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I am using Power MOSFET from Vishay part number Sie882DF to operate at linear region, Following are the steps I have followed to determine the switching speed,

Datasheet electrical spec: enter image description here

Sie882DF MOSFET are driving with a 5V source with 100 ohms series resistance. Gate threshold is specified in datasheet at between 1 and 2.2 V. This means that the gate charging current would be 28-40mA.

Rs = 100E; Vgs(th) = 1-2.2V; OPAMP O/P max (Vd)=5V Gate charge current Ig = (Vd-Vgs(th))/R Ig = (5-1V)/100 = 4/100= 40mA

                Ig = (5-2.2V)/100 = 2.8/100 = 28mA

Total gate charge (Qg) = 145nC (as per datasheet)[condition -> VDS = 12.5 V, VGS = 10 V, ID = 20 A]

MOSFET maximum Switching time (Tsw) = QgMax/Imin = 145nC/28mA = 5.1785uSec

Total gate charge (Qg) = 145nC (as per datasheet)[condition -> VDS = 12.5 V, VGS = 4.5 V, ID = 20 A]

MOSFET maximum Switching time (Tsw) = QgMax/Imin = 70nC/28mA = 2.5uSec

Kindly verify above calculation is correct or not, provide your valuable feedback thanks in Advance

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1 Answer 1

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Sie882DF MOSFET are driving with a 5V source with 100 ohms series resistance. Gate threshold is specified in datasheet at between 1 and 2.2 V. This means that the gate charging current would be 28-40mA.

No that is incorrect. You are charging a gate capacitor and that means it begins charging from 0 volts. At 0 volts, the current is simply 5 volts / 100 Ω = 50 mA.

The charging current diminishes exponentially to zero as the gate rises towards 5 volts. Hence, I'd model it as an RC time constant and see how many time constants are needed to reach a gate-source voltage that is sufficient to produce the drain current you want at the on-resistance you need.

But it's a little more complex than that. If you look how the gate-capacitance varies (from the data sheet) with applied voltage you'll see what I mean: -

enter image description here

So, gate capacitance starts at 8000 pF and diminishes to about 6800 pF at 5 volts so, you could roughly call it 7400 pF (average) and, because the graph is "typical" you might choose to add an extra factor of about 1.5 on top of that (11 nF).

Note that the gate charge figures of max to typical are about 1.5 times bigger

So, I'd be considering 11000 pF in an equivalent RC circuit. But how far up the RC charge curve do you need to reach? Only you can answer that but, here's an example: -

enter image description here

I've put my red marker on 20 amps drain current and 0.1 volts \$V_{DS}\$. This is an \$R_{DS(ON)}\$ of 5 mΩ.

This nicely falls on the 3 volt curve of gate voltage so, I would say as a minimum, the RC charge curve needs to reach 3 volts for this scenario. That's roughly 1 RC time constant hence, the rise time would be 11 nF × 100 Ω = 1.1 μs. But, Id be tempted to assume 1.5× this figure to account for the graph immediately above being typical only (1.65 μs).

Then, after saying all the above I'd want to model the circuit in a simulator and double check things.

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  • \$\begingroup\$ During the miller plateau (the phase one could interpret as turning on/off), the gate current is constant as stated by the OP. \$\endgroup\$
    – tobalt
    Oct 4, 2023 at 9:17
  • \$\begingroup\$ He also uses a charge value that is when VGS is 10 volts. \$\endgroup\$
    – Andy aka
    Oct 4, 2023 at 9:27
  • \$\begingroup\$ That mislabelled graph is capacitance vs drain-source voltage, not gate-source voltage. Ciss changes a little bit with gate voltage, but not by much, and it's the variation with drain voltage that matters. Datasheets almost never include plots of Ciss vs Vgs. \$\endgroup\$
    – Hearth
    Oct 4, 2023 at 13:34

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