Why is mosfet turning off slowly?

[ZeroResistance]

I am feed a 100Khz signal to the gate of a mosfet from an arduino and my gate drive is 5V with 50% duty

The circuit is similar to this


except my transistors are 2N3904 and 2N2907 and the base resistor is 1K and I don't have a RC network as shown for the gate driver nor I have used the gate resistor for the mosfet.
The mosfet is IRF540 and I have connected a 1K ohm load from drain to supply. The mosfet supply is 15V.

The voltage at drain is 15V.

However the mosfet is turning off quite slowly although turn on is sharp.

[Zero999]

What's the time base setting on the oscilloscope?

Have you looked at the gate voltage?

You do realise you don't need a base resistor. It will work without one but then a gate resistor is a good idea to suppress high frequency oscillations due to the gate's inductance. Try 10R on the gate and no base resistor.

[ZeroResistance]

What's the time base setting on the oscilloscope?

Have you looked at the gate voltage?

You do realise you don't need a base resistor. It will work without one but then a gate resistor is a good idea to suppress high frequency oscillations due to the gate's inductance. Try 10R on the gate and no base resistor.

Time base is 5uS / div.

New waveforms attached with no base resistor and gate resistor 10E

The top waveform is at the drain.
The bottom waveform  is at the gate.

[Dave_PT]

In the first graph, the voltage is up to ~ 15V.

Should not it be 12V?

I think the gate is taking a long time to charge (1.7nF) ...


PS: Are you switching the MOSFET at 200KHz? Try to reduce the frequency to 20KHz.

[ZeroResistance]

In the first graph, the voltage is up to ~ 15V.

Should not it be 12V?

I think the gate is taking a long time to load ...

The drain is connected to a 1K resistor and the other end of the resistor is given to +15V.

[snarkysparky]

switch transistor to 2n5088, 2n5087.  You need a lot of gain at low current for fast switching.

[Zero999]

The oscillograms suggests a switching frequency of 125kHz, which isn't too high for the MOSFET specified.

What's the time base setting on the oscilloscope?

Have you looked at the gate voltage?

You do realise you don't need a base resistor. It will work without one but then a gate resistor is a good idea to suppress high frequency oscillations due to the gate's inductance. Try 10R on the gate and no base resistor.

Time base is 5uS / div.

New waveforms attached with no base resistor and gate resistor 10E

The top waveform is at the drain.
The bottom waveform  is at the gate.

The gate waveform is fine, so that implies there is something else, other than the gate driver which is responsible for this.

Where did you put the probes to measure the drain waveform? You should connect the oscilloscope's earth/ground to the source and the probe to the drain.

Can you probe both the gate and the drain simultaneously?

[ZeroResistance]

switch transistor to 2n5088, 2n5087.  You need a lot of gain at low current for fast switching.

I don't have those transistors with me at the moment. The one's I have use have a gain of around 300 max.

[bktemp]

Get a proper mosfet driver: 5V is too low for driving a IRF540. And you are wasting ~0.6V at the driver stage, so you are driving a mosfet designed for 10V gate voltage with only 4.4V!

The mosfet isn't turning off too slowly, it is simply your large pullup resistor at the drain being unable to charge the parasitic drain capacitance quickly. Therefore you see a ramp instead of a square wave.
Use 100 ohms or less and you will see a much better rise time.

[ZeroResistance]

I think the gate is taking a long time to charge (1.7nF) ...


PS: Are you switching the MOSFET at 200KHz? Try to reduce the frequency to 20KHz.

The switching frequency is slightly above 100Khz.
This is the screenshot of switching at 20Khz, its improved but not quite but really need this to work at 100Khz.

[Zero999]

It's actually no better. The rise time is the same as it was before. It's closer to an ideal square wave because it's on for longer so the rise time forms a much smaller part of the waveform.

Have you tried adding a 1k resistor between the gate and source?

[ZeroResistance]

It's actually no better. The rise time is the same as it was before. It's closer to an ideal square wave because it's on for longer so the rise time forms a much smaller part of the waveform.

Have you tried adding a 1k resistor between the gate and source?

No noticeable change with 1K resistor between gate and source rise time still around 3.5uS. But it seems what bktemp is saying makes sense probably 4.4 volt may not be sufficient to drive the mosfet. I have some old TC427 lying around will check out with that.

[Kalvin]

The 1 kohm resistor at the drain is pretty large as IRF540 has some capacitance between the drain and gate as well... You could try smaller than 1kohm load resistors and see how it will affect the rise time.

[alsetalokin4017]

Get a proper mosfet driver: 5V is too low for driving a IRF540. And you are wasting ~0.6V at the driver stage, so you are driving a mosfet designed for 10V gate voltage with only 4.4V!

The mosfet isn't turning off too slowly, it is simply your large pullup resistor at the drain being unable to charge the parasitic drain capacitance quickly. Therefore you see a ramp instead of a square wave.
Use 100 ohms or less and you will see a much better rise time.

This. ^^

[ZeroResistance]

Where did you put the probes to measure the drain waveform? You should connect the oscilloscope's earth/ground to the source and the probe to the drain.

Can you probe both the gate and the drain simultaneously?

Oscilloscope is between drain (body) and source.

Simultaneous attached. Gate waveform is 2V / div.

[alsetalokin4017]

In addition to using insufficient gate drive voltage, now it looks like your PNP transistor is not turning on quickly enough to discharge the mosfet gate. When you boost your gate voltage to something reasonable (10V or so) and you use a proper gate driver chip instead of the totempole, you should be OK.

