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Let say if i have a transistor switching circuit as in the diagram (attached image), by switch on the J1, the voltage drop across Q3 will cause damage to the transistor. Is there any way to add a control circuit to turn off the Q2 so that Q3 will off as well.

Thank for the opinion as i start learning designing electronics circuit now.


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Hi Monster,
Don't short the output, turn off the input instead. The input needs the added current limiting 1.8k ohm resistor anyway to stop the base of the BD135 from being blown up.

Why does the BD135 have such a high current in R1? The current in the BD136 is a max of only 24mA so its base current needs to be only about 1mA. Then a 22k resistor is fine for R1.

R2 is used to turn off Q3. Instead of having 23.2V across it which wastes power, it should be across the base-emitter of Q3.

Why use power transistors when the currents are so low?

Look at my corrections:


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Actually the R3 shd be the load

Then my fixed circuit is fine since it has only 24mA load current.

a max of 1A must be able to supply to the load from another current driver circuit.

This circuit has a 24mA load current. Smaller resistor values are needed for a 1A load current.

A protection circuit to turn off the switching circuit is needed when user accidentally short the output (load removed and short to ground).

A short circuit protection circuit is complicated:
1) a small value resistor senses the load current.
2) A comparator amplifies the small voltage from the current sense resistor.
3) Another transistor turns off the first transistor when the comparator senses that the current is too high.
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