Build the circuit below. Explain the function of the circuit. The diode requires 5mA to emit visible light. How much time it takes for a fully-
discharged capacitor to reach the required voltage, so the diode can emit visible light?
+5V
+5V
20k
2.9V
100
2.9V
LM324
SW
100µF
220
TIP: the voltage follower op-amp follows the voltage on the capacitor. When the voltage on the capacitor exceeds
For the diode to emit visible light, a current of 5mA is required. 5mA * 0.22K = 1.1 V required on the 220R resistor.
The diode takes 1.7V. This means that the LM324 must deliver 1.1V + 1.7V = 2.8V. To conclude, when the capacitor reaches 2.8V, the diode sees a
5mA current (required for the emitted light to be considered visible)
For a fully-discharged capacitor, it takes 1640ms = 1.64s to charge from OV to 2.8V