Push-pull output stage and driving circuit function

simulate this circuit – Schematic created using CircuitLab Hello, what is the exact purpose of the C3 and C4 capacitors, two diodes and R3 and R4 resistors in this circuit used to drive the output stage. The input signal is a 1kHz square wave with 5.2V on and 0V off provided from an opamp. Vref is 6V. Thank you

• The diodes allow stored charge to get "shorted" to the rails during transitions. It's not perfect, but it gets much of the charge dumped quickly. The two resistors keep the two MOSFETs off when they aren't being driven. The two capacitors allow the two MOSFETs to be off when there's insufficient activity (rate of voltage change per unit time) at the input. – jonk Apr 7 at 19:27

1 Answer

This requires a crystal ball to imagine why the circuit was designed this way without you telling us where you found it. I don't think it is a great design but it could be useful as a floating high side Pch level shifting pulse driver if the 6 Ohm load was AC coupled to a high voltage gate for a negative spike to trigger an IGBT or FET

I understand what waveforms to expect as the input square wave switches off each Vt=2~4V threshold FET with a DC restorer diode to each rail. Each gate turns on only for a pulse of a few μs due to T=C5*R5 = 470 nF*6R = 2.8 μs.

M2 is current-limited only by R5 in ground switching yet M1 is current-limited by R2 = 50 Ω shunted the 6 Ω AC coupled load so the positive spike will be useless for triggering FETs at only 6/56*6 ~= +0.64V while the negative spike can be -6V.

So because of R2, I don't see this as being a useful bidirectional pulse output with a lot of redundant parts for just a negative AC coupled pulse.