Guru
Joined: 02/02/2017
Location: AustraliaPosts: 1483
| Posted: 02:58am 26 Mar 2017 |
This is growing to be a good topic. I think we needed one where we can talk about details of this and that. I hated potentially derailing other's posts.
Part 4: Closely watched mosfet gate voltages
I have been interested in what is happening to the mosfet gate when switching on and off.
Some test setup facts, first.
Mosfets are IRF3808, 3 per leg of the bridge.
gate drive ICs are IR21844 1/2 bridge rated at 1.4A source current, 1.8A sink current.
The outputs of the driver chips have 4.3 Ohm gate resistors in series, 10K pull down and 1N4148 diodes in parallel with the 4.3 Ohm resistors.
Only a couple of DSO traces this time.
This shows the low side 20kHz mosfet switch ON event.
Light blue is voltage on driver side of the 4.3R resistor
Purple is the mosfet side of the 4.3R resistor.
Blue is the differential voltage across the 4.3R.
The peak voltage drop across 4.3R is not quite 6V. This corresponds to a peak current
of 6V/4.3R = 1.4A
So far so good, The driver IC can only source 1.4A from the specs. so it seems that for a little period the driver IC is in some constant current mode.
Or maybe the gate drive resistor value was chosen well so that the 1.4A max is not exceeded.
Next, I can see the plateau in Vgs when it attains Vgs-threshhold.
Specs say it's 2 to 4V, use the purple trace to see this, it is the mosfet gate voltage, not the driver IC output.
I read about this multi-stage Vgs behavior from many sources but here it is.
It seems to plateau out for about 40ns.
Finally, looking at the blue trace, it appears to me to be just as if we were
charging a capacitor. (we really are when switching a mosfet..)
Anyway, it's really nice to see this in Real Life(tm)
Your thoughts on this?
Just to see if I am on the right planet, let's work out the capacitance at a particular point in time - right after the purple trace plateau finishes and it rises nearly linearly.
Let C (capacitance) = Current x time / voltage change
(this is the same thing as the Inductance formula I used earlier in this topic.)
I will use
1.4A for current which I will make it remain constant for the below calcs.
2 ticks of the 200ns time div, or 80ns
voltage rise will be something like 4 ticks of the 5V, or 4V
so C = (1.4 x 80E-9)/4 = 28nF
I'm probably making a stupid error here but when I look at
It seems to make sense. The drain to source voltage is the inverter supply and it's approx 30V.
Here is the switch OFF event, same trace legend as above.
The blue trace is negative now, the gate drive IC is now sinking current, with a max sink of about -0.46 A
It seems we can drive the gate down faster, the drive IC is rated at 1.8A sink.
I wonder if the 1N4148 is limiting the sink current?
Anyway, the whole point of this is show to myself that what we are doing is charging and discharging a capacitor when switching a mosfet.
I find it necessary to see for myself rather than simply read what it is from textbooks.
And this has lead me to the next question:
what is going on with the 1N4148 diode during mosfet turn OFF.
Is there a big voltage drop across it? Or maybe it has a slow reverse recovery time.
Are these diodes needed? Could I use faster ones or those with less forward voltage drop?
wronger than a phone book full of wrong phone numbers