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

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canarybird33

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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 :cry:)

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
 

According to your simulation Vcb = .2V, Ic = 1A
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)

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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