[ZeroResistance]

Get a proper mosfet driver: 5V is too low for driving a IRF540. And you are wasting ~0.6V at the driver stage, so you are driving a mosfet designed for 10V gate voltage with only 4.4V!

The mosfet isn't turning off too slowly, it is simply your large pullup resistor at the drain being unable to charge the parasitic drain capacitance quickly. Therefore you see a ramp instead of a square wave.
Use 100 ohms or less and you will see a much better rise time.

Yess!! 100ohm load makes a huge difference. Drain waveform attached @ 5uS / div

[T3sl4co1l]

The MOSFET is switching as fast as you are driving it.

But the drain voltage is not dependent only on drain current.  There is capacitance, which slows the rise, particularly at low load currents.

If you zoom in on the delay between gate voltage falling, and drain voltage just beginning to tick up, you'll see that the MOSFET is working quite quickly, under 100ns.

(You may even see the voltage reversing momentarily -- gate-drain capacitance pulls the drain down slightly!)

Tim

[Zero999]

Get a proper mosfet driver: 5V is too low for driving a IRF540. And you are wasting ~0.6V at the driver stage, so you are driving a mosfet designed for 10V gate voltage with only 4.4V!

The mosfet isn't turning off too slowly, it is simply your large pullup resistor at the drain being unable to charge the parasitic drain capacitance quickly. Therefore you see a ramp instead of a square wave.
Use 100 ohms or less and you will see a much better rise time.

Normally I'd agree but the load is only 15mA and it is turning on quickly enough. The problem is turn off.

Get a proper mosfet driver: 5V is too low for driving a IRF540. And you are wasting ~0.6V at the driver stage, so you are driving a mosfet designed for 10V gate voltage with only 4.4V!

The mosfet isn't turning off too slowly, it is simply your large pullup resistor at the drain being unable to charge the parasitic drain capacitance quickly. Therefore you see a ramp instead of a square wave.
Use 100 ohms or less and you will see a much better rise time.

Yess!! 100ohm load makes a huge difference. Drain waveform attached @ 5uS / div

One thing you could do is use another transistor to pull it down.

[snarkysparky]

Another thing is that LED below their ON current will supply very little charging current for the mosfet capacitance.  Put a 1k resistive load in parallel with the LED to see if it squares it up any.

[TheoB]

If this is a led driver, why are you watching the voltage? You should connect a LED and measure the current. The current will switch on/off as fast as you can deliver charge to the gate. A slowly changing drain voltage is on it's own no problem (if no current flows through the mos channel).
So a gate resistor makes it slower as it limits the current, but that is not what you measure. You measure an off device and see a Rload*Cdg time constant. If you are going to drive a big current, you need to make sure the gate voltage is high enough to guarantee a low channel resistance. Otherwise it might become hot. (10-20W LED?). Also think about the temperature range you are going to use it for. At the end switching losses can also play a role. During every switch on/off event power dissipated depends on the switch speed (and frequency). I normally limit the gate drive current to 1A or so by a 10 Ohm resistor. Get it on/off as fast as you can.

[neil t]

I'm not quite sure why you using bjt to drive the gate surely an op amp with gain of 2 would drive the gate adequately maybe a resistor to pull the gate down to ground, considering that the Arduino is 0 - 5v you would then have 10v at the gate through perhaps 100 ohms, just make sure the op amp goes to ground (data sheet).

p.s just watch for oscillation, plenty of info on youtube about that.

[Zero999]

Get a proper mosfet driver: 5V is too low for driving a IRF540. And you are wasting ~0.6V at the driver stage, so you are driving a mosfet designed for 10V gate voltage with only 4.4V!

The mosfet isn't turning off too slowly, it is simply your large pullup resistor at the drain being unable to charge the parasitic drain capacitance quickly. Therefore you see a ramp instead of a square wave.
Use 100 ohms or less and you will see a much better rise time.

Yess!! 100ohm load makes a huge difference. Drain waveform attached @ 5uS / div

Reading again: I understand it was the MOSFET's drain-source capacitance which was holding the charge when it it went open circuit, which was responsible for the slow rise time.

You could use the most perfect drive circuit in the world, but it won't discharge the MOSFET's internal drain-source capacitance, if the impedance of the load is too high. See simulation attached, which shows the MOSFET driven with a 10V square wave with 1ns rise/fall times and a 10R gate resistor.

[ZeroResistance]

Another thing is that LED below their ON current will supply very little charging current for the mosfet capacitance.  Put a 1k resistive load in parallel with the LED to see if it squares it up any.

Not really driving an led I want to drive a coil finally that circuit was just a reference circuit  I found on the web.

If this is a led driver, why are you watching the voltage?

Same as above, but point taken!

I'm not quite sure why you using bjt to drive the gate surely an op amp with gain of 2

Never really thought of an op amp, and I never thought an opamp could deliver the charge required for a mosfet running at 100Khz.

[ZeroResistance]

Reading again: I understand it was the MOSFET's drain-source capacitance which was holding the charge when it it went open circuit, which was responsible for the slow rise time.

You could use the most perfect drive circuit in the world, but it won't discharge the MOSFET's internal drain-source capacitance, if the impedance of the load is too high. See simulation attached, which shows the MOSFET driven with a 10V square wave with 1ns rise/fall times and a 10R gate resistor.

Excellent deduction and simulation Hero999 

What would those small negative going sections be caused by?!

The datasheet states that the Vgs Threshold of the IRF540 is 4V max for (Vds = Vgs) and Id = 250uA so probably I'm too close to that region since finally I want to use the mosfet with Id of between 1A to 2A.

I'm just trying to understand is if the mosfet is taking time to charge the Cds is is actually in the linear region during that point of time?