http://imgur.com/a/FmrJg

This is when both A and B are at 5V

The base voltage of Q5 is 83 mV, the base voltage of Q6 is 1.9V, and the voltage at the emitter of Q5 and the collector of Q6 is 1.16V which makes absolutely no sense to me.

Any help would be appreciated!

 

Since Q5 is off, Q6 has no collector voltage. The Base-Emitter junction is acting like a simple diode, and the 'highs' on Q3/Q4 emitters passes through R9 and Q6 B-E junction.

 

Edited schematic for convenience:

 

• Q5 is OFF so the collector of Q6 is an open circuit.
• Q3 and Q4 are ON with an emitter voltage of about 4.3V at the left of R9.
• Q6 has no collector current so R9's current goes through the base-emitter junction of Q6 and through R8 to ground, with a Vbe drop of about 0.7V, giving about 3.6V left over across R9 and R8.
• Due to the voltage divider action, about 1/3 of this voltage appears across R8 or 1.2V (close to your observed 1.16V).
• This voltage also appears at Q6's collector since the transistor is ON, making its collector and emitter voltages essentially equal (a transistor can conduct in either direction when ON).
• Q6's base voltage is about 0.7V higher than its emitter or 1.9V, as you observed.

Make sense now?

So the circuit is working as built, but perhaps not as you intended.

 

• Q5 is OFF so the collector of Q6 is an open circuit.
• Q3 and Q4 are ON with an emitter voltage of about 4.3V at the left of R9.
• Q6 has no collector current so R9's current goes through the base-emitter junction of Q6 and through R8 to ground, with a Vbe drop of about 0.7V, giving about 3.6V left over across R9 and R8.
• Due to the voltage divider action, about 1/3 of this voltage appears across R8 or 1.2V (close to your observed 1.16V).
• This voltage also appears at Q6's collector since the transistor is ON, making its collector and emitter voltages essentially equal (a transistor can conduct in either direction when ON).
• Q6's base voltage is about 0.7V higher than its emitter or 1.9V, as you observed.

Make sense now?

So the circuit is working as built, but perhaps not as you intended.

Since Q5 is off, Q6 has no collector voltage. The Base-Emitter junction is acting like a simple diode, and the 'highs' on Q3/Q4 emitters passes through R9 and Q6 B-E junction.

Hey, thanks for the replies! What you guys are saying makes sense, but how is {Q5, Q6, R6, R8, R9, output} any different from an AND gate?

 

Hey, thanks for the replies! What you guys are saying makes sense, but how is {Q5, Q6, R6, R8, R9, output} any different from an AND gate?

Those components don't really make a gate, so I don't understand your question.

 

Maybe there are just some missing components?

 

See

 

Those components don't really make a gate, so I don't understand your question.

What I'm saying is this looks exactly like what I have on the right where my circuit isn't working correctly.

 

A=0 and B=1 ==> 0<Q<1
BCq2-close, BEq2-open

 

A=0 and B=1 ==> 0<Q<1
BCq2-close, BEq2-open

But I'm gett

A=0 and B=1 ==> 0<Q<1
BCq2-close, BEq2-open

Yeah, I can see how this gate wouldn't really work correctly either when B = 1 and A = 0. What is the standard way to make an AND gate then?

 

What I'm saying is this looks exactly like what I have on the right where my circuit isn't working correctly.

That will sort of work as an AND gate if the value of R is much larger than 1kΩ. That will minimize the feedthrough voltage with A is low and B is high.
The upper limit on R is determined by the current gain of the transistor.

 

But I'm gett


Yeah, I can see how this gate wouldn't really work correctly either when B = 1 and A = 0. What is the standard way to make an AND gate then?

Here's a better version:

 

Here's a better version:

Oh wow that actually works perfectly, thanks. One thing I noticed with my circuit was that the voltage dropped to about 3.3V after going through the gate. Why does the same thing not happen with this AND gate?

 

Oh wow that actually works perfectly, thanks. One thing I noticed with my circuit was that the voltage dropped to about 3.3V after going through the gate. Why does the same thing not happen with this AND gate?

Because there are no base-emitter voltage drops in the output path.
The output is a common-emitter transistor that is either fully-on saturated (output within a few mV of ground) or full-off (output pulled to the supply voltage by R4).