- Joined Feb 1, 2018
You're making this too easy. The Base Emitter voltage depends on the current. This voltage for the second transistor will be less than the third one by 120 mV. That's the voltage for the first transistor, which will be even lower. Also, the voltage on the first transistor may be greater than the power supply.If the LED were in the emitter path the voltage available to drive it would be 3V- 3 x Vbe = 3V - 3 x 0.6V = 1.2V, which would not be enough to light it. (Vbe is the base-emitter junction voltage drop). As indicated above, there should be at least one current-limiting resistor in the circuit.
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