I might be confusing myself, but from what I am reading so far, a BJT is a current-controlled device and a MOSFET is a voltage-controlled device, which to me implies that a MOSFET requires very low input current.

If that is correct, than why, in say an H-bridge, do we need to use a high side driver to supply 2A? Is it because of the Miller capacitance that we need to charge the gate capacitance for a split second? Please explain.


3 Answers 3


Exactly what you suspect. The effect of the gate capacitance is to slow down the switching. (The Miller effect multiplies the 'effective' gate capacitance.)

If we want the H-bridge to switch only occasionally (let's say at 1 Hz) a low gate current is (in most cases) fine, because the thermal effects of the switching are spread over 1 second, which is a relatively long period.

But if PWM is used, for instance at 300 Hz, there is only ~ 3ms to spread the heat dissipated in a switch (on+off), so this heat must be minimized.


The simple answer is that \$I_c=C \dfrac{dv}{dt}\$ and there is only current demand during fast switching.

It is my experience that the best rapid switching bridge commutators from 5 to 500 Amps will use a current gain of around 50 to 200 depending on component selection. For BJT's it ranges from 10 to 100 or 5 to 10% of the linear hFE or beta. Depending on dopants and geometry of junction.

  • and we were always taught FETs were just voltage controlled high impedance, but due to gate-drain or base-collector Miller capacitance, it acts like a current controlled rise time switch during transition, then voltage controlled resistance after.

  • it is wise to learn about dead-band commutation in Bridge drivers too, in the microsecond range to prevent frying drivers.

  • -

Techie Stuff

Charge time for the Miller capacitance is larger than that for the gate to source capacitance Cgs due to the rapidly changing drain voltage for entire duration of Vgs transition (current = C dv/dt).

Once both of the capacitances Cgs and Cgd are fully charged, gate voltage (VGs) starts increasing again until it reaches the supply voltage, where Ig drops near zero.

  • \$\begingroup\$ Rashid if someone said you need 5A gate current, then your output switched current must be >50A. \$\endgroup\$
    – user40708
    Apr 25, 2014 at 0:06

You need a high side driver to drive N channel FETs that are connected to the \$V_{\text{in}}\$ rail. The gate needs be driven higher than the source, which will be at \$V_{\text{in}}\$ when the switch is on.

Gate drive current doesn't necessarily need to be 2A. There are many high side drives that only drive ~500mA, and that's just fine. Usually the amount of drive current ends up being determined by \$R_g\$ (gate resistance), some of which is internal to the FET and can't be reduced. Often \$R_g\$ will end up being 10 or 20 Ohms for an optimal (non ringing) drive signal.

As a crude description, the gate is capacitively coupled to the channel via \$C_{\text{iss}}\$, which is made up of \$C_{\text{dg}}\$ (the Miller capacitance) and \$C_{\text{gs}}\$. These are both charged when the FET is switched. All the switching happens as the drain voltage changes (I'm not trying to be too obvious). As the drain voltage changes, charge on \$C_{\text{dg}}\$ changes, and the gate voltage is effectively stuck at a (near) constant level during that time. That's called the Miller plateau. So, that means that the switching time is defined by how quickly the gate drive circuit can manage the charge in \$C_{\text{dg}}\$. That is determined by \$R_g\$, \$Q_{\text{dg}}\$, Miller plateau voltage, and gate drive voltage.

  • \$\begingroup\$ but the gate drive current depends on the gate charge right? please correct me if i am wrong. \$\endgroup\$
    – rashid
    Apr 24, 2014 at 21:10
  • \$\begingroup\$ It depends on the charge yes, but also gate resistance and drive voltage and to a degree the drain current. \$\endgroup\$
    – gsills
    Apr 24, 2014 at 21:14
  • \$\begingroup\$ Really sorry about that. I knew that the \text{} was there for a reason, I just didn't know why. The downside of it is that it is really a pain to render the correct notation with LaTeX if it requires all that markup. Thanks for letting me know, anyway. In any case, I know there's a simple way to roll back those kinds of changes, but I don't know how. If you want, I can edit it all back. \$\endgroup\$
    – Ricardo
    Apr 26, 2014 at 3:21
  • \$\begingroup\$ @Ricardo ... It's OK, you didn't know. Everyone learns stuff on this site. I know I have. I don't know how to roll back either (heh), so I just edited it back. But, I did leave Vin with subscript. \$\endgroup\$
    – gsills
    Apr 26, 2014 at 3:28

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