So I was looking through the MAX713 datasheet and came across the schematic below. In short, MAX713 is a NiCd battery charging IC and below is shown the schematic for it in the the switching mode. The DRV pin is a 30mA current sink PWM output. Sinking the current turns the PNP Q1 on and the NPN Q2 transistor off, thus pulling the P-channel M1 transistor gate to the power and turning it off, so in a way it inverses the output.

I have two questions about this schematic.

1) Why is the Q1-Q2 pair needed after all? Won't just the R2 resistor be enough?

2) Can the Q1 and Q2 be just switched places to put a much more common N-Channel MOSFET instead of the P-Channel M1?

3) Is there any reason to use a P-Channel MOSFET here after all?

Switching-mode MAX713 schematic


1 Answer 1


answer 1: the Q1/Q2 provide high currents to rapidly charge and discharge the gate capacitance, even with Cmiller (drain-gate capacitance)

answer 2: using an Nchannel, without a boost capacitor, will introduce a 5 or 10 volt drop in the switch output swing.

answer 3: If you introduce a BOOST capacitor, you can use a Nchannel. Without boosting, and if you insist on good efficiency, then the Pchan or a PNP is required.

  • \$\begingroup\$ Are the currents flowing through the gate really that large that the schematic won't work without this push-pull driver? \$\endgroup\$
    – sx107
    Commented May 19, 2018 at 0:13
  • \$\begingroup\$ What is the time constant of 5,000 ohms and the Cgate + Cmiller of your chosen FET? 5,000 ohms and 2,000pF has TAU of 10 microseconds. That puts the FET into risk of Safe Operating Area failure-mode, perhaps. To be blunt, you need to avoid slow ( longer than 1 microsecond) turnoff and turnoff times. \$\endgroup\$ Commented May 20, 2018 at 2:26

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