Vol. DIY Electronics Projects

Chapter 8 555 Timer Circuit Projects

555 Lab - Blue LED Flasher With Voltage Doubler

In this hands-on electronics experiment, build a blue LED flasher using a 555 timer and learn the basic theory of a voltage doubler using a transistor switch and a capacitor.

Project Overview

The circuit of Figure 1 used in this project adds a voltage doubler feature to the circuit and concepts from these two experiments:

Schematic diagram of the 555 timer LED flasher with high voltage LED drive.

Figure 1. Schematic diagram of the 555 timer LED flasher with high voltage LED drive.

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.

Parts and Materials

Two AAA batteries
Battery clip
U1 - complementary metal-oxide semiconductor (CMOS) TLC555 timer IC (or equivalent).
- Note: This is a static-sensitive part. If you do not use protection as described in the electrostatic discharge (ESD) chapter, you run the risk of destroying it.
Q1 - 2N3906 PNP transistor or equivalent
Q2 - 2N2222 NPN transistor or equivalent
CR1 - 1N914 diode or equivalent
D1 - Blue LED
R1 - 1.5 MΩ 1/4W 5% resistor
R2 - 47 kΩ 1/4W 5% resistor
R3 - 2.2 kΩ 1/4W 5% resistor
R4 - 620 Ω 1/4W 5% resistor
R5 - 82 Ω 1/4W 5% resistor
C1 - 1 µF tantalum capacitor
C2 - 100 µF electrolytic capacitor
C3 - 470 µF electrolytic capacitor

Learning Objectives

Learn a practical application for an RC time constant
Learn one of several 555 timer astable multivibrator configurations
Working knowledge of duty cycle
How to handle ESD-sensitive parts
How to use transistors to improve current gain
How to use a capacitor to double voltage with a switch

Instructions

Step 1: Build the 555 oscillator circuit shown in Figures 1 and 2.

Breadboard implementation of the 555 timer LED flasher with high voltage LED drive.

Figure 2. Breadboard implementation of the 555 timer LED flasher with high voltage LED drive.

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.

Voltage Doubler Theory of Operation

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.

When transistor switch Q2 is open, capacitor C3 charges.

Figure 3. When transistor switch Q2 is open, capacitor C3 charges.

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.

When transistor switch Q2 is closed, the voltage across the diode doubles, turning it on

Figure 4. When transistor switch Q2 is closed, the voltage across the diode doubles, turning it 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.