# Gate Drive Current Measurement

#### BlackMelon

Joined Mar 19, 2015
168
Hi

Please refer to the attached circuit. It shows my gate driver (+8V/-23V, 100kHz output). I would like to know how much average current it supplies the MOSFET's gate during the rise/fall time of Vgs. (Not the periodic average, because it's 0 A obviously). I have no current probe with me because it's f_ _ king expensive...
So, here is my plan to measure. (Let's do it for the rise time)

1. Hook the oscilloscope's probe across Vgs.
2. Calculate dVgs/dt at each point during the rise time. Let's say we got dVgs/dt@t1...... dVgs/dt@t10
3. Calculate the equivalent gate capacitance from the total gate charge curve of my MOSFET: K2837

Qg(rise time) = Qg(-23V to 0V) + Qg(0V to 8V)
Ceq = Qg(rise time) / [8 - (-23)]

4. Calculate Ig ----> Ig@t1 = Ceq(dVgs/dt@t1) ... Ig@t10 = Ceq(dVgs/dt@t10)
5. Ig average = (Ig@t1 + Ig@t2 + ... + Ig@t10)/10

I would like to know if hooking the oscilloscope across Vgs would affect its operation or not?
Also, is there any key point that I forgot according to my plan?

BlackMelon

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#### BobTPH

Joined Jun 5, 2013
8,665
You probably have a resistor between the driver and the gate. Just measure the voltage across it on the scope. If not, add a small one (maybe 1Ω.)

#### BlackMelon

Joined Mar 19, 2015
168
You probably have a resistor between the driver and the gate. Just measure the voltage across it on the scope. If not, add a small one (maybe 1Ω.)
Great idea! I will try it out

#### Sensacell

Joined Jun 19, 2012
3,400
Remember that the GND lead of your scope is connected to the chassis of the scope and probably earth ground too via the power cord.
You cannot just connect a probe across the resistor.

Use 2 channels on the scope and use the MATH- SUBTRACT function to measure the voltage over the gate resistor.

You probably need to use super short ground connections, even the 4" typical ground lead will give you lot's of measurement artifacts at those speeds.

#### Alec_t

Joined Sep 17, 2013
14,223
Just curious. Why are you driving the gate Vgs at +8V/-23V? At +8V the FET may not have its rated minimum Rds(on). And -23V seems unnecessarily close to the rated absolute max of -30V.

#### ci139

Joined Jul 11, 2016
1,898
you need to keep in mind that :
Depending on (behavior of) the actual load, the ON/OFF driving losses may vary ..
Depending on the circuit / specifics (including the one for the G drive), the driving losses may vary ..
Depending on desired ON/OFF transition times in combination to a specific MOS-FET model, the losses may vary ,,
/// .. erraticly

#### BlackMelon

Joined Mar 19, 2015
168
Just curious. Why are you driving the gate Vgs at +8V/-23V? At +8V the FET may not have its rated minimum Rds(on). And -23V seems unnecessarily close to the rated absolute max of -30V.
From the Vgs vs Qg curve, if we increase Vgs beyond the flat region (5.1V in my case), the Rds(on) will be at its minimum and not decrease anymore. However, driving at 5.1999V can be interfered by noise. So let's do 8V.

Turning the MOSFET off by -23V has two advantages:
1. Turn-off speed is very fast, due to high magnitude of Vgs
2. Harder to turn-on.
3. Because of 1 and 2, two MOSFETs in the half bridge configuration will not shoot through.