# Can somebody tell me the calaculations of this 2 transistor circuit?

can somebody tell me this circuits calculations, how come 10mA

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Just a short hint:
When you assume a "normal" (classical) value for the voltage across the B-E path of Q6 (linear operation of Q6), it is a very simple task to get a rough estimate for the current through a resistor which is in parallel to this B-E junctiun.

CWatters
why you make it so difficult?????
it easy, i put the problem in other forum
this is the answer for there
The base-emitter voltage of the 2N5401 Q6 is about 0.62V when it begins to turn on. When Q6 turns on it begins to cutoff Q7 so the current in R9 and Q7 is constant at 0.62V/62 ohms= 10mA.

donpacino
Gold Member
why you make it so difficult?????
it easy, i put the problem in other forum
this is the answer for there
The base-emitter voltage of the 2N5401 Q6 is about 0.62V when it begins to turn on. When Q6 turns on it begins to cutoff Q7 so the current in R9 and Q7 is constant at 0.62V/62 ohms= 10mA.
The goal of this forum is to share knowledge and grow the community and members, not to just blindly give you answers. In order to properly help you learn we need to know what level of experience you have, and why you are stuck on a problem. We won't spoon feed you answers if you don't respond to our questions.

LvW
berkeman
Mentor
why you make it so difficult?????
Why do you need to be given answers, when it is so easy to figure them out with our hints. You will not get very far in life with this approach, I'm afraid.