Q1 has 10.4V across it at 75mA, so its going to be pretty hot (well above its rating in fact), but Q2 has
nothing to limit the load current so its pulling a huge current - Vbe is over a volt, in reality its blown.
LEDs need either a constant current drive, or a current limiting resistor.
You also need to limit the base current of Q2
Hi,
Is this burning in REAL life or in your SIMULATION?
Please post your circuit, not the simulation graphic, get rid of the meters all over the place and draw proper battery symbols.
A quick circuit evaluation, WITHOUT the simulator will tell you what you need to do to get this circuit working.
Thanks.. Tom.. 
PS, Picture of a pen/pencil drawing will do.
1.634V across the base-emitter junction of Q2 is a clue here!
Its just an NPN driving a PNP, with some elementary circuit errors (missing limiting resistors, R2 too large)
There is a clearer version. Its burning up in real life.
This is what the circuit ought to be I think:
The second 1k resistor might need to be lower if the load current is really high. R sets the current for the load.
BTW its the second 1k resistor here that stops the transistors frying themselves.
Oh great you've removed the diagram I commented on and make my postings meaningless. Thanks.
Thanks for the info MarkT. You must have posted in between my update. Ill try that out.
Hi,
Please do not go back to previous posts and change them, it makes the thread of no use to anyone.
Other people may want to use this thread to help with their problem, updating/changing the diagram in post #1 has now made EVERYTHING following it untraceable.
This thread has helped you alone, which is not just why the forum exists and we give help, it is here to help others.
Tom...
Thanks for the information Tom. The update was really just a removal of the meters so its still a good learning point for others. Well noted though.
Thanks MarkT for the suggested resistor, it made all the difference. I can keep my finger on the transistor without it getting hot at all.
The circuit seems to work fine now this way. Any reasons to change the other values?
Thanks again for all your help.
So to sum it up , the NPN transistor Q1 was just getting to hot without the R4 resistor.
Although it appeared to be rated for what amperage it would see, in reality it cooked it pretty quick.
slickpuss:
So to sum it up , the NPN transistor Q1 was just getting to hot without the R4 resistor.Although it appeared to be rated for what amperage it would see, in reality it cooked it pretty quick.
It's the power rating on Q1 that was exceeded in the original circuit. Roughly, assuming no more than about 1 V drop across Emitter to Base of Q2, then the voltage across Q1 Collector to Emitter is over 10 V. The current might be something on the order of 0.5 A, so power was on the order of 0.5 A * 10 V = 5 Watts or about 10x the specified maximum. This won't end well.
Adding the R4 resistor limits the current through Q1 to about 10 mA. R4 could be about 5x larger (thus 5k) with the presumption that Q2 gives a minimum current gain of 10.
Referencing the circuit in post #9, the typical maximum current for a generic LED is 20 mA, which indicates that R3 needs to be 510 Ohms or larger.
Unless I'm missing something subtle with regard to power sequencing, R1 isn't doing anything useful and can be deleted from the circuit.
R1 keeps the base from floating when Q1 is off.
Mandatory for a MOSFET, shouldn't be necessary for a BJT, but I've seen the suggestion to add it nonetheless as apparently BJT transistors have the ability to switch themselves on.
Now the real simplification of the circuit would be to use Q1 as low side switch for the LED. Then Q2, R1 and R4 are unnecessary.
I think for high speed switching R1 would help Q2 to get out of saturation. Probably not important here.
When you string multiple transistors together and they are running hot, then the leakage currents get amplified
unless their are base-pulldowns - this can lead to thermal runaway in dire cases. The base pulldown will
usefully speed up switch-off in a BJT by draining the stored charge in the base due to saturation.
MarkT:
The base pulldown will usefully speed up switch-off in a BJT by draining the stored charge in the base due to saturation.
Slightly.
