I need to trigger the mosfet on and off for the circuit shown in the picture.

 

That might be a challenge. The device number is not cleat - it seems to be IRF x505. The symbol seems to indicate a P channel hexfet operating in depletion mode. The odd part is the zener diode symbol superimposed connecting source to drain.

At any rate, you should be able to find a data sheet by using the device number as a search term. That will describe what voltage on the gate control the device.

 

i'm using irf4905, the gate to source voltage is -20 to+20 v.

 

How do you want to trigger it? Do you have another circuit this will be hooked up to? Do you want some sort of clock input or timer input?

The more info you can give the more likely someone can offer a suggestion. To simply answer your question, there are dozens of ways to trigger the MOSFET.

 

I just want the mosfet to act like a switch meaning that there will be a pulse to switch it on and off.

 

You need to take the gate to negative to turn it on, positive to turn it off. The switching transition should be as short as possible.

Reduce the 1KΩ resistor on the gate to 100Ω (this is meant to prevent oscillation) and connect it as close to the gate lead as possible. Since a MOSFET is a voltage controlled device it will not affect current draw.

I could not make out your power supply voltage. It should be 10V or more.

 

The IRF4905 is a standard-level P-ch MOSFET rated for Vdss=-55v, Rds(on)=20mOhm, Id=74A, Qg=180nC. That's a really large gate charge. You will need a pretty capable driver to switch the MOSFET on and off.

Standard-level P-ch MOSFETS are fully OFF when Vgs=0v, and fully ON when Vgs=-10v. Vgs is the voltage on the gate, using the source terminal as a reference point.

Have a look at the attached schematic and simulation on the right. The green trace is a logic-level pulse coming in to the circuit. The yellow trace is the voltage on the gate, measured in respect to the source terminal.

The values of R1 and R2 will need to be adjusted depending upon your supply voltage.
In order for it to really work properly, calculate:
R2 = (+V-12v)/20mA (use the closest standard value of resistance)
R1 = R2 * 6 (approximate; 5 to 7 times as much should work well. If R1 is too low in value, turn-off times will be slow)

Q1 and Q2 make up a common-emitter voltage follower circuit. The relatively small base current is multiplied by the gains of the transistors.

R4 is there to "snub" the tendency of the gate to "ring". Wiring has inductance, and a MOSFET gate is (on a basic level) a capacitor. That makes a series LC circuit that if not controlled, will oscillate for a considerable period of time when the charge on the gate changes states rapidly.

In order to turn the MOSFET fully ON with this circuit, +V must be at least 11 volts. If +V is less than ~18v, D1 is not necessary, and R2 can be replaced by a piece of wire.

 

You need to take the gate to negative to turn it on, positive to turn it off. The switching transition should be as short as possible.

Reduce the 1KΩ resistor on the gate to 100Ω (this is meant to prevent oscillation) and connect it as close to the gate lead as possible. Since a MOSFET is a voltage controlled device it will not affect current draw.

I could not make out your power supply voltage. It should be 10V or more.

I had tried using smaller value resistor but it still can't work.

 

The IRF4905 is a standard-level P-ch MOSFET rated for Vdss=-55v, Rds(on)=20mOhm, Id=74A, Qg=180nC. That's a really large gate charge. You will need a pretty capable driver to switch the MOSFET on and off.

Standard-level P-ch MOSFETS are fully OFF when Vgs=0v, and fully ON when Vgs=-10v. Vgs is the voltage on the gate, using the source terminal as a reference point.

Have a look at the attached schematic and simulation on the right. The green trace is a logic-level pulse coming in to the circuit. The yellow trace is the voltage on the gate, measured in respect to the source terminal.

The values of R1 and R2 will need to be adjusted depending upon your supply voltage.
In order for it to really work properly, calculate:
R2 = (+V-12v)/20mA (use the closest standard value of resistance)
R1 = R2 * 6 (approximate; 5 to 7 times as much should work well. If R1 is too low in value, turn-off times will be slow)

Q1 and Q2 make up a common-emitter voltage follower circuit. The relatively small base current is multiplied by the gains of the transistors.

R4 is there to "snub" the tendency of the gate to "ring". Wiring has inductance, and a MOSFET gate is (on a basic level) a capacitor. That makes a series LC circuit that if not controlled, will oscillate for a considerable period of time when the charge on the gate changes states rapidly.

In order to turn the MOSFET fully ON with this circuit, +V must be at least 11 volts. If +V is less than ~18v, D1 is not necessary, and R2 can be replaced by a piece of wire.

Do you mean i need to add additional components to trigger the mosfet? Actually, i had tried out with 2N6660 which is a n channel mosfet. I can get the result but when i change to IRF4905, it totally can't work. Is it due to the extra diode inside the IRF4905?

