Hi readers, is this the correct way to power a PMOS mosfet? I am unfamiliar with PMOS but quite familiar with NMOS.
Question: How to use PMOS to power a 10 Ohm Load? Thank you
I attached LT Spice files below. Help me understand by attaching or posting a new schematic
Thank you and have a nice day ahead

 

See

 

View attachment 252017
Hi readers, is this the correct way to power a PMOS mosfet? I am unfamiliar with PMOS but quite familiar with NMOS.
Question: How to use PMOS to power a 10 Ohm Load? Thank you
I attached LT Spice files below. Help me understand by attaching or posting a new schematic
Thank you and have a nice day ahead

For PMOS, the load connect to the Drain, the source connects to +V. A voltage on the gate less than VGS(th) turns it on.
So...the connections in your schematic are backwards.

 

Don't forget that the "gate voltage" is the voltage between gate and source, not the voltage between the gate and 0V; and that SOURCE goes to the supply, just like it does with an N-channel device, but it goes to the POSITIVE supply.

A gate voltage more negative than Vgs(th) turns the device ON.

 

SeeView attachment 252020

Hi Bordo, thank you for your kind guidance. I have tried placing an inductor near the R1 (10Ohms) Load in order to smoothen out the average voltage (100V)/2 = 50V. However, after i simulated it using an inductor connected together with the load, I get strange signals.

 

You didn't do the right thing and you didn't set the right goal. And I was in a hurry. You should have waited until you found out what you really needed. I have, of course, tweaked my scheme. But it was not optimal. Now I will not interfere in the conversation until I find out what you really need.

 

Hi readers ,
This is the schematic that help to amplify square wave input signal pulse using the p-channel mosfet. The circuit that is without inductor or capacitor is working fine with good efficiency. However, when i included inductor and capacitor in the load, the circuit becomes inefficient and the mosfet has high power dissipation or heat. The reason I want to include capacitor and inductor is to smoothen out the ripple current (0 to 100V) so that it becomes 50V , 50% duty cycle.

Question: How to average out the ripple current (0 to 100V) of load efficiently. I have tried using capacitor and inductor in this case. I am trying to power a very high power device with smooth dc voltage. Thank you for reading . Schematics are below for your kind reference.

 

It won't smooth out the current unless there is a path for the current when the MOSFET is OFF.
That is why @Bordodynov added D3.

 

MrsssSu, you are demonstrating a complete dullness of circuit design. It is not terrible, as long as you are willing to understand the problem. When designing a circuit and selecting circuit elements, you need to consider the allowable voltages of these elements. If you implemented your circuit, the first time you turned it on, the transistors would go up in smoke or explode. Check the allowable voltages of your transistors. Don't rely on the simulator to do this for you. They usually do not simulate the breakdown voltage of transistors. And even if you use a bipolar transistor with a breakdown voltage of 30 volts in a circuit with an effective voltage of 1000 volts on that transistor, the simulator will be fine!

 

The tiny Mosfet in the first post has a maximum allowed drain source voltage of 50V and a maximum allowed heating of 0.36W.
Its on-resistance is a maximum of 10 ohms when cold.
Then if it survives the 100V, the current when it is turned on will be 5A and its heating will be 250W.

 

It won't smooth out the current unless there is a path for the current when the MOSFET is OFF.
That is why @Bordodynov added D3.

Hi Ian, But the diode will absorb a lot of current and it will get extremely hot making the design impratical

 

I'll start at the end. The absence of a diode will cause the switching transistor to fail. This will happen because of inductive rejection. You are a beginner and you should read textbooks about inductance. Although there is an easier way, electrical circuit simulators. Simulate your circuit. See what happens at the drain of the transistor. Also see how much power the transistor is dissipating. Then add a diode and look again. Keep in mind how much current will flow through the diode and what voltage will act on it during operation. It also didn't hurt you to see how step-down regulators work. They use diodes. Are the designers of such stabilizers stupid and dissipate power on a diode for nothing.

 

I'll start at the end. The absence of a diode will cause the switching transistor to fail. This will happen because of inductive rejection. You are a beginner and you should read textbooks about inductance. Although there is an easier way, electrical circuit simulators. Simulate your circuit. See what happens at the drain of the transistor. Also see how much power the transistor is dissipating. Then add a diode and look again. Keep in mind how much current will flow through the diode and what voltage will act on it during operation. It also didn't hurt you to see how step-down regulators work. They use diodes. Are the designers of such stabilizers stupid and dissipate power on a diode for nothing.

Hi, thank you for your time. Now i understand. I am sorry if I bothered you with my beginner knowledge. I am still in high school and just trying to learn all I could.