Can MOSFET be used in PWM circuit

[grant fair]

Can a p-channel power mosfet be used instead of the PNP Darlington in this circuit?<p>http://www.eio.com/crntctrl.htm<p>What circuit mods must be added to do this (if it is practical in the first place)?<p>Grant

[Chris Smith]

Mos fets are the BEST for PMW. Keep in mind that a voltage present at the gate is all it needs to turn on a FET. No actual current draw is required, so you can terminate the feed to ground through a resistor, and tap off the resistor like a voltage divider to feed the gate. Do read up on fets before you start though, they have other considerations if your switching them very fast.

[grant fair]

Thanks, Chris-<p>How fast is fast? I understand that the faster the rise and falltime of the voltage driving the gate, the better, in order to lower power dissipation. But I suspect you mean the switching frequency? The circuit is set to run at 20 kHz, is that "fast" for a mosfet?<p>Grant

[analogee]

Wow, that schematic appears to have multiple errors. At least two at first glance. It looks like they are trying to do a classic buck regulator. The base of the BDW84 is grounded, which means the PWM chip can't do anything to affect it. D1 should be grounded, but not the transistor base. Also, looks like the BDW84 is in backwards (collector and emitter reversed). So, be aware of that in any case.<p>Regards,
Todd

[Chris Smith]

That would be considered slow. Fast runs in to Capacitance problems, the reactive speed of the gate and its delay, and other problems. MHZ is fast.

[grant fair]

Thanks Todd-<p>Not to be disrespectful, I am so ignorant of electronic design that I would like someone else to confirm your observations.<p>As well, has anyone seen additional mistakes in the schematic?<p>I have made a PCB pattern but haven't etched it. It would be nice to fix the errors before I make the board. <p>Thanks in advance,
Grant

[analogee]

A healthy dose of skepticism will serve you well in electronics.<p>However, in this case, if you do know some fundamentals, you might consider that the way the schematic is drawn, there is 24 V forward biasing the collector-base junction of the BDW84. Of course, it is a Darlington, but still there is a single junction in that path that is forward biased. That is clear evidence of a mistake. The base of this transistor should not be grounded. The resulting smoke if you build this will be confirmation!<p>Secondly, follow the arrow in the emitter of the transistor; it indicates conventional current flow. You want current to flow from +24V to the load. The transistor, the way it is drawn, shows current flowing from the load toward +24V.<p>Regards,
Todd

[grant fair]

Hi Todd- thanks for your explanations. Several others on other lists agree completely with you, including one I know who knows what he is talking about - and as I can see now, you also do.<p>I can change the PCB layout easily, and thanks for saving me from smoking a $2.50 BDX66 (which I intended to use for the PNP darlington). It is good for 16 amps so there would have been considerable smoke.<p>One comment suggested that the circuit would be horrible at high speeds (but no explanation for this was made). I don't know if something can be modified to improve this - any thoughts?<p>Thanks for you time and knowledge.<p>Grant

[analogee]

Grant,<p>Switching time I'm not all that experienced with. However, I believe Darlingtons are pretty lousy at switching, unless you were able to bring out the internal connection between the emitter of one transistor and the base of the other one (you can't, in a pre-packaged Darlington). Otherwise, the base of the output transistor essentially floats at a high impedance when you're trying to switching it off rapidly. So, I'd expect turn-off times in the microsecond, or even greater, range. The reason to use a Darlington, of course, is that it doesn't require so much base current to turn it on, which improves efficiency of the overall regulator. Especially with a 24 V supply. 10 mA wasted means 240 mW of dissipation.<p>You could conceivably go with a single switching PNP transistor (and probably lower the 10k and 1k base resistors a bit). Problem with that is you might need really high base current to turn the transistor on hard enough. Besides efficiency, can the regulator IC supply enough current?<p>Or, a p-channel MOSFET could work, in which case I would probably reduce the 10k resistor, depending on how much gate drive the MOSFET needs to turn on fully. As it is shown, there would be roughly 2 V of gate drive; most MOSFETs will require more than that. Rough guess: 4 or 5 V might be enough -- change the 10k to 3.9k. Now you're taking 5 mA from the +24V supply whenever the switch is on. That may or may not be a significant efficiency killer. I don't know how much total supply current is being used.<p>Or, as someone else said, at 20 kHz switching rate, maybe you are okay as it is.<p>Regards,
Todd

[Ron H]

Grant, I have edited the schematic to show a typical way to use a P-channel MOSFET in place of the PNP Darlington. Click here to see it. I simulated the driver and MOSFET, but not the entire circuit. It has sub-microsecond gate rise and fall times. You can change the two 10k resistors to lower values - as low as 1k each or even lower if you want really fast transition times. Keep the values equal. This gives you about 10 or 11 volts of gate drive. You can use a different MOSFET, but be sure it can handle 4 amps continuous, 30 volts source-to-drain, and has sufficiently low ON resistance to be efficient.
Obviously, I can't guarantee that the entire circuit will work as advertised.<p>Ron<p>[ July 23, 2003: Message edited by: RonH ]</p>

[grant fair]

Thanks to both Todd and Ron.<p>The comment on another list about problems at higher speeds has since been explained as the need to supply a path with diodes to dump coil current back to the power supply when a coil is switched off. A diode from the collector of each of the NPN motor driver transistors to the common (i.e. emitter and ground) was the suggested solution, which is easy enough.<p>Ron thanks for the typical mosfet circuit, that's exactly what I need, at least as a starting point, I will try it out and look for smoke. Now to find a high current low RDs on P channel mosfet.<p>Grant

[Ron H]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by Grant Fair:
Thanks to both Todd and Ron.<p>The comment on another list about problems at higher speeds has since been explained as the need to supply a path with diodes to dump coil current back to the power supply when a coil is switched off. A diode from the collector of each of the NPN motor driver transistors to the common (i.e. emitter and ground) was the suggested solution, which is easy enough.<p>
Ron thanks for the typical mosfet circuit, that's exactly what I need, at least as a starting point, I will try it out and look for smoke. Now to find a high current low RDs on P channel mosfet.<p>Grant<hr></blockquote><p>Connect the diodes from collectors to the junction of L1 and C1, not to GND, i.e., across the coils. Anodes connect to the collectors.<p>[ July 25, 2003: Message edited by: RonH ]</p>

[grant fair]

Thanks Ron, that diode hookup info was my mistake. I am thinking that an IRF4905 will work in this circuit. It has an R DSon of .02 ohms at 25 deg C and has a max continuous drain current of -74 amps at V GS @ -10 volts, at 25 deg. C. DO you have any other suggestions for as P channel mosfet?<p>Grant

[Ron H]

IRF4905 looks like it should work, but it may be overkill. The gate capacitance is really high due to its size, so the gate and output transition times will be slower. If you use it, you might want to change the gate drive transistors to 2N2222 and 2N2907. In fact, they are probably better choices in any case. I think I might use IRF5305, which has 1/3 the gate capacitance but 3 times the on resistance of IRF4905.

[grant fair]

Many thanks, Ron.<p>Grant