I saw a schematic for building a NOT gate, it worked, but there's something I don't understand.

Here is a pic of what I did (sorry about my finger hiding the push button)

enter image description here

If I push the button, the LED at the bottom gets turned off, even though there is a path between Resistor 2, the LED and ground (which is a bit strange to me).

I played around and realized that, if I put a resistor between the emitter terminal and the ground terminal, the NOT gate stops working, so I came to the conclusion that the path between Resistor 2, Collector, Emitter and ground, has less total resistance than if it had to go from Resistor 2 to the LED and ground.

My question:

I read that the transistor causes a drop of 0.7V, just like the LED, so why wouldn't the led shine even just a little ?



2 Answers 2


The transistor when saturated has a base-emitter voltage of 0.7V, but the collector-emitter voltage is about 0.2V. That why your led doesn't shine.

If you add an emitter resistor, the current running through it causes a voltage drop, so the emitter (and collector as well) are on a higher voltage.

  • \$\begingroup\$ Ahh! It makes sense now. That's an important detail I couldn't find in my search. \$\endgroup\$
    – user101874
    Commented Feb 28, 2016 at 21:53
  • 1
    \$\begingroup\$ Also, the forward drop of a red LED is well over 1 volt. 1.6 - 1.7 V is typical. \$\endgroup\$
    – Dave Tweed
    Commented Feb 28, 2016 at 22:07


simulate this circuit – Schematic created using CircuitLab

Schematics are so much better for understanding what's going on. If I understand your photo correctly Figure 1 is working and Figure 2 is not.

Figure 1

  • With SW1 open there is no base current for Q1 so it is off.
  • D1 lights with current from R2.
  • The voltage at NODE1 should be about 1.8 V for a red LED.
  • When SW1 is pressed Q1 turns on, the voltage at NODE1 drops to less than 0.5 V and the LED turns off.

Figure 2

  • When Q2 is turned on this time some of the current flows through it and R5. From Ohm's Law we know that this will cause the voltage at the top of R5 to rise. This in turn will raise the voltage on the collector (NODE2) and the LED will remain on - although it may dim somewhat as the current is shared with R5.

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