The gate voltage is applied with respect to the source pin. I.E. Gate = +12V = MOSFET off, and gate = +2V = MOSFET on.

I didn't understand that. How's the gate gonna be off if its at 12V?

 

Is the circuit element labelled "Is1" an ideal current generator? If so, it will force 1.41A into Q1 source irrespective of the gate voltage. The drain current would have to be 1.41A.

 

If i remove the current source, i get a much lower valve of current there. But the MOSFET still doesn't work as a switch.

 

Hi,

Supply the circuit with a voltage source, then limit the current through the solenoid (R1) with a resistor (R2).

Refer to attachment. The IRF7406 is a suitable choice for this, but there are many other choices in different packages. The Vgs voltage should be at least 5V for your application. The attachment shows a 10V Vgs.

Regards,
Ifixit

 

That worked . Thank you . But the PWM is gonna be 0-5V and 0-12V as its the output of a PIC microcontroller . That circuit works only for 0-12V...

 

Your original FET was in the circuit backward, and as they always contain a diode, you just had the conductivity of a diode there. It would never have been possible to turn that off.

What you want is an N-channel FET with logic-level drive (i.e. will turn on fully with a 5V gate signal) set up as an open-drain circuit with the source pin connected to Gnd and the valve/resistor between +12V and the drain pin. You can drive the gate directly off the processor, with just a small resistor between the two. It's best for safety to add a pulldown resistor on the gate, just in case the processor pin goes into high-impedance state, which it will briefly at startup anyway.

The valve coil is undoubtedly quite inductive, so you also need a freewheeling diode in parallel with the coil (with the anode connected to the FET drain and the cathode at the positive side--a "fast" diode or a Schottky diode is best for this).

Then you need a 2:1 voltage divider to get your 10V input into the 5V range of the PIC's A/D converter. After that, it's just software (grin).

Yes, it would be better to have an op-amp circuit to provide current feedback, but you should get reasonable performance even without it.

By the way, the accuracy of this design depends on the matching of the "10V max" input and the 5V power supply that the PIC runs on. If 10V is not 2 * 5V, you won't get an accurate result. I hope that's OK.

 

But the current flowing into the coil is too less... literally lower milliAmps and i need 1.4Amps there .

 

I used multisim for simulation and got the desired outputs. But i used a signal generator there to generate PWM. The signal generator like we know uses 2 pins which is connected across gate and source of the MOSFET for switching. But the microcontroller has only one pin for PWM output. How do i solve this problem?

 

You are using an "isolated" the signal generator . The common (-) of the generator is isolated from the common of the MOSFET circuit. To use the microcontroller, it's common and the MOSFET's circuit common must be the same. You need to use a "common source" configuration for the MOSFET.
Move R1/R3 so they are between +12V and the MOSFET's drain. Connect the MOSFET's source to the ground. The microcontroller's common also goes to ground. The single output goes to the MOSFET's gate.

Ken

 

Actually i need 12V across the solenoid coil and a current of 1.5A passing through it at 100% duty cycle. I can manage to get only one of those at a time . How do i approach this problem?