3
\$\begingroup\$

enter image description here

As a practice question for our exam, we were given this circuit. The question was to find out which transistors have current flowing between the collector and the emitter and which do not. The problem is, we were not taught much except how a transistor works.

How I approached the problem was:

  1. Seeing that the base and the emitters of T1 have the same potential, the transistor will not be opened.
  2. The emitter on T4 is connected to the ground; its potential is 0.
  3. The emitter on T2 is connected to the ground; its potential should be 0 as well (not sure about this one).
  4. Y will be 0, as it is a NAND gate.

But that's about as far as I got.

I have no idea how to solve circuits like this, plus the resistors surely do make a significant change in how the current flows in this circuit. Can anyone point me the right way as to how to solve similar circuits?

\$\endgroup\$
8
  • \$\begingroup\$ @Finbarr Thank you for your answer. My bad, I meant to type in "emitter". The fact that it is off does not help me much, because I have no way of knowing whether current passes between the base and the collector of T1. It is because I don't know the potential at the base of T2. If the potential is greater, the current does not pass. And as the base of T2 is not connected to ground or to the source, I have no way of knowing its potential. That's where I am stuck \$\endgroup\$
    – ampersander
    Jan 13, 2021 at 15:53
  • \$\begingroup\$ Still, I don't know how that helps me with T2 as I can't be sure whether the current passes through or not. Is there a precise way to tell? A rule? \$\endgroup\$
    – ampersander
    Jan 13, 2021 at 15:57
  • 2
    \$\begingroup\$ T1 is off in the conventional sense, i.e. no current is flowing from base to emitter and from collector to emitter. But the base-collector junction is now forward biased, so current can flow into the base of T2. \$\endgroup\$
    – Finbarr
    Jan 13, 2021 at 16:01
  • \$\begingroup\$ @Finbarr My problem is probably with understanding this: current flows if and only if there is a difference in potential. However, I have no way of knowing the potential on the base of T2. I don't know whether T1 is forward biased if I don't know the potential on the other side. \$\endgroup\$
    – ampersander
    Jan 13, 2021 at 16:07
  • 1
    \$\begingroup\$ Don't use hydraulic valve terminology like this: the transistor will not be opened - because when a transistor is activated we tend to say it is "closed" like a contact closure and is therefore capable of passing current. Alternatively, when the transistor is deactivated we say it's open-circuit. See the confusion? \$\endgroup\$
    – Andy aka
    Jan 13, 2021 at 16:09

1 Answer 1

Reset to default
2
\$\begingroup\$

It's mostly about what T2 is doing. It works as a 'phase splitter', providing signals of opposite phase on collector and emitter. These opposite phases control T3 and T4 independently.

  • T2 off: T3 is on, T4 is off -> output high
  • T2 on: T3 is off, T4 is on -> output low

This phase-splitter arrangement can be found in most TTL gate implementations.

For the NAND function, what may be less than obvious is what T1 is doing. T1 dual-emitter NPN in the input behaves as follows:

  • Either emitter pulled low: brings T1 collector low, turning off T2 and bringing the output high.

  • Both T1 inputs high or unconnected: T1 base-collector junction is forward biased through T2 base, turning T2 on and making the output go low.

The bolded point is important: a transistor can behave as a pair of diodes connected together. The B-E junction is one diode, and the B-C junction is another.

In an NPN, positive B-C junction voltage will be in forward bias, allowing current to flow. And that's exactly what happens when the two emitters are disconnected: T1's B-C is forward biased, and conducts through T2 base, turning T2 on.

Try a simulation, here:

enter image description here

I've modeled the T1 dual-emitter NPN as two separate NPNs with their bases and collectors tied together, but the operation is the same: both emitters high or unconnected, base-collector junction is forward biased through T2.

Related: TTL Logic Gate Resistor Values

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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