# Current through a transistor

I have a problem related with a transistor circuit. Please tell me whether my approach is correct or not.

I need to find the current through the 94.6ohms transistor.

My approach (Assuming PN voltage is 0.6V)

1. Since the NPN transitor has 5.6V between the base and emitter, it will be biased. So, the voltage at the base of NPN will be 0.6V with respect to ground.
2. So, current through 1K resistors will be (15V-0.2V)/2kohms = 7.4mA. (Vce = 0.2V)
3. Since the voltage at the top of 1k emitter resistor is at 0.6V, , the PNP transistor now, will turn as, the voltage between the base and the emitter is greater than 0.6V (As 15V-0.6V > 0.6V)

4. So,Current through the 94.6ohms is (15V-0.2V)/94.6Ohms = 156mA.

Is my calculation right?

• The circuit is interesting because the PNP emitter is at the same voltage as the NPN base, idealistically speaking. (It is also the basis for a practical circuit, though this isn't a good example of one.) This would imply about 100 mA in the resistor you mentioned (99 mA would be closer.)
– jonk
Commented Nov 2, 2019 at 4:09
• Could you explain your 100mA calculation ?
– user220456
Commented Nov 2, 2019 at 4:10
• And the voltage at the base of NPN is 0.6V and PNP is also 0.6V with respect to ground, right ?
– user220456
Commented Nov 2, 2019 at 4:11
• No. I see 5.6 V at the base of the NPN, relative to the wire at the bottom of the image, which is also what I'm assuming the 15 V supply rail is also relative to.
– jonk
Commented Nov 2, 2019 at 4:12
• No. I don't think so. When the NPN has more than 0.6V drop, the transistor will turn ON, right. So , it will drop only the required Vbe , right?
– user220456
Commented Nov 2, 2019 at 4:13

Assuming $$\V_{BE} = 0.6V\$$ and $$\I_C = I_E\$$