I have a very dumb question in understanding p-MOSFET driver. simple BJT voltage divider or totem pole driver.

first i will ask Simple BJT driver.

enter image description here

During turn ON, Q1 is ON, MOSFET gate charges to v1/2 through the path in violet(from V1 current flows to gate to GND).

During Turn OFF, Q1 is OFF, MOSFET gate should discharge. discharge current has to flow in R3. we expect potential difference for current to flow, Here from where to where current flows because only V1 is connected there is no GND connection in discharge path.

Similarly for totem pole driver also. during discharge path upper transistor will be ON for short time which gives path for MOSFET discharge but there also no potential difference.

Please correct if i am not clear..

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    \$\begingroup\$ related: On-going quest to understand p-channel MOSFETs \$\endgroup\$ Commented Feb 5, 2018 at 1:39
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    \$\begingroup\$ If there is no potential difference then what does that tell you about the charge? If there was a gate charge there would be a potential difference. If there is no potential difference there must be no gate charge. \$\endgroup\$ Commented Feb 5, 2018 at 3:05
  • \$\begingroup\$ @immibis: During charging there is potential difference gate to source cap will charge through path V1,MOSFET G-S,R2, Q1 (C-E) to GND. But for discharge(when Q1 OFF), discharge current has to flow from V1 and into V1 through R3 for which there should be potential difference (or) Gate to Source and R3 loop (V1 is not involved), where gate capacitor forces current to flow through R3, but charge will flow back again to gate, then how we can say gate charge is discharged. \$\endgroup\$
    – user19579
    Commented Feb 5, 2018 at 5:43
  • \$\begingroup\$ @user19579 Current "flowing to itself" in a loop is exactly how a capacitor discharges! Think about some simple capacitor circuits. \$\endgroup\$ Commented Feb 5, 2018 at 10:16

1 Answer 1


During turn off, gate is charged to V1 through R3. The resulting gate voltage is V1, so \$V_{GS} = 0\$

During turn on gate is discharged through R2. The resulting gate voltage is V1 / 2, so \$V_{GS} = -\frac{V_1}{2}\$

  • \$\begingroup\$ This was flagged as a LQ answer, even though it is correct. Please add some details so the severs do not flag single-line answers. \$\endgroup\$
    – user105652
    Commented Feb 5, 2018 at 1:35

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