I found this image of a transistor-based and gate:
and I can't understand what the purpose of the 4.7K resistor is. If the component controlled by the gate is placed between "Out" and ground, then that resistor is just a waste of energy since the circuit should work with or without it.
What am I not understanding?
Best Answer
First, let me tell you that this AND gate is one of the worst AND gate implementations. It has the several problems. Two of them:
1) It lacks regeneration (ok, this is a common issue on other discrete-component implementations too).
2) Under certain circumstances, it could not even work. If B is high (6V in your circuit), then the output will be: ( Vb - Vbe ) * R2 / (R1 + R2), where R2 is 4.7k and R1 = 10k. In other words, 1.69V. In some cases that value could be too high to be considered a "0".
Similar approach (just with a higher input current) could be achieved with a diode-resistor AND. Still, this would not have the issue of point 2.
That said, the 4.7 kOhm serves as pull down resistor, i.e. to create a "0" output signal, when A or B is "0". Furthermore, a "low" resistor value (few kOhms) is mandatory for the case of B = "1" and A = "0" to work correctly (see point 2). In fact, if it were, let's say, 470 kOhm, with B="1" and A = "0" (Vb = 6V, Va = 0V) you would get Vout= 5.2 V, as the base-emitter junction would act like a diode...