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I am simulating a simple NPN BJT switch for PWM in LTspice, and I observe some current transients on the rising and falling edges that I do not understand. The circuit is here: Circuit schematic for NPN BJT switch

Picture of the falling edge: Falling edge

Picture of the rising edge: Rising edge

On the falling edge, why does the current through base resistor R3 go negative by ~1 mA? Is there a way to eliminate this such that the PWM voltage source does not have to sink this current?

On the rising edge, why does the load (R6) current go negative by ~1 mA? Is there a way to eliminate this such that the voltage source V3 does not have to sink this current?

EDIT: I put D1 there to prevent Q1 from entering saturation (Baker clamp with a single Schottky diode). The current through D1 on the falling edge mimics the current through R3, but slightly smaller in magnitude. On the rising edge, the current through D1 is plotted in the following picture:

Rising edge with D1

Edit 2: Removing D1 only serves to increase the turn-off delay, as expected. With D1, adding a 22 pF capacitor across ce junction of Q1 has no effect. Why the value of 22 pF anyways? Adding 100pF or more across R6 appears to help reduce the negative current draw on the rising edge.

user181297
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Unless you over dampen the circuit with 100pF capacitors across R6, fast rise and fall times will always cause some type of overshoot and undershoot. If it is bad enough it becomes a decaying ring wave. D1 may play some part in the negative current, as it does bypass the cb junction of Q1. But Schottky diodes do help with overall rise and fall times by lowering the effect of base capacitance. You do not want to over-dampen circuits that switch power, as this can increase overall power consumption itself.

If you look at the output waveform of fast logic IC's or PWM supplies you will notice at least some amount of overshoot and undershoot. The best you can do is make it small enough not to cause ringing or exceed the voltage rating of the driver (Q1 in this case) or cause the next stage (logic gate, etc) to behave in the wrong way.

Run the same test without D1 and see the results. Run the same test with D1 and a 22pF capacitor across the ce junction of Q1. You may find that no compromise is perfect, as having one issue reduced brings up another issue.

NOTE: If you can slow down the rise and fall times of your signal source then overshoot and undershoot problems should go away, but your design using a Schottky diode implies you are doing this intentionally?

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    Removing D1 only serves to increase the turn-off delay, as expected. With D1, adding a 22 pF capacitor across ce junction of Q1 has no effect. Why the value of 22 pF anyways? Adding 100pF or more across R6 appears to help reduce the negative current draw on the rising edge. Added this info as edits. – user181297 Nov 13 '18 at 20:54
  • @user181297. I picked 22pF, a tiny value, to avoid over dampening the circuit. In a way it is a tuned circuit in that any bypass capacitor will have a range that helps *some*, but no more than that. You can try 100pF+ next, but at some point Q1 will get warm driving a capacitive load. The trick is to find the right value for 'C' that helps but does not over-dampen. –  Nov 13 '18 at 21:00
  • Ok, that makes sense. It appears to solve the problem on the rising edge. – user181297 Nov 13 '18 at 21:02
  • Build your circuit (input PWM, the bipolar, the output/return path, on a GND plane. That pushes the inductance very low and the Fring quite high, and life should be very good for you. – analogsystemsrf Nov 14 '18 at 03:28
  • I am trying to minimize turn-on/off delays and get fast rise/fall times because I want good linearity (in terms of power delivered to the load) across a wide range of PWM duty cycles (~2-98%). The Schottky diode helps reduce the turn-off delay by preventing the transistor from entering saturation. I could likely afford to slow down the rise/fall times a bit. I'll play around with that a bit. – user181297 Nov 14 '18 at 16:43