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I'm following along with Ben Eater's digital electronics playlist on YouTube. In this video, he demonstrates two circuits each with the same voltage and resistance but with one circuit the LED is emitting more light. This is explained because each circuit provides a different amount of current. I understand that less current means the LED is not as bright, however as I = V / R, I don't understand how the current can be different if the voltage and resistance is the same.

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    \$\begingroup\$ Questions on stack exchange sites must stand on their own, they may not rely on an external link (especially not to a video!) to provide the most essential information. To ask this question here, you will need to edit it and make it so that it is clear exactly what you are asking about without clicking any of your links. Eg, include schematic diagrams of the two circuits being compared, and the alleged currents through them. Note that any circuit with an LED is one with more than a simple voltage and resistance, the current through an LED has a non-linear relationship to the voltage. \$\endgroup\$ Sep 30 '20 at 5:23
  • \$\begingroup\$ two circuits each with the same voltage ... that is not quite true ... the dim LED power supply voltage is only 1.8 V ... you can see it at 4:00 \$\endgroup\$
    – jsotola
    Sep 30 '20 at 7:11
  • \$\begingroup\$ @jsotola The video before shows the output voltage as 5V. Shouldn't both LEDs have the same voltage drop in this case? \$\endgroup\$ Sep 30 '20 at 7:35
  • \$\begingroup\$ that means nothing ... the voltage, at the power supply terminals when the LED is on, is the actual voltage being supplied to the circuit ... the two LEDs should have the same voltage drop if the two power supplies were the same, but they are not the same \$\endgroup\$
    – jsotola
    Sep 30 '20 at 7:40
  • \$\begingroup\$ I = V / R is an equation that only applies to the resistor that should be connected in series with the LED. The LED itself doesn't work by Ohm's Law. \$\endgroup\$
    – mguima
    Sep 30 '20 at 14:28
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In this video Ben Eater is explaining the need to apply a signal to a load through a driver (a transistor in his example.)

The upper signal is provided by a source that has a high internal output resistor. The lower signal is provided by a power supply with a low output resistor (the output voltage is regulated so we may assume an adaptable output resistor.)

So you must consider the internal resistor like in the following:

schematic

simulate this circuit – Schematic created using CircuitLab

R1 is the internal resistor of the source applied to the upper circuit. R2 is the internal resistor of the power supply applied to the lower circuit (assumed null.)

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  • \$\begingroup\$ The video before shows the voltage as 5V. As a result, I don't think your diagram is accurate. \$\endgroup\$ Sep 30 '20 at 7:38
  • \$\begingroup\$ 5V when the circuit is open. I am showing the case when the circuit is closed. (when the circuit is open the voltage drop on R1 is 0 and hence the voltage will be 5V) \$\endgroup\$ Sep 30 '20 at 7:40
  • \$\begingroup\$ Oh yes, I see that now! Still getting my head around parallel circuits on breadboards. Thanks \$\endgroup\$ Sep 30 '20 at 7:47

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