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In his video Making logic gates from transistors Ben Eater draws a following scheme

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch is on, the LED stops emitting light (to demonstrate that there is no current on the wire the LED is connected to). I wonder why there is no current on the LED wire. In theory, there should be at least SOME current flowing through LED.

Is it assumed that the resistance of wire going through the transistor is negligible as compared to a huge resistance that is provided by LED and therefore the current going through LED is negligible too?

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  • \$\begingroup\$ If we were talking about the transistor and LED as pure resistors then yes, there would be some current flow through the LED. But we're not. Take a look at the current v voltage graph of a typical LED and see if you can understand why. \$\endgroup\$ – Finbarr Oct 13 '19 at 12:54
  • \$\begingroup\$ Yes, the resistance of the LED (or any diode) is strongly dependent on the voltage across it. At 0.2V or lower (as here) the diode is extremely high resistance. \$\endgroup\$ – Brian Drummond Oct 13 '19 at 13:09
  • \$\begingroup\$ "In theory, there should be at least SOME current flowing through LED." - Yes you are correct, some current will flow through the LED. "and therefore the current going through LED is negligible too?" and that may also be correct, depending on your criteria for 'negligible'. If Ben Eater said that no current will flow then he is technically incorrect - even if that current is only 1 fA (0.000000000000001 Amps). \$\endgroup\$ – Bruce Abbott Oct 13 '19 at 16:32
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When the transistor is turned on in that manner it can be switched into saturation if the base current is sufficient. The collector-emitter emitter voltage will drop to about 0.2 V. This voltage will be applied to the LED.

enter image description here

Figure 1. Current through various colours of LED as a function of forward voltage. Image source IV curves.

Figure 1 shows that none of the LEDs from infrared to ultra-violet will pass any significant current at 0.2 V. There just isn't enough voltage to get the charge carriers to jump the P-N junction.

enter image description here

Figure 2. A water check-valve analogy. Image source: What is an LED?.

If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.

In a similar manner the PN junction in an LED causes a voltage drop. For a red LED it is about 1.5 V to 2.0 V. You need to exceed the Vf to get enough current to flow and light the LED.

The links are to articles by me and may help you further.

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  • \$\begingroup\$ This is a much more thorough answer \$\endgroup\$ – DerStrom8 Oct 13 '19 at 12:57
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Is it assumed that the resistance of wire going through the transistor is negligible as compared to a huge resistance that is provided by LED and therefore the current going through LED is negligible too?

Not exactly. Don't forget that diodes have a minimum voltage required in order to conduct. In normal silicon diodes this voltage is usually around 0.6-0.7 volts. In LEDs it's higher, often between 2 and 4 volts. When you place a transistor across the LED, the Vce of the transistor when saturated is assumed to be less than this, so the LED will not see a high enough voltage to turn on.

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I am replying from mobile so I can't show the schematic I am intending. I would welcome the schematics illustrating my answer.

The transistor acts like a electronic switch. The mechanical switch is controlled by mechanical force, but this electronic switch is controlled by the voltage applied at the base. I won't go further deep into the transistor theory assuming you know the basics.

When the switch is open, there is no voltage at the base. So the transistor acts like a open switch. When the switch is open, the LED'S anode is at the forward voltage level of the LED. The resistor does the job of limiting the current. The cathode is at ground potential. So the potential difference exists across the LED and it lights.

When the switch is closed, the transistor acts like a closed switch. So the LED'S anode is at same potential as the transistor's collector is.

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    \$\begingroup\$ "So the LED'S anode and cathode are both at ground potential meaning no potential difference across the LED." This is not quite correct. The voltage at the collector will be close to but not equal to zero. The lowest it can be is the transistor's saturation voltage. "When the switch is open, the LED'S anode is at the potential determined by the supply voltage." Nope. The LED's anode is at the voltage determined by the LED's forward voltage at the current determined by the current limiting resistor. See my answer. \$\endgroup\$ – Transistor Oct 13 '19 at 20:52

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