Now we'll re-visit another of our earlier experiments, this time we'll look again at the three-voltage LED board. This breadboard used 3 different voltages (3.3V, 5.0V and 9.0V) and the same resistor. As you recall, the higher the voltage, the brighter the LED, even with the same resistor.
I bet you know what's coming next!
Quick Quiz!
Quick Quiz!
Lets start with the 9V-powered LED. The forward voltage of the LED is 2.2V, what is the voltage across the resistor?
This is a KVL question. 9V = 2.2V + Vresistor so Vresistor = 6.8V
Now that you know the voltage, how much current (I) is going through the 1,000 ohm resistor (R)?
We determined the voltage (V) is 6.8V, we use Ohm's Law I = V/R = 6.8/1000 = 0.0068 Amps = 6.8 milliAmps.
Next, we will examine the 5V-powered LED. The forward voltage is 2.2V, what is the voltage across the resistor?
Another KVL! 5V = 2.2V + Vresistor. Vresistor = 2.8V
Now that you know the voltage, how much current (I) is going through the 1,000 ohm resistor (R)?
We determined the voltage (V) is 2.8V, we use Ohm's Law I = V/R = 2.8/1000 = 0.0028 Amps = 2.8 milliAmps.
Finally, we will examine the 3.3V-powered LED. The forward voltage is 2.2V, what is the voltage across the resistor?
Another KVL! 3.3V = 2.2V + Vresistor. Vresistor = 1.1V
Now that you know the voltage, how much current (I) is going through the 1,000 ohm resistor (R)?
We determined the voltage (V) is 1.1V, we use Ohm's Law I = V/R = 1.1/1000 = 0.0011 Amps = 1.1 milliAmps.
As you have seen with the calculations, increasing the voltage powering both resistor and LED increases the voltage across the resistor which causes more current to flow.
Page last edited July 18, 2025
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