I've started studying basic electronics, and in this video I noticed he had a similar example as in my textbook: https://youtu.be/sTu3LwpF6XI?t=3m28senter image description here

Above is a screenshot if you don't bother to click on the link.

So, in essence.. when the switch is off, the LED to the right on. This means, there's a voltage drop across the second resistor and a voltage drop across the LED.

Now, by turning the switch on, we'll have voltage U across both first two paths. The first resistor should make sure not too much current is flowing through the base and thus, turning the transistor on. Now when the transistor is on, the voltage will be lowered to 0 after the second resistor.

Why the LED turns off at this point is because there's no (or at least not enough) current flowing through it (the potential is zero so there can't be). This should be due to the resistance of the LED itself. Or is there anything else going on here?

  • \$\begingroup\$ The LED turns off because the voltage drops below the threshold and then no current flows through it. Research I-V curve \$\endgroup\$
    – Voltage Spike
    Jul 5 '17 at 22:02

Your understanding is pretty much correct.

When the transistor is turned on it is driven into saturation and becomes a very low resistance in parallel with the LED. This pulls the voltage down towards zero - maybe 0.2 V and this isn't enough to light the LED.

enter image description here

Figure 1. LED I versus V curves.

If you pick the colour of your LED and have a look at the graph you can see that at 0.2 V virtually no current will go through the LED.

Your example circuit is using the transistor to "shunt" the current away from the LED. This is a bit wasteful of energy as it draws more current when the LED is off than when it's on. The more usual arrangement is to put the LED in series with the resistor.


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