# Fire Engine Flashing LED Project

See Typical Pinout Diagrams below:

For the purposes of this website, the detailed operation of the chip will not be discussed, since it is actually part of an Electronics Engineering curriculum and thus beyond the scope of this paragraph. However, enough detail will be provided to package a set of Flashing LEDs for various model railroad projects.

**Fire Engine Flashing LEDs**

The following photograph depicts a Fire Engine (HO Scale) with two 2 mm rectangular clear white lens LEDs that illuminate to a red color in place of the original non-functioning placed over the cab area. It is also possible to replace two other original non-functioning LEDs at the rear of the truck. We can use the 556 Chip to implement the four LEDs to flash at a periodic rate.

See the following schematic diagram for information reflecting the A side of the chip:

As shown in the above circuit diagram, you must use a power supply between + 4.5 and +15 vdc. The 556 chip Section A develops a square wave at output pin 5 as shown. The square wave out will swing between + Vs and zero volts dc at some periodic rate. Since it is a square wave, the amount of time it is at + Vs volts is equal to the amount of time it is at zero volts. This is achieved by the proper selection of resistors and capacitors as described below. Note that if the output is equal to +Vs then LED CR1A will illuminate, and while the output square wave is at zero volts then LED CR2A will illuminate. Hence, the LEDs will flash on and off as the square wave transitions from +Vs to zero at a rate that is determined below.

Given that we use a +12 vdc Power Supply, and assuming that 2 vdc will appear across each of the LEDs, then 10 vdc will appear each of the 1.5K Resistors. Hence, using Ohms Law the current passed by each LED will be calculated by the following formula:

Current = Voltage/Resistance or Current = 10 vdc/1500 ohms = 6.6 milliamps

**LED Timing Determination**

There are three components (resistors R1A and R2A and capacitor C1A) that determine the rate at which the LEDs flash on and off which is determined by the period of the square wave.

More specifically the time that the output is positive is equal to t1 = .7(R1A + R2A)C1A

The time that the output is positive is at zero volts is equal to t2 = .7(R2A)C1A

Therefore the total period of the square wave is t1 + t2 = T

Now lets assume that R2A is much, much larger than R1A, then we get,

T = t1 + t2 = .7 (R1A + R2A)C1A + .7(R2A)C1A Generally, R2A is 10 times larger than R1A.

Now let R1A = 10k; R2A = 100k; C1A = 1 uf

We now get T = (0.7)(10k + 100k)(1xE-6) + (0.7)(100k)(1xE-6) = (0.7)(110k)(1xE-6)+ (0.7)(100k)(1xE-6) = 0.147 sec.

Translating T into a Frequency we get Frequency = 1/T = 1/0.147 = 6.8 Hz

Therefore the LEDs will flash at 6.8 Hz or 6.8 cycles/sec

See the following schematic diagram for information reflecting the B side of the chip:

As shown in the above circuit diagram, you must use a power supply between + 4.5 and +15 vdc. The 556 chip Section B develops a square wave at output pin 9 as shown. The square wave out will swing between + Vs and zero volts dc at some periodic rate. Since it is a square wave, the amount of time it is at + Vs volts is equal to the amount of time it is at zero volts. This is achieved by the proper selection of resistors and capacitors as described below. Note that if the output is equal to +Vs then LED CR1B will illuminate, and while the output square wave is at zero volts then LED CR2B will illuminate. Hence, the LEDs will flash on and off as the square wave transitions from +Vs to zero at a rate that is determined below.

Given that we use a +12 vdc Power Supply, and assuming that 2 vdc will appear across each of the LEDs, then 10 vdc will appear each of the 1.5K Resistors. Hence, using Ohms Law the current passed by each LED will be calculated by the following formula:

Current = Voltage/Resistance or Current = 10 vdc/1500 ohms = 6.6 milliamps

**LED Timing Determination**

There are three components (resistors R1B and R2B and capacitor C1B) that determine the rate at which the LEDs flash on and off which is determined by the period of the square wave.

More specifically the time that the output is positive is equal to t1 = .7(R1B + R2A)C1B

The time that the output is positive is at zero volts is equal to t2 = .7(R2B)C1B

Therefore the total period of the square wave is t1 + t2 = T

Now lets assume that R2B is much, much larger than R1B, then we get,

T = t1 + t2 = .7 (R1B + R2B)C1B + .7(R2B)C1B Generally, R2B is 10 times larger than R1B.

Now let R1B = 10k; R2B = 100k; C1B = 1 uf

We now get T = (0.7)(10k + 100k)(1xE-6) + (0.7)(100k)(1xE-6) = (0.7)(110k)(1xE-6)+ (0.7)(100k)(1xE-6) = 0.147 sec.

Translating T into a Frequency we get Frequency = 1/T = 1/0.147 = 6.8 Hz

Therefore the LEDs will flash at 6.8 Hz or 6.8 cycles/sec

In general, the following vendors are recommended for any of the above electronic components. :

Radio Shack – Local Store or on-line at www.radioshack.com

Futurelec at **www.futurlec.com**

**Additional Improvement**

There must be a 2200 uf Capacitor between the power supply input between pins 14 and 7 to reduce ripple. It should be physically placed as close as possible to the 556 chip.

A datasheet for the 556 Timer is available at the following link: http://www.national.com/ds/LM/LM556.pdf

If you wish to acquire a better understanding of Electronics Theory, I suggest you go to the following link: Electronics