Hi ,

i am implemented Switching circuit for 50V@10A .and i am controlling on and off by micro controller.but i am getting 50V@2A maximum.
and let me know what should i change in below attached circuit to get 50v@10A.


Thanks,
Ashok

 

Hello,

In the configuration you have now, the gate voltage will be 2/3 X 50 Volts = 33.33 Volts, wich will exeed the mosfets specs.
Exchange R1 and R2 to get 1/3 X 50 Volts = 16.666 Volts.

Bertus

 

hi a,
You say that you getting 50V at 2A, what is the value and wattage rating of the load resistor.?
E

 

hi a,
You say that you getting 50V at 2A, what is the value and wattage rating of the load resistor.?
E

Hi,

6 OHM(550W)

Thanks,
Ashok

 

Hello,

In the configuration you have now, the gate voltage will be 2/3 X 50 Volts = 33.33 Volts, wich will exeed the mosfets specs.
Exchange R1 and R2 to get 1/3 X 50 Volts = 16.666 Volts.

Bertus

Hi,
In the current values what i implemented in circuit i am getting gate voltage of 16.66V. if i exchange the resistor values i will get 33V at gate.


Thanks,
Ashok

 

It's the Gate → Source voltage that is too high in the #1 drawing, not the absolute voltage measured from Ground.

 

Hi,
In the current values what i implemented in circuit i am getting gate voltage of 16.66V. if i exchange the resistor values i will get 33V at gate.

That is the voltage from the gate to GND. But you are using a p-channel MOSFET, with the source tied to the 50 V. The correct measurement always is from the gate to the source. Follow Bertus' advice.

ak

 

How can you possibly get 50 volts and 2 amperes across the load?

If what you are saying is that the supply is 50 V and the current is 2 A, then that is possible, but it also means that with a 6 ohm load the voltage across the FET will be 38 volts and its dissipation 76 watts, which will cause it to heat to high temperature very quickly unless it is on a good heatsink - and given that is is a surface mount package that is next to impossible. On one square inch of foil the predicted die temperature would be 3800°C! (50°C/W)

 

50 V / 6 ohms = 8.33 Amps- that's all you will ever get.

ohms law.

 

Here is a much simpler way to do it.
The NPN acts as a current sink, it sinks about 1.2 mA, which gives you about 12V Gate to Source, over the gate resistor.
Works as long as the supply voltage is over about 15 volts.

 

If load inductive place a transient suppression diode across load for
reverse V protection.

Regards, Dana.