case.

I built the circuit but having no motor handy I just had a quick look on a scope using a resistor as load. I'll do some proper test with a motor tomorrow.

Thanks for all the people that helped here!

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  • \$\begingroup\$ If you have a well regulated power supply you can get away without C1 and C2. However, those caps make sure you have a steady DC voltage in case your supply voltage isn't that stable. The larger capacitor (C1) works against the low frequency ripple and acts as a reservoir. The smaller cap (probably a ceramic one) works against the high frequency noise on the DC supply. It's my understanding that R2 is there to control the voltage level at the base and therefore the current going into the base of the BJT transistor. Otherwise, if you didn't have R2, the voltage at the base could be too big. \$\endgroup\$
    – Big6
    Nov 16, 2016 at 21:19
  • \$\begingroup\$ So, what does your friend say when you ask him. Always best to consult the designer first. \$\endgroup\$
    – Andy aka
    Nov 16, 2016 at 21:25
  • \$\begingroup\$ @Andyaka: he said something about good practices for noise/spikes reduction but couldn't explain it better :) \$\endgroup\$
    – dodod
    Nov 16, 2016 at 21:30
  • \$\begingroup\$ @SixtoCabrera, I get why you put two such capacitors on PS but my question is are they needed here? There's no other load but the motor. R2 limits the voltage but is it really needed? Without it base current is simply (12 - 0,7)/470 and Vbe is always 0,7 \$\endgroup\$
    – dodod
    Nov 16, 2016 at 21:33
  • \$\begingroup\$ Sorry dude but you should ask him properly, There are many reasons why the resistor is there and the circuit doesn't give a clue so maybe the wider picture inside his head needs a little forceful extraction. \$\endgroup\$
    – Andy aka
    Nov 16, 2016 at 21:37

1 Answer 1

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C1 and C2 are needed to smooth out the the supply for low frequency noise and high frequency noise respectively.

I cant think of any reason to add a capacitor to the motor leads. Would probably just act as a lag for the motor to turn off and on.

Any voltage spike from the motor will be dissipated through the diode

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    \$\begingroup\$ small capacitors (1nF?) across motor leads are useful for noise suppression cause by DC motor switching on at its commutator \$\endgroup\$
    – dodod
    Nov 16, 2016 at 21:40

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