The circuit of Figure 1 used in this project adds a voltage doubler feature to the circuit and concepts from these two experiments:
Blue and white light-emitting diodes (LEDs) have a higher Vf (forward dropping voltage) than other colors, typically around 3.6 V. Therefore, 3 V batteries can’t drive them without help. Thus, the need for extra circuitry will be added to this experiment.
The capacitor, C3, is used to double the voltage of this pulse across the blue LED, but it can only do this for a short time. Measuring the current through the LED is impractical with this circuit because of the short pulse duration.
Step 1: Build the 555 oscillator circuit shown in Figures 1 and 2.
The LED should flash approximately every 1 second. As in the previous circuits, the LED is given a 0.03-second (30 ms) pulse.
Step 2 (optional): This particular design can also be used with a single 1.5 V battery. If you want to try a 1.5 V battery, change R5 to 10 Ω and use a red LED with a better CR1 diode. CR1 is not the best choice for this component, it was selected because it is a common part and it works. Almost any diode will work in this application.
Schottky and germanium diodes drop much less voltage than silicon diodes. A silicon diode drops 0.6-0.7 V, while a Schottky diode drops 0.1-0.2 V, and a germanium diode drops 0.2 V-0.3 V. If Schottky or germanium diodes are used, the reduced voltage drop would translate into a higher voltage across the LED and increased intensity, as the circuit's efficiency is increased.
In this circuit, Q2 is used as a switch. When Q2 is off, C3 is charged to the battery voltage minus the diode voltage drop of CR1, as shown in Figure 3.
Since the forward voltage (Vf) of a blue LED is 3.4 to 3.6 V, it is not conducting and is effectively out of the circuit.
Figure 4 illustrates what happens when Q2 turns on.
The capacitor C3's positive (+) side is grounded, which moves the negative (-) side to -2.4 V. The diode CR1 is now reverse-biased and is effectively out of the circuit. The -2.4 V is discharged through R5 and D1 to the +3.0 V of the batteries. The total of 5.4 V provides enough voltage to light the blue LED. The 5.4 V value is not quite double the supply voltage because of the loss of diode CR1. Long before C3 is discharged, the circuit switches back, and C3 starts charging again.
You may notice a dim blue glow in the blue LED even when it is off. This demonstrates the difference between theory and practice, 3 V is enough to cause some leakage through the blue LED, even though it is not fully conducting. If you were to measure this current, it would be very small.
Learn more about the fundamentals behind this project in the resources below.
Textbook:
Calculator:
Projects:
Worksheet:
In Partnership with Würth Elektronik eiSos GmbH & Co. KG
by Dale Wilson
by Harwin
by onsemi