P-channel FET won't turn off from micro

Question

I'm having an issue with what I think is a standard high-side PWM LED dimmer circuit.

It uses a P-channel FET ATP304-TL-H and gets an input voltage of anywhere from 9 V to 16 V.
The Zener diode across the gate and source is 16 V to prevent exceeding V_GS(max) and the two diodes across drain-source are actually one clamp: SMBJ24CA.

The BJT has the gate and pull-down resistor internally, the emitter resistor is there in case I want to mess with the values.

The LED strip is not power hungry, it's internally limited at 30 mA.

What is happening is that at rest with no control the light is fully off. Scoping the output shows about 600 mV, about a diode drop, which I expect. When it is run however, at minimum duty cycle (1/256) the output off state is around 2 V. This causes the LEDs to stay relatively bright at their lowest dim setting. Normally I would consider the BJT is not fast enough to drain the current but the PWM frequency is only 400 Hz, so I really can't believe it can't drain whatever is in the FET residually.

Any help is appreciated I really am stumped.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

From the datasheet \$Q_G - V_{GS}\$ chart, the total gate charge for the 5.1V gate voltage supplied by the zener diode os about 140nC.

This charge must be applied during the pulse rise time and removed during the fall time. Short rise and fall times reduce FET power dissipation but increases the current required to charge the gate capacitance.

The period for a 400Hz pwm signal is 2.5ms. The pulse width of 1/256 is about 10 micro-seconds. Although short rise times are desirable, the worst case that is just enough to show a flat top is 45% of the pulse width, about 4.5 micro-seconds.

So the average urrent required during the rise or fall is: $$I_{ave}=\frac{Q_G}{t_f}=\frac{140nC}{4.5\mu s}\approx 31mA$$

So \$R_2\$ must be about \$82\Omega\$.

This reveals that the gate current required to provide barely acceptable pulses is more than the load current being switched. So the drive circuit dissipates power as well. More well defined pulses require more gate current.

It comes from the very high input capacitance of 13nF. The time constant with the 10k resistor is 130 micro-seconds. The FET will turn off in 1 to 2 time constants. 1 time constant is 13 times longer than the minimum pulse width.

You should try find some FETs that can handle a drain current 1/2A< ID < 2A. Then sort them for gate capacitance. You should be able to find some < 2nF. This would decrease the gate current to a few milliamps.

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1. A current limiting resistor should be placed in series with the BJT collector. During the turn-on transient the gate-source capacitance acts as a short.

2. The zener is useful here to limit the gate-charge. To lower RDSon it can be removed. This will increase the gate charge thus trading gate current for RDSon.

Answer 2

That is a fairly massive power MOSFET, with a lot of gate charge ~250nC.

I would expect it to take the better part of a millisecond to turn off with only a 10kΩ resistor to drain that charge away. The problem isn't the BJT, which is probably fast enough for this application, but rather the resistor.

The transistor turns the MOSFET 'on' but the 10kΩ resistor has to turn it 'off' so you will get very asymmetrical ton vs. toff. So your 1/256 is stretched.

This circuit is adequate (perhaps) for static switching but not for PWM with a 100A power MOSFET. You'll need a better gate driver circuit. A band-aid solution is to reduce the 10kΩ R2 to maybe 1/5 or 1/10 of that (but watch the resistor power dissipation).

Alternatively, replace that power MOSFET with a BSS84 or something like that (~1/300 the gate charge), using a 100A MOSFET to switch 30mA is not ideal.