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Old 03-01-2009, 11:01 PM
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Default Submission, CMOS 555 Long Duration Blue LED Flasher

Article: Proof Read
Illustrations: Complete
Experiment: Schematic Verified
Duration Test Started 8/1/09

*******************

CMOS 555 LONG DURATION BLUE LED FLASHER



PARTS AND MATERIALS
  • Two AAA Batteries
  • Battery Clip (Radio Shack catalog # 270-398B)
  • One CMOS TLC555 timer IC (Radio Shack catalog # 276-1718 or equivalent)
  • Q1 – 2N3906 PNP Transistor (Radio Shack catalog #276-1604 (15 pack) or equivalent)
  • Q2 – 2N2222 NPN Transistor (Radio Shack catalog #276-1617 (15 pack) or equivalent)
  • CR1 – 1N914 Diode (Radio Shack catalog #276-1122 (10 pack) or equivalent, see Instructions)
  • D1 – Blue light-emitting diode (Radio Shack catalog # 276-311 or equivalent)
  • C1 – 1 µF Tantalum Capacitor (Radio Shack catalog 272-1025 or equivalent)
  • C2 – 100 µF Electrolytic Capacitor (Radio Shack catalog 272-1028 or equivalent)
  • C3 – 470 µF Electrolytic Capacitor (Radio Shack catalog 272-1030 or equivalent)
  • R1 – 1.5 MΩ ¼W 5% Resistor
  • R2 – 47 KΩ ¼W 5% Resistor
  • R3 – 2.2 KΩ ¼W 5% Resistor
  • R4 – 620 Ω ¼W 5% Resistor
  • R5 – 82 Ω ¼W 5% Resistor

CROSS-REFERENCES

Lessons In Electric Circuits, Volume 1, chapter 16: Voltage and current calculations
Lessons In Electric Circuits, Volume 1, chapter 16: Solving for unknown time
Lessons In Electric Circuits, Volume 3, chapter 4 : Bipolar Junction Transistors
Lessons In Electric Circuits, Volume 3, chapter 9 : ElectroStatic Discharge
Lessons In Electric Circuits, Volume 4, chapter 10: Multivibrators



LEARNING OBJECTIVES
  • Learn a practical application for a 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

SCHEMATIC DIAGRAM





ILLUSTRATION





INSTRUCTIONS

NOTE! This project uses a static sensitive part, the CMOS 555. If you do not use protection as described in Volume 3, Chapter 9, ElectroStatic Discharge, you run the risk of destroying it.

This circuit builds on the previous two experiments, using their features and adding to them. Blue and white LEDs have a higher Vf (forward dropping voltage) than most, around 3.6V. 3V batteries can’t drive them without help, so extra circuitry is required.

As in the previous circuits, the LED is given a 0.03 second (30ms) pulse. C3 is used to double the voltage of this pulse, but it can only do this for a short time. Measuring the current though the LED is impractical with this circuit because of this short duration, but blue LEDs are generally more predictable because they were invented later.

This particular design can also be used with a single 1½V battery. The base concept was created with a now obsolete IC, the LM3909, which used a red LED, the IC, and a capacitor. As with this circuit, it could flash a red LED for over a year with a single D cell. When newer red LEDs increased their Vf from 1.5V to 2.5V this old chip was no longer practical, and is still missed by many hobbyists. If you want to try a 1½V battery change R5 to 10Ω and use a red LED with a better CR1 (see next paragraph).

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, a silicon diode drops 0.6-0.7V, while a Schottky diode drops 0.1-0.2V, and a germanium diode drops 0.2V-0.3V. If these components are used the reduced voltage drop would translate into brighter LED intensity, as the circuits efficiency is increased.



THEORY OF OPERATION

Q2 is a switch, which this circuit uses. When Q2 is off C3 is charged to the battery voltage, minus the diode drop, as shown in Figure 1. Since the blue LED Vf is 3.4V to 3.6V it is effectively out of the circuit.



Figure 2 shows what happens when Q2 turns on. The capacitor C3 + side is grounded, which moves the – side to -2.4V. The diode CR1 is now back biased, and is out of the circuit. The -2.4V is discharged through R5 and D1 to the +3.0V of the batteries. The 5.4V provides lots of extra voltage to light the blue LED. Long before C3 is discharged the circuit switches back and C3 starts charging again.



In the LM3909 CR1 was a resistor. The diode was used to minimize current, by allowing R4 to be its maximum value.

You may notice a dim blue glow in the blue LED when it is off. This demonstrates the difference between theory and practice, 3V is enough to cause some leakage through the blue LED, even though it is not conducting. If you were to measure this current it would be very small.
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Last edited by Bill_Marsden; 08-09-2009 at 06:34 PM.
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Old 03-01-2009, 11:02 PM
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Old 03-01-2009, 11:02 PM
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Attached Images
File Type: gif 1.GIF (2.3 KB, 112 views)
File Type: gif 2.GIF (2.2 KB, 113 views)
File Type: gif Schematic.GIF (7.5 KB, 94 views)
File Type: png Illustration.PNG (23.2 KB, 90 views)
Attached Files
File Type: doc LDBLED Flasher.doc (52.0 KB, 6 views)
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Last edited by Bill_Marsden; 08-09-2009 at 06:35 PM.
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Old 03-01-2009, 11:03 PM
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Old 08-02-2009, 11:35 PM
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OK, this experiment is basically completed. The blue LED is a high brightness unit that lights up the entire living room. Now to see how long it lasts, the duration test is now running.
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Old 09-25-2009, 01:58 PM
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This one is also going strong. It is also noticably dimmer, but still blinding. I suspect the basic concept is plain more efficient than the other designs.
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Old 11-01-2009, 04:59 AM
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Well, that was easy, around 7:00PM tonight, 10/31/09, it dimmed dramatically. No gradual fade out here,though it is still flashing dimly. I like it when an experiment is so neat and clean like this. I'm debating whether to modify it with a simple constant current source and use a red LED, for maximimum life. Whatcha think?
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Old 11-07-2009, 10:58 AM
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Hello Bill,

The current source will also take some current for itself.
This will shorten battery life a bit.

You could try a fet as current source, this will only drop some voltage.



The current will be depandend on the type of fet used.

Greetings,
Bertus
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Old 11-07-2009, 12:40 PM
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I'm aware of the fet variaty, but you can't program them (I think). I've got another design that will work, not a totally flat current curve, but it will work. When I get a chance I'll put it up.
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Old 11-07-2009, 01:59 PM
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Hello,

The current source with a FET can be regulated by adding a resistor.
See the attached application note for more info.

Greetings,
Bertus
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