Turning off the transistor feeding LOAD (When the Load beacoms Short Circuit)

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Turning off the transistor feeding LOAD (When the Load becomes Short Circuit)

hello
there is a current limiting circuit.It has been designed to diagnose the short circuit.a Potentiometer has been used as a Load
problem: When short circuit is occurred; Q1 gets hot (P=Vce * Ic = 2.2Watt )

How could I refine the circuit?(I don't want to use Heat sink)



 

According to your simulation Vcb = .2V, Ic = 1A, so Pc = .2 W. I do not believe this because you only have 24V/10K = 2.5 mA of base drive, which to saturate Q1, would mean it has a Hfe of 1.0/2.5mA ~ 400. As 40 would be a more realistic figure I would decrease the 10K to 1 K.
Frank
 

chuckey said:
According to your simulation Vcb = .2V, Ic = 1A
Click to expand...
I think you read it wrong!!!

In first circuit which is short circuit(Load is short) I have problem because:
Vbe= 24.3 volts and Ic=95.2 mA then p=2.2watt

and in the second circuit there is no problem(because load current is not short)
 

I didn't understand your circuit...where is the load connected..where is the short circuit...?
 

How could I refine the circuit?(I don't want to use Heat sink)
Click to expand...

Counter question, what's the intended behaviour? Current limiting generates power dissipation by nature.

Possible option with lower continuous power are
- current limiting with foldback characteristic
- shutdown with delayed restart ("hiccup" current limiting)
 
You may not want to use a heat sink but that's what you need to get rid of heat when a linear current limit circuit is limiting the short circuit current. It a simple matter of volts (across the limit circuit) times the amps.

One way to minimize the power is to use a fold-back type limiter which reduces the current when the voltage across the load drops to a low level (indicating a short or near short).
 
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