Hello,

I would like to understand more clearly the limitations and design requirements regarding with FET's gate charge.

1. Do i need to consider the magnitude of the gate drive current?
2. Is it only comes into consideration when switching the FET in high frequency?
3. How shall i calculate the needed current per gate capacitance\charge?

Thank you for your support.

 

1) yes
2) No.
3) The data sheet describes the capacitance of the gate. I'm looking at a data sheet that says, 85 nanocoulombs and 10 volts to turn on. You have to look at the definition of a capacitor and figure how much current you have to shove into the gate in order to get 85 nanocoulombs into it, and the time you will allow the MOSFET to be in between off and on. It's basically a voltage that you provide, driving through a resistor and charging the gate capacitor.

Suppose you have 20 volts to work with and the gate must get to 10 volts.

One problem with this is that most IC's won't conduct a lot of amps. That's why it is definitely a consideration in designing with MOSFETs.

Ahh...there it is...time = - RC Ln Vcap/Vin
Lets try 100 ohms
time = - 100 (85e-9) [Log natural 10/20]
5.9 microseconds

But you'll need a momentary .2 amps capability to do that. (That's what MOSFET drivers are for.)

Anyway, there's the math. Play with it.

 

Thank you,but let me see if i got it right:
Q=85 nC
V =20V
R = 100ohm
I = 0.2amp

----------
Q=I x t = C x V

==> t= (C x V)/I

1. Does it make any sense?
2. What i miss here is the dependancy in Vcap (Vgate=Vthr), can you help me?

One more thing:

Lets say i drive a nFET with a NE555 which can sink and source currents up to 200mA.
I connect to the 555 output (pin 3) a 100ohm resistor in series with the FET's gate.
I supply the 555 with a P.S of 12V\500mA.

==> meaning that the 555 output is:12V\120mA

The FET was conncet in common source configuration (source to gnd).
I measured the voltage at the gate and i saw 4V, but as i disconnected the FET the voltage was 12V.

CAN you explain why does the voltage droped that much?

Thank you for your support.

 

The FET was conncet in common source configuration (source to gnd).
I measured the voltage at the gate and i saw 4V, but as i disconnected the FET the voltage was 12V.

CAN you explain why does the voltage droped that much?

A schematic would be very useful. It's just guesswork, otherwise.

 

The QITCV kind of almost works if you are using a constant current to charge the capacitor, but a resistor into a charging capacitor is not a constant current. That's what the complicated formula is about. You can do that with a $20 "scientific" calculator. In fact, you can download a scientific calculator and not spend the money!

Vthr is the voltage when the gate is just starting to turn the MOSFET on.
10 volts is the gate voltage that will fully turn on the MOSFET that I have...not the one you have. Read the data sheet. Look at the graphs until they make sense.
The GATE IS A CAPACITOR. You can treat the gate as nothing but a capacitor and get the answers right.

As for the voltage changing when you disconnect a MOSFET from an active circuit, don't do that! Turn the volts off, first...and while you're at it, send us a drawing of your circuit.