relationship between Switching charge (Qsw) and MOSFET loss??

[forever074]

If I have two Mosfet A and Mosfet B with:
- The same Coss (Output capacitance), Gate threshold voltage (Vgs_th), Gate plateau voltage (V plateau).
- Switching charge (Qsw) of mosfet A is 25nC, switching charge (Qsw) of mosfet B is 50nC.
Both Mosfets are hard switching with driver voltage=15V, Uds_off=150V, I_DS=20A.
1. So if we adjust gate resistor so that switching time of Mosfet A equal to B, the switching loss of mosfet A will equal to mosfet B ?
2. A smaller gate resistor will result to a less switching loss ?
3. What is the minimum gate resistor we can have to drive MOSFET ? What about 0?
Thank you!

 

[BobK]

The main thing is how much current you can supply to the gate. For fast switching of high powered MOSFETs more than 1A can be necessary to keep switching times low. Zero resistance if okay if the driver can handle it.

Bob

 

[Arouse1973]

When driving inductive loads a resistor is sometimes used to slow down the turn on time to reduce shoot through current from diodes that have clamped the flyback energy. Especially important in PWM motor drive in H-Bridge configuration when the Mosfets intrinsic diode is used. There will always be a trade off between shoot through current and switching times when choosing the correct value
Thanks
Adam

 

[forever074]

@BobK , @Arouse1973 . If we have two MOSFET with: The same R_on, Coss, I_DS, U_DS, they have the same quality? (We adjust gate resistor to have the same switching time, so MOSFET loss is the same). How to evaluate the quality of these two Mosfet? Which is better?

 

[(*steve*)]

If all other specifications are identical except for the gate capacitance, and you arrange the switching conditions (gate current essentially) so that both switch in the same time, then the losses in the channel during switching must also be identical if one assumes that the entire switching event happens while V_gs is held by the gate charge at the plateau voltage.

However there will be losses due to the internal gate resistance. This is not given in the question, but we must assume that it is identical in both MOSFETs. Whatever this non-zero resistance is, the gate losses will be I²R_g. We don't know any of these figures, but we do know that I varies in a ratio of 2:1 for the two mosfets.

Thus gate dissipation will be four times greater in the MOSFET with the highest gate charge.

Typically gate dissipation is low compared with other switching losses.

Whilst the lower limit on external R_G is zero, the gate will always have some finite (and often unspecified) internal resistance and also a maximum (also often unspecified) gate dissipation. So the actual R_g can never actually be zero because the resulting infinite gate current would cause an infinite power dissipation for no time at all. This would probably result in Max Planck leaping out of his grave to warn you about doing anything in less than 5×10^-44 seconds and/or a black hole being formed.

There are other issues as well, are you interested in reverse transfer capacitance?

 

[forever074]

@(*steve*) I think power lost due to charging the gate of the MOSFET include gate resistance loss, It equals to
P_GATE=Q_G*V_driver*fs. It only depend on Q_G (Total Gate Charge) not depend on Rgate. With I_DS=30A, U_DS=100V, P_GATE is smaller than other loss.
Example. With a buck converter, I_ouput=60A, U_in=100V, U_out=60V. I can set external gate resistor equal to 0OHM to get minimum switching loss ?

 

[(*steve*)]

What you're interested in is charging the gate capacitance as fast as possible, consistent with any limits the device may have. Typically that might entail the minimum additional gate resistance.

Depending on the method you use to drive the gate, additional resistance may be required, and even if you calculate that you are charging up the gate capacitance faster than the device can switch, there are still limits to the switching speed.

At some point you get into the region of diminishing returns.