I'm doing an AND gate in a circuit using 2 transistors BC337-40. I'm also using an LED in series and a 680 Ohms resistor.

I expected the LED to light up only when both transistor bases were HIGH. But when the second transistor is short-circuited it closes the circuit with the base of Q2 and current passes from the Base to the Emitter.

How can I go around this with BJT transistors?

enter image description here


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


No way to get around the base current flowing through the terminal biased in forward direction, the emitter if not in reverse operation.
Limit the base current to sane values - TTL used almost 1 mA, if memory serves.
Design the rest of the circuit such that this base current does not impede operation of the circuit.
Here, base resistors of about 4.7 kΩ and a 680 Ω resistor from Q2 emitter to ground should do the trick.
If you place R1 in series with the LED, you even get a TTL level output, but the input will not be compatible to any 5 V logic family.
For lack of voltage amplification, such "gates" will not work in deep networks: to build a 42 input gate, you would need 6 levels at least.


Not the answer you're looking for? Browse other questions tagged or ask your own question.