Transistors and boolean expression

[charlies1902]

Hi.
If I could get some clarification on the attached circuits that would be great. The question asks to find the boolean expression for C.

I already know the answers, but I don't quite see how they got it.
For the first circuit: C=A_bar
The truth table looks like this:
A C
0 1
1 0

This is where I'm getting completely lost. Isn't C basically connected to ground?? Wouldn't that mean C is always 0?

 

[NascentOxygen]

In the first circuit, C is solidly connected to ground. C is permanently 0.

I think your lecturer must have had a slightly different circuit in mind. Or else he has a wicked sense of humour.

What is the output of your second circuit?

 

[charlies1902]

In the first circuit, C is solidly connected to ground. C is permanently 0.

I think your lecturer must have had a slightly different circuit in mind. Or else he has a wicked sense of humour.

What is the output of your second circuit?

Thanks.

I actually think he was being serious because I remember him justifying it. He said something about the arrow in the transistor represented a diode, so C would be the inverse of A. I guess that makes sense, but C=0 makes sense as well. So I think maybe the circuit was drawn wrong?

 

[NascentOxygen]

I guess he intended that C be at the collector, not the emitter.

 

[charlies1902]

I guess he intended that C be at the collector, not the emitter.

Isn't the diode from the base to the emitter?



I don't quite understand the second circuit.
This is what I 'think' happens:
If A or B is =1, then there would be current flowing through the circuit, thus causing a voltage drop over the top resistor (the one right next to the 5V symbol).
If both A and B are on, then the above statement holds true as well.
Otherwise (A and B are both off), then the 2 transistors act as gaps, thus no current would flow in the circuit. Thus, the voltage drop across the top resistor is 0V. Then C=5V.
This matches the truth table they have:
A B C
0 0 1
0 1 0
1 0 0
1 1 0

 

[NascentOxygen]

Isn't the diode from the base to the emitter?

There is a PN junction there, yes.

I don't quite understand the second circuit.
This is what I 'think' happens:
If A or B is =1, then there would be current flowing through the circuit, thus causing a voltage drop over the top resistor (the one right next to the 5V symbol).
If both A and B are on, then the above statement holds true as well.
Otherwise (A and B are both off), then the 2 transistors act as gaps, thus no current would flow in the circuit. Thus, the voltage drop across the top resistor is 0V. Then C=5V.
This matches the truth table they have:
A B C
0 0 1
0 1 0
1 0 0
1 1 0
✔

So what Boolean operation does this represent? AND, OR, or what?

 

[charlies1902]

There is a PN junction there, yes.

So what Boolean operation does this represent? AND, OR, or what?

NAND.

Are my justifications in the previous post correct?

 

[NascentOxygen]

NAND.

Yes

Are my justifications in the previous post correct?

They sound right.

 

[charlies1902]

For the first circuit, if we placed the output C right between the transistor and resistor, would that mean C is the inverse of A?
This is why I think it's that.
If A is low (0V), then the transistor is OFF, then the circuit becomes an open circuit, so the voltage drop across the resistor would be 0V due to no current flow. Thus C is =5V.
Else, transistor is ON=>current flows=>V drop across resistor=>C is low.

 

[NascentOxygen]

For the first circuit, if we placed the output C right between the transistor and resistor, would that mean C is the inverse of A?
This is why I think it's that.

Correct, with the collector being point C.

 

[charlies1902]

Correct, with the collector being point C.

If it was at C wouldn't that mean C is always HIGH(1)=5V?

 

[NascentOxygen]

If the output is taken from the collector, it will be 5V when there is no current through the resistor & transistor.

When current flows into the base (to the emitter), the voltage between collector and emitter drops to approx. 0V.

 

[charlies1902]

If the output is taken from the collector, it will be 5V when there is no current through the resistor & transistor.

When current flows into the base (to the emitter), the voltage between collector and emitter drops to approx. 0V.

Sorry, I got my notations wrong. I kept thinking that the collector was above the resistor. Normally, I think of the collector as the very top of the circuit in a transistor, but the addition of that resistor changed it.