Gate Drive CIRCUIT Power loss for switching FET?

[grizedale]

Hi,

When driving a switching N mosfet with GS capacitance "C".

The energy to switch the fet on is 1/2CV^2, where V = 10V say.

Therefore the power consumed by the gate drive circuit is 1/2CV^2.f

This equals 1/2.(Qg)V.f

...so why does the equation at the top right of page 8 of the following call it (Qg).V.f ??

..Surely they have missed out the "1/2"?

**broken link removed**

 

[mtwieg]

1/2CV^2 is the energy delivered to the gate, but in the process of charging it through a resistor you will also dissipate 1/2CV^2 (this is independent of the resistor value). Then when turning the FET off, the 1/2CV^2 stored in the gate capacitance will be dissipated in the driver. So your circuit overall consumes CV^2 per switching cycle.

 

[grizedale]

i see what you mean mtwieg, but the energy is only pulled from the bias supply when the gate capacitance is charged up.......there is no energy pulled from the bias supply when it is being discharged?

 

[mtwieg]

Right, and I didn't say the bias supply provides energy when discharging the gate. When charging up the capacitance, you draw C*V^2 from the supply. Half of this is immediately dissipated in the driver, and half is delivered to the gate capacitance. While discharging the gate, the 1/2CV*2 in the capacitor is dissipated in the driver again; none of the energy is recovered. So the draw from the supply in a full switching cycle is CV^2.

 

[grizedale]

the amount of energy drawn from the supply when the gate is charged up cannot always be CV^2.

....because this amount varies with the gate resistance.

If its zero gate resistance, then its 1/2CV^2 that gets pulled from the supply.

if its enormous gate resistance, then its much more than CV^2 energy that gets taken from the supply.?

 

[mtwieg]

the amount of energy drawn from the supply when the gate is charged up cannot always be CV^2.

....because this amount varies with the gate resistance.

If its zero gate resistance, then its 1/2CV^2 that gets pulled from the supply.

if its enormous gate resistance, then its much more than CV^2 energy that gets taken from the supply.?

Nope, it's always CV^2, so long as you allow the capacitor voltage to reach steady state after each switching event. It sounds silly, but it's true, and doing the math proves it (or you could do transient simulations). If you were to find a function for losses in a single switching transition (that is, one rising edge), as a function of resistance, you would find that it is always equal to 1/2CV^2. If R=0 then the function will give a 0/0 result, but its limit as you approach R=0 will still be 1/2CV^2.