# The input to the gate of an N-channel MOSFET is a pulsed signal but I am getting a ramp signal

I am trying to generate an output voltage of minimum 30V and maximum 50V, current of minimum 10A for a load of 2-10ohms using a MOSFET.

I am connecting an NPN transistor's collector to the gate of the MOSFET. I am applying a 5V pulsed input to the base of the transistor. The input to the gate of the IRF530NS MOSFET must be a pulsed signal but I am getting a ramp signal.

The frequency of the pulse needs to be 100KHz. The supply to the MOSFET drain is 50V. I want the output voltage of the MOSFET to be 50V. 30V also is fine but 50V would be better.

At the output of the MOSFET, I am able to generate 10A only upto 5 ohms. What may be the reason for this? I want to generate minimum 10A for 10ohms of load too.

Please suggest any changes in the circuit so that I can achieve my desired output.

IRF530 has RdsON of 160 mOhms which corresponds to a dissipation of 16W at 10A, which is unnecessary. Here's an example thru-hole MOSFET with 10mOhm RdsON, although I believe SMD would be better suited to a 100kHz switching frequency due to lower inductance.

So with 10 mOhms RdsON you're looking at 1W losses due to RdsON.

Switching losses can be approximated by Frequency * Switching time * I * V. So for 50V 10A, we could have:

• With 100ns switching time, 5W switching losses: OK for thru-hole package FET.

• With 20ns switching time, 1W switching losses: OK for SMD FET.

Appropriate switching time also depends on inductance and wiring to the load: if actual wires are involved, 100kHz and 100ns are already on the "too fast" side. So let's go with 100ns.

The thru-hole MOSFET has a gate charge of: 24nC at VDD ≈ 64 V, VGS = 10 V which are close enough to 50V that we can use this value.

In order to pump q=24nC into the gate in t=100ns we need a current of i=q/t=240mA.

R13 limits current to a much lower value, which is why the gate is charging so slowly.

You need a push-pull driver that is able to source and sink 240mA.

Gate voltage should also be limited to a safe value like 12V. With your current setup, nothing stops Vgs from reaching 50V, and the maximum allowed is 20V, which means the MOSFET gate oxide will puncture, destroying the MOSFET.

You've got two separate problems.

The first is the low output current. You simply cannot get 10A through a 10 ohm resistor with a power supply of only 50V. Ohm's law (E=IR, rewritten to I=E/R) says you can get 5A through a 10 ohm resistor with a 50V supply (I = 50V/10 ohm = 5A.)

If you want 10A through a 10 ohm resistor, then you'll need a 100V power supply.

The second problem is that your gate voltage is a ramp. That comes from the way you are driving the gate.

The NPN transistor pulls the gate down to ground (0 V) quickly, but pulling the gate back up is done through R13. That's a 4k resistor.

The gate of a MOSFET is a capacitor. You have to "fill" that capacitor to turn the MOSFET on - that is, to let current flow from the drain to the source.

The datasheet for the IRF530NS says there's about 920pF of capacitance gate capacitance when the gate starts from 0 V.

That's parameter $$\C_{ISS}\$$.

With the 4k resistor, that means that it will take several microseconds to charge the gate. You are sending pulses at 10 microsecond intervals, so the charge time for the gate is as long as (or longer than) the pulse interval.

You need to actively pull the gate high through another transistor or with a gate driver. What you really need to do is to actively pull the gate up and down with a gate driver. Charge and discharge of the gate voltage both take time.

In your simulation, you could simply change the pulse generator to the correct voltage values and drive the gate directly from the pulse. I expect your real circuit will need a gate driver since you probably can't change the parameters of the pulse. There are ICs available that do the job.

Alternativell, this answer gives an example of a MOSFET gate driver made with discrete components:

Vi is the pulse, Vo goes to the MOSFET gate. Note that the circuit is powered from 10-15V.

That brings us to a third point: You are lucky that your circuit works as poorly as it does. The slow charge time of the gate through resistor and the rapid pulse rate prevents the gate voltage from getting above 20V. You are trying to pull the gate to 50V, but that exceeds the maximum gate to source voltage ($$\V_{GS}\$$ as given in the datasheet:

If your gate driver circuit were working properly, it would slap the gate with 50V, and destroy it.

You need to use a lower voltage to drive the gate. You need to stay well below 20V. 12V is a typical value.

• Thanks for the reply. I also have a challenge to get 10% duty cycle at my output Commented Nov 14, 2023 at 11:55
• Can you explain what exactly is the purpose of the diode in your given circuit. Is it a normal PN-junction diode? Commented Nov 14, 2023 at 12:04
• @SahasraVaiishnavi: Check the linked answer for more details. It isn't my circuit. D1 keeps R1 from pulling up the gate signal. It ought to be a normal signal diode.
– JRE
Commented Nov 14, 2023 at 12:30