PWM circuit...

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

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

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

If you're running the PWM at a kHz or lower the specified diode is fine -- but a Schottkey wouldn't hurt.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

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.

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John Popelish

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.

--
John Popelish

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.

--
John Popelish

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

Sure. I would replace the 555 with the CMOS version (LMC555) and use a low turn on voltage mosfet (like an IRLD024

or IRL3202 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 30BQ015 95SQ015

--
John Popelish

Could this circuit be modified to run on 6V (for a 6V motor) ?

Regards

Andrew

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--

I built the circuit last night... works great!

Just wondering, is there a beefier MOSFET than the IRF530? Say, 30 amps? I was thinking of powering a motor for an electric bicycle, and I wouldn't want the surge current to fry my MOSFET.

Thanks,

Mike Darrett

Could you recommend some larger FETs, say 30 amps?

Could I simply hook up a second FET in parallel (gate-to-gate, source-to-source, drain-to-drain)?

(I am kind of new to these kind of circuits ;)

Thanks,

Mike

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To increase the maximum rated current from 15A to 30A, would it be a good idea to simply hook up a second IRF530 in parallel (gate-to-gate, source-to-source, drain-to-drain)?

If so, would the resistor R2 need to be changed? I'm currently using

50 ohms (two 100-ohm resistors in parallel).

Mike

there are larger fets out there and you could also use multiple fets to drive the load.

Mosfets are resistor type devices via the drain and source you can go to digikey and look up Hexfets how about an IRF2804-ND 40 Vds, 280 Amps cont! $3.33 each from an older cat.

anyways, i am sure that gives you some idea's

"good idea" is probably overstating it, but it might work. A bigger device would be more reliable, but for a one off, the important thing is the tab temperature. It shouldn't burn your knuckle. Remember that every watt of heat showing up in the parts, is one less watt getting ot the motor, but still draining out of the battery.

I would separate the gates, and run one resistor to each.

You will also need a diode rated for a fair fraction of peak motor current. The big schottky I listed a while back might be close. But you can get bigger ones in the same tab package as the fets which are easier to heat sink.

--
John Popelish

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using

My setup with one IRF530 gets uncomfortably hot at only 7 amps (12VDC). Then again, it's not connected to a heatsink. I'm afraid to connect the load (large DC motor) for more than 2 seconds at a time.

Is this rate of heat generation at the MOSFET normal?

Mik

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Ok... but, how do you calculate the 35 watts? IIRC, the calculation goes 7A x 6V = 42 watts... even if I divide this by sqrt(2) this is still only about 30 watts (rms)... or is there something here I'm not quite understanding...?

I'm estimating between 60 and 80 degrees Centigrade, after running for about 2-3 seconds. (I was being cautious with the run times, because if I made a mistake in the wiring, I didn't want the electronic components to explode in my face.) I was using the 12V battery in my car, and a 12VDC surplus kiddie car motor which draws 7A on startup (about 1/2 second), about 4A continuous, with no load.

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So, if the gate drive delivers more current, the MOSFET operates more efficiently? (This implies, remove the resistor?)

Thanks,

Mike Darrett

It is expected. There are two parts to the loss (heating) problem. There is the simple ohmic loss when the transistor is on that is calculated by load current squared times the on resistance. And there are switching losses that can approach the supply voltage times the load current. fortunately, these come in short spikes and are minimized by getting through the switching process quickly.

But with a 7 amp load and a 6 volt supply, you are controlling about

35 watts of output power. A single watt will heat an un heat sunk TO220 tab enough to burn your finger. But that watt is only 1/(1+35)=2.8% of the total power from the source, so you are operating a quite efficient regulator.

How hot does the diode get when you run a 50% duty cycle (diode carrying the load half the time)?

A 30 amp regulator will probably need at least a small heat sink, because even if you find a huge, low resistance device to keep the DC losses down, its high gate capacitance will slow the switching process and generate heat there. At lease it will unless you also beef up the gate drive to something that can deliver more current than a C555 and a 47 ohm resistor.

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John Popelish

That is the answer you get if you can multiply correctly. I hadn't had my tea yet when I wrote that. Sorry.

I would want a bigger diode than that, if it is getting that hot.

Or reduce it. But you will quickly run into the limited output current capability of the 555. Switching a really big fet very quickly takes an amp or so for a fraction of a microsecond.

--
John Popelish