# Why is my IRL540 slow to turn off?

I'm trying to switch 12V with an IRL540 from a 5V micro. Right now I have it a 1k resistor between 12V and the drain, the source to ground, and the gate has a 100kHz 5V square wave on it. That's the whole circuit, this was just to validate the switching. If I go faster than this it starts to not even reach 12V before I turn on the FET and pull it down again. As we can see from the capture there's quite a lag in rise time, I suspect by the charge needing to return to the drain of the FET. But should it really take this long? It's at least an order of magnitude longer than what's specified in the data sheet (given that's with Vdd at 50V, but this seems very slow). My square wave seems fine. What can I do to fix this?

• Rapidly changing the charge on a MOSFET can take amps of current in less than a microsecond. Gate resistors are seldom over twenty ohms, and are often parallel with a 1N4148 diode used to speed up the transition to OFF. – user105652 Jun 14 '20 at 2:26

## 2 Answers

The IRL540 has a parasitic capacitor between drain and source so, when you "turn off" the MOSFET, that capacitor ($$\C_{OSS}\$$) takes time to charge via the 1 kohm drain pull-up resistor. To add a little misery/complication, $$\C_{OSS}\$$ changes with $$\V_{DS}\$$ as per this graph: - So, when you initially turn-off the gate voltage ($$\V_{GS}\$$), $$\V_{DS}\$$ starts at 0 volts and $$\C_{OSS}\$$ is about 2200 pF - this has an RC time-constant of 2.2 μs (1 kohm pull-up) and so the output rises relatively slowly initially. As $$\V_{DS}\$$ rises to (say) 3 volts, $$\C_{OSS}\$$ drops to around 1400 pF and things start to speed up but, it's a law of diminishing returns when charging a capacitor from a resistor and, despite $$\C_{OSS}\$$ falling to about 800 pF when $$\V_{DS}\$$ is 10 volts, there is still an overall time period of several micro seconds involved.

If you used a 100 ohm load you would see $$\C_{OSS}\$$ charge much more quickly.

Even though $$\C_{ISS}\$$ (gate capacitance) is circa twice the value of $$\C_{OSS}\$$, I suspect that your gate driving impedance is no more than 50 ohms therefore, $$\C_{ISS}\$$ isn't a significant issue. Its effect will be about ten times less than a 1 kohm drain resistor and $$\C_{OSS}\$$.

• Thanks, this was very informative. So my options then are to either select a different FET or lower that resistor value? – brenzo Jun 13 '20 at 18:35
• Take a step back and tell me what you are trying to accomplish and ultimately, what load are you trying to drive? – Andy aka Jun 13 '20 at 19:01
• I think the other poster answered this, but I'm just trying to use this as a level shifter essentially. I'm not going to be driving a load with it (or a very small one, it's switching the gate voltage to a larger power MOSFET). So I think I'm better off choosing something like a 2N7000. This was the only FET I had handy but I'm going to stock up on a few different options. – brenzo Jun 13 '20 at 19:06
• Yes, a mosfet that isn’t a power type will have smaller drain source capacitance. – Andy aka Jun 13 '20 at 19:10
• You should also be able to model this using a simulator. Do you have a sim tool? – Andy aka Jun 14 '20 at 7:54

Your micro output has to supply the gate charge, so maybe it behaves like a 100 ohm resistor, and 30-40mA flows, so 64nC (max) will take 1.8usec.

If you want it to switch snappily at 100kHz you will need a gate driver capable of much more peak current. There are many gate driver chips available, or you could use a couple BJTs for less performance but lower cost.

• I'm going to try this as well but the other post points out the parasitic capacitor Coss which might not be helped by this. For what it's worth I did try implementing a very simple gate driver by using a BJT to switch the gate voltage on the FET, but that seemed to have little effect. Could be my implementation, though. – brenzo Jun 13 '20 at 18:37
• Yes, the pull-up high value is a major factor, if that’s your final load you’d be better using a 2N7000 etc, but I assume with such a large mosfet you’ll be putting a much heavier load on there. – Spehro Pefhany Jun 13 '20 at 18:39
• Actually no, and I apologize - I should've clarified that in the question. This was the only FET I had handy and was confused about its behavior - it's just being used effectively as a level shifter. I was confused about its performance but am realizing now it's not appropriate for my use case. Yes I'm looking at digikey now and just found the 2N7000, which seems like a better solution for me. – brenzo Jun 13 '20 at 18:50