1
\$\begingroup\$

I've read several so-called explanations on the web of how a NOT gate works, but they all explain WHAT it does, not HOW it actually works. I know what it does.

Consider this schematic of a NOT gate:

NOT gate

When A is low (0), switch T1 is open, and OUT is high (1). That I understand: Current passes from the positive voltage (+Vcc) through R2 to OUT. Is this correct?

What I don't understand is what happens when A is high (1) and T1 is closed. OUT is supposed to be low (0) in this case, but I don't see what's actually happening in terms of current. There is a path through T1 from +Vcc to ground, but how does that result in OUT being low (0)?

Can someone who understands this please explain in terms of voltages and current flow exactly what's happening?

\$\endgroup\$
  • 1
    \$\begingroup\$ The collector current causes a large voltage drop across the collector resistor. The collector becomes ~~ zero volts. \$\endgroup\$ – analogsystemsrf May 12 at 4:00
2
\$\begingroup\$

Let's take a simpler approach.

Remove "R" resistor from the circuit and replace "T1" transistor with a resistor "R3".

Now, also assume that when "A" is LOW(0), "R3 >> R2". (R3 is much bigger than R2, we will pick R2 as such)

Also assume that when "A" is HIGH(1), "R3 << R2". (R2 is much bigger than R3)

As you can see, "R2" and "R3" are just a basic voltage divider circuit.

You can assume those because T1 can be approximated by a resistor which changes its resistance based on the current applied on its base.

So to answer your questions:

When A is low (0), switch T1 is open, and OUT is high (1). That I understand: Current passes from the positive voltage (+Vcc) through R2 to OUT. Is this correct?

Yes. Because T1 has very high resistance since it is open. It still passes some current though, i.e. leakage currents.

What I don't understand is what happens when A is high (1) and T1 is closed. OUT is supposed to be low (0) in this case, but I don't see what's actually happening in terms of current. There is a path through T1 from +Vcc to ground, but how does that result in OUT being low (0)?

It results in low(0) output because now that the T1 switch is closed, it has very low ON resistance, a.k.a R3 is very low compared to R2.

This is of course all possible because you have to pick the right R2 and T1. If you pick for example T1 with on resistance 0.5 ohms and R2=0.5 ohms, then you'll see that the output is half Vcc, when A is HIGH(1).

\$\endgroup\$
  • 1
    \$\begingroup\$ Thanks, Maximus... your use of the term "voltage divider" led me to the Wikipedia page by that name, where I found the formula supplied by @Kalsi. I see now that an open T1 has a very high resistance, and a closed T1 has a very low resistance. Thanks to you both! Wish I could make both of your answers "correct." \$\endgroup\$ – Marc Rochkind May 12 at 16:34
  • 1
    \$\begingroup\$ +1. And since the input voltage is transferred to a change in output resistance R3, we could call T1 a TRANsferred reSISTOR ... or TRANSISTOR. \$\endgroup\$ – Brian Drummond May 12 at 18:28
2
\$\begingroup\$

When A is low (0), switch T1 is open, and OUT is high (1). That I understand: Current passes from the positive voltage (+Vcc) through R2 to OUT. Is this correct?

Correct. And the maximum current that can be drawn is limited by the resistor.

What I don't understand is what happens when A is high (1) and T1 is closed. OUT is supposed to be low (0) in this case, but I don't see what's actually happening in terms of current. There is a path through T1 from +Vcc to ground, but how does that result in OUT being low (0)?

I will be similar to the circuit below.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. An equivalent circuit using a relay.

Here it should be clear that with the base voltage low the relay will not be energised, the contact will be open and Vc will be pulled high by R1.

When the base voltage is high the relay will be energised, the contact will close and Vc will be pulled low.

Notice that when the relay contact closes we have a short to ground and can pass high currents with no voltage drop. As pointed out already the high state current is limited by R1. This means that the circuit works as a far superior current sink than it does as a current source.

The beauty of the circuit is that Vcc can be a different voltage to the base driving circuit so the circuit can act as a level shifter between two different voltages of logic.

Note that the transistor circuit won't be quite as good as the relay as it will have some voltage drop between the collector and emitter and this will vary with temperature and current.

\$\endgroup\$
1
\$\begingroup\$

For the circuit to work like an inverter (when logic ‘1’ is applied at base of BJT), the ON state resistance (emitter to collector resistance) of Transistor (BJT) say Rt1, has to be very low compared to that of R2.
So this basically becomes a voltage divider circuit. As a result most of the voltage falls across the larger resistor R2. Thus we get typically very low voltage at output node, which we consider as ‘0’ logic.

Vout = (Rt1/(Rt1 + R2))*Vcc
assuming R2 >> Rt1
Vout = 0 (approx.)

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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