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I am using a circuit shown below. it is designed to sense 1.5 A whenever the load takes current >= 1.5 A. Rsen senses the related voltage and activates the transistor. Here I use a Rsen = 0.46 R(0.7V/1.5A).

Whenever 1.5 A flows the BC547 turns ON and the gate of MOSFET gets turned off (Vgs less than Vth).

But our problem is source of MOSFET provide a voltage and MOSFET getting over heated.

Can someone provide a solution for this?

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Your circuit is in fact a current regulator: it won't let the current exceed the threshold, but it will never cut off the load in case of over-current. And power MOSFETs make very poor linear regulators because they tend to go into thermal runaway and die.

If you want over-current protection, you need to latch the over-current signal either temporarily or until a reset. If you PWM comes from an MCU, you can route your over-current signal to its input pin and program the protection the way you want.

Another form of protection is to install a polyfuse in series with the load: polyfuses have the latch-up behaviour built-in. Getting back into a working state would require power-down and cooldown delay, so it's a less flexible solution.

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You have built a current limiting circuit.

If the load resistance falls below \$ R_L = \frac {V}{I} = \frac {24}{1.5} = 16 \ \Omega \$ then the circuit goes into current limit.

For example, if you reduce RL to 12 Ω then 2 A would flow if your current limiter was not in place. What should happen is:

  • When the voltage at the BC547 base gets close to 0.7 V it starts to turn on.
  • The BC547 "steals" some current from the 1k resistor and this causes the gate voltage to fall. The IRF540 starts to turn off a little.
  • The circuit stabilises with 1.5 A through RL, the MOSFET and the sense resistor.
  • The voltages will be 0.7 V across RSENSE, 12 x 1.5 = 18 V across RL which leaves 24 - 18 - 0.7 = 5.3 V across the MOSFET.
  • The power dissipated in the MOSFET is given by \$ P = VI = 5.3 \times 1.5 = 8\ \text W \$. This is why the MOSFET gets overheated.

This is always the problem with linear voltage and current regulators. The excess power is burned off as heat. You will need to heatsink the MOSFET properly and make sure that it is rated for this power.

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