# Transistor circuits: NOR and short

I've two short questions regarding circuits with a transistors or diodes:

1) The following circuit with two diodes: Does it produce an electrical short for B=1 (5V) and for B=0 there is V[out]=1? Is this correct? 2) Does the following transistor circuit represent a logical NOR (all resistors are equal)? • Sounds like an assignment to me. What have you done so far? Did you try it out in a circuit simulator? – shimofuri Nov 30 '12 at 12:46
• Draw schematics properly with junction dots where there are supposed to be connections. You should be able to see for yourself that the top schematic is ambiguous. If the line crossing above the resistor is supposed to be connected, then how are you going to show unconnected ones? – Olin Lathrop Nov 30 '12 at 13:13
• No, don't know any circuit simulator. I've already written what I've done: I've come to these conclusions (my assumptions above) but I'm not sure if they are correct. – caw Nov 30 '12 at 13:15
• @MarcoW. try circuitlab.com – vicatcu Dec 1 '12 at 0:06

Does it produce an electrical short for B=1 (5V)

I'm afraid this is an example of muddled thinking about short circuits.

For clarity's sake, assume the diodes are ideal, then, the input B can never be greater than zero volts; B will never = 1.

The input B is a short circuit for current into the anode and an open circuit otherwise.

With a real diode, B may rise to a few hundred millivolts.

Regardless, B cannot affect the rest of the circuit. Vout is independent of whatever is connected to B.

Does the following transistor circuit represent a logical NOR

I'll answer with a question: If, as is shown, the base of the right most transistor is connected through a resistor to ground (0V) along with the emitter, is there any way to turn on that transistor with A and/or B?

Think of a diode as a current controlled switch. if no forward current, then leakage might be effectively >10 Mohm near zero current. If it is a small signal diode using silicon, consider it has a 0.6V drop with an internal series resistance such as 100ohms. The Vf at rated current will indicate how much Rs exists in the diode. For a high power white LED it may be Vf=3.4V at 1 A this is equivalent to Vf of 2.9V and an Rs of 0.5 ohm and a leakage equivalent to >>1 Mohm near 0V.

• Oh BTW B input on 1st schematic is wrong going to ground.
• 2nd schematic has error with 1st emitter being grounded.
• To make it simple remember that emitter output is non-inverting and collector output is inverting. so you have OR input with inverting output = NOR.