 # MOSFET as a switch, try to understand transient behaviour

Hi everyone

I wonder, if someone can explain to me the transient behaviour of a MOSFET in a specific situation.

I want to discharge a capacitor over a solenoid. The capacitor has 1000 muF and is charged to 20V. The solenoid has a DC resistance of 0.6 ohm.

When I discharge the capacitor manually (by just connecting + of capacitor to one end of the solenoid and - to the other by hand), I get an expected discharge curve. It looks like a regular discharge curve of a capacitor. With my DSO I captured the voltage drop across the solenoid, it‘s at max 20V.

When I use a push button to close the circuit, I get some bouncing effects but otherwise the curve looks the same. Max voltage drop is slightly below 20V.

When I use a NPN MOSFET IRF530N (low side switch), I see a completely different behaviour. The maximum voltage drop is only 12V and the curve is completely different:

Can anyone explain me what‘s going on and what I could do to get a regular discharge curve?

Thanks for your help.

What is that really? milli-micro Farad? That does not make sense.

"NPN MOSFET IRF530N"
IRF530N is an N-channel MOSFET. NPN is a BJT (bipolar junction transistor).
https://media.digikey.com/pdf/Data%20Sheets/Fairchild%20PDFs/IRF530N.pdf
As you can see from the datasheet, it needs 10V on the gate to really turn on fully. What are you driving the gate with?

The capcitor has 1E-3F (1000μF), sorry, did not relize it‘s confusing in english.

The gate threshold voltage is 4V, therefore I thought I‘m fine using 5V.

Ok, I will try with 10V. But nevertheless I would like to understand what‘s going on.

What is it that you try to do ?
Discharge the capacitor trough the mosfet ?
I mean, are you controlling the mosfet's gate through some circuit and then you want it to close the circuit and discharge the cap ?

I think you should have posedt the circuit diagram first

Maybe discharge the capacitor through the mosfet with a series resistor of some calculated value. then do the math T = R*C
Again cant tell for sure if you dont post the diagram. sorry.

I think you'll find with Vgs = 4v, the part is barely turned on and has high Rds, nowhere near the Rds of 0.064 ohm when Vgs = 10V.

I don‘t have the circuit sketch on my phone right now. I thought it‘s not required because it is so simple.

The MOSFET is on the low side of the solenoid. By turning it on, the capacitor is discharged over the solenoid and MOSFET.

@CrossRoads

A high R_DS alone will not explain the completely different shape of the curve?

Okay, then the signature on the Oscope is normal i guess. The solenoid, being a coil it will act as a charge reservoir. That will cause the capacitor to have an irregular discharge curve. Remember that a normal discharge curve will only occur with a resistor not an inductor.

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With the very same solenoid (inductor) I get a conpletely different curve when I exchange the MOSFET by a push button switch. Then, it’s a regular discharge curve over a resistor. Therefore I have to assume that the shape of the curve is somehow related to the MOSFET, but I don‘t understand how and why.

You should show the "right" curve when using a switch as well as the circuit diagram (and photo of the actual circuit would be great too).
Note the voltage does not drop to zero but to some small positive voltage. Try to measure the voltage over the MOSFET, it may be interesting.
The peak is only 10us long. It may be a clue.

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Nor do I, but as has already been suggested, it is most likely to be a combination of the MOSFET not being switched fully on, along with some kind of charge coupling back to the gate.

@Smajdalf

What you ca not see in the figure: voltage dropped to 0 after approximately 4ms.

I will try with a higher gate voltage and I will post the circuit diagram and the other curves later on (don‘t have access right now).

If you zoom out to see the whole 4 ms the decay is exponential?

A circuit diagramm is needed - for sure. Simply to see if reason for obervered current is coming form the MOSFET or your circuit around the MOSFET. And, even then, if the circuit is not feasible for your purpose it should be adjusted to it rather than trying to understand why a not-well designed circuit behaves, well, not-well.
Nevertheless, calculation of internal dynamics of MOSFETs can be quite difficult ... I was able to do this quite a lot of years ago but can't do this anymore... You want to use the MOSFET as a switch, so drive it like this, and share circuit. Thanks.

My guess what is going on with the limited information I have:

The initial voltage rise is caused by opening of the MOSFET. At the peak after about 2.5 us it is open at its maximum (considering the limited Gate voltage). Since the solenoid is an inductor the current lags voltage and is increasing only slowly. Most of the voltage drop over the solenoid is due to its inductance; resistance plays only a minor role.
As the current increases more and more voltage is dropped over the partially open MOSFET and so the voltage over the solenoid decreases again. At about 5 us the current reaches its maximum when the voltage over solenoid is nearly zero. The resulting oscillation afterwards is due to interwinding capacitance of the solenoid and Gate-Drain capacitance of the MOSFET - I don't know what is more important. Anyway the current is nearly constant from this point. After a while the current and voltage stabilize at a "steady state", all voltage drop over the solenoid is only due to its resistance. The MOSFET is in the linear (or saturation) region - it may be considered a constant current sink of about 4 A (!). I expect the solenoid voltage staying nearly constant for large part of the 4 ms while the cap is discharging and the Drain-Source voltage (and capacitor voltage) will decrease about linearly. At the end when the cap voltage decreases sufficiently the current will start to drop, the MOSFET will get to the ohmic region and you may see the expected exponential.

Note that due to insufficient Gate drive the MOSFET dissipates most of the energy stored in the capacitor. According to the datasheet linked previously it stays in the safe operating area so it should not be damaged by your experiments. The solenoid would perform poorly if it had to do some real work.

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Hi,

The gate threshold voltage is the gate-source voltage that the gate BEGINS to make the N-CH MOSFET conduct.
At this point there is still significant Drain to Source resistance.
A Logic level MOSFET has a much lower Vgs and the data will possibly state Logic Level.

Tom...    Thanks for your help. Yes, the problem was indeed caused by the low gate voltage.

To summarize:

Here is the circuit diagram:

If I use the MOSFET with 5V gate voltage to discharge, the curve looks like this:

If I use 20V gate voltage (I only have one power supply with 1 output) it loooks like expected:

I was looking for MOSFETs with very low RDS_on. I will search for one which has logic level gate voltage as well.

Question regarding the low side MOSFET: is there any easy way to drive this N-channel MOSFET with an NPN transistor? I know how to switch a high side P-channel MOSFET with a NPN transistor, but this won‘t work since the gate of the MOSFET would be pulled high when the transistor is off…

Low into PNP turns it on, sources high voltage into N-channel MOSFET to turn it on and drain current from the load.

The PNP transistor needs more than 11.3volt on it's base to turn off.
How is an Arduino pin going to do this.

Study this page.
Diagram#3 could work.
Leo..

Correct, theArduino will never be able to turn the PNP off as shown.

If +5v on the Arduino was connected to +12v and there was no common GND between the Arduino and the 12v supply then the Arduino would be able to turn the PNP ON and OFF.

But there should be a large WARNING sign placed on the circuit as the GNDs are floating, not for the general faint of heart noob.