 

I recommend using a much clearer schematic making software:
1) The text should be larger and be very clear.
2) The background should be white, not dark grey.
3) The chicken-pox dots all over the place should be turned off.

 

Do you mean i need to add additional components to trigger the mosfet?

Did you even look at the circuit I posted?

Or did I just waste that time?

Actually, i had tried out with 2N6660 which is a n channel mosfet. I can get the result but when i change to IRF4905, it totally can't work. Is it due to the extra diode inside the IRF4905?

It's because the 2N6660 is a very different MOSFET. It has a very high Rds(on), and the gate charge isn't even specified.

The IRF4905 has a VERY LARGE gate charge requirement.

If you would try it out with the circuit I posted, you would have better luck with it.

 

sorry for making the circuit so small. I had attached a clearer picture.

 

Did you even look at the circuit I posted?

Or did I just waste that time?

Yea, I did see the circuit and read all the info u had given me. Thank you very much. I just used a 1k resistor to operate the triggering, so what I mean is other than the resistor, I need other components as well if according to your circuit.

 

The IRF4905 is a standard-level P-ch MOSFET rated for Vdss=-55v, Rds(on)=20mOhm, Id=74A, Qg=180nC. That's a really large gate charge. You will need a pretty capable driver to switch the MOSFET on and off.

Standard-level P-ch MOSFETS are fully OFF when Vgs=0v, and fully ON when Vgs=-10v. Vgs is the voltage on the gate, using the source terminal as a reference point.

Have a look at the attached schematic and simulation on the right. The green trace is a logic-level pulse coming in to the circuit. The yellow trace is the voltage on the gate, measured in respect to the source terminal.

The values of R1 and R2 will need to be adjusted depending upon your supply voltage.
In order for it to really work properly, calculate:
R2 = (+V-12v)/20mA (use the closest standard value of resistance)
R1 = R2 * 6 (approximate; 5 to 7 times as much should work well. If R1 is too low in value, turn-off times will be slow)

Q1 and Q2 make up a common-emitter voltage follower circuit. The relatively small base current is multiplied by the gains of the transistors.

R4 is there to "snub" the tendency of the gate to "ring". Wiring has inductance, and a MOSFET gate is (on a basic level) a capacitor. That makes a series LC circuit that if not controlled, will oscillate for a considerable period of time when the charge on the gate changes states rapidly.

In order to turn the MOSFET fully ON with this circuit, +V must be at least 11 volts. If +V is less than ~18v, D1 is not necessary, and R2 can be replaced by a piece of wire.

Hi, i had tried out the circuit. But i cant get the simulation done when i put the schottky rectifier to the circuit. My software does not have the model that stated in the diagram. It works when i replace with a diode. I had also obtained the same waveform as given in the diagram but my gate voltage is positive not negative.

 

Hi, i had tried out the circuit. But i cant get the simulation done when i put the schottky rectifier to the circuit. My software does not have the model that stated in the diagram. It works when i replace with a diode. I had also obtained the same waveform as given in the diagram but my gate voltage is positive not negative.

It's OK to substitute parts if you don't have the exact models available.

The reason my simulation shows the gate voltage as negative is because I referenced the probe to the MOSFET's source terminal. See the yellow V(Gate,Source)? That means the yellow trace is the voltage at the node labeled Gate, referenced to the node labeled Source.

 

It's OK to substitute parts if you don't have the exact models available.

The reason my simulation shows the gate voltage as negative is because I referenced the probe to the MOSFET's source terminal. See the yellow V(Gate,Source)? That means the yellow trace is the voltage at the node labeled Gate, referenced to the node labeled Source.

Thank you very much. I had obtained the negative values. For the voltage at L1, I keep getting almost the same value which is roughly 22V. My battery is 12V so I'm thinking that the value should be changing from 10V to 22V.

 

For the voltage at L1, I keep getting almost the same value which is roughly 22V.

That doesn't make sense. How are you measuring the voltage?

How fast are you switching the gate of the MOSFET?

When you initially apply a voltage across an inductor (time=0), there is no current through the inductor. The current flow through the inductor increases over time.

My battery is 12V so I'm thinking that the value should be changing from 10V to 22V.

That doesn't make sense either.

 

That doesn't make sense. How are you measuring the voltage?

How fast are you switching the gate of the MOSFET?

When you initially apply a voltage across an inductor (time=0), there is no current through the inductor. The current flow through the inductor increases over time.

I measured the voltage across the diode. My MOSFET frequency is 1K and pulse width is 10%.
My current increases from negative to positive then back to negative but with a very small value.

Since this circuit is a chopper, the voltage across the diode when the mosfet is on is Va=kVs. Then, k value is 0.1 so Va should be 2.2V?