Developing A Safety Light Systems For A Crane

Real cranes will have complex electrical circuits that include microcontrollers, surge protectors, and relays circuits to control large high-powered lighting systems; however a scaled down version control circuit can be created using a 555 timer. 

The 555 timer is one of the most widely used Integrated Circuits (IC). It was designed in 1971 by Hans Camenzind under contract to Signetics. The versatility and functionality of the 555 timer quickly made it one of the most widely used ICs on the market with over 1 billion units being sold annually as of 2017. The 555 timer can be configured to operate in one of 4 different configurations:

1. Astable:In astable mode the 555 can operate as an electronic oscillator. Uses include LED and lamp flashers.
2. Monostable:In monostable mode the 555 functions as a “one-shot” pulse generator. Applications include timers
3. Bistable:In bistable mode the 555 functions as a flip-flop circuit. Uses include bounce-free latched switches.
4. Schmitt Trigger:As a Schmitt trigger the the 555 operates as an inverter gate which converts a noisy input into a clean digital output.

In “Mode 1  – Astable” the 555 timer can create an oscillating square wave pulse. This configuration is very useful when creating a wide variety of digital circuits, but more importantly it can provide a consistent output that can control the rate that safety lights on a crane will flash. Since the pulse rate of the circuit can be calculated in advance an electoral engineer to create a very specific interval for the flashing of the lights. That is how long the lights are ON (Time High) and OFF (Time low)

Simple 555 Timer Circuit

This simple 555 timer circuit can be used to produce a square wave in which the T1 (High Time) and the T2 (Low Time) can be calculated. This method of determining specific values ahead of time is used by electrical engineers to control the rate that an LED will blink. The final output obtained from Pin 3 (Output) is by the graph “555 Timer Circuit Duty Cycles”: 

555 Timer Circuit Duty Cycles

Time (T) is measured in seconds (s) and the Voltage measured in Volts (Vs). 

This graph shows what is happening in this simple 555 timer circuit. In very simple terms the capacitor (C1) is constantly charging and discharging. This causes
Pin 6 (the trigger) to be tripped. This then turns PIN 3 (the output) ON and then OFF again. This process of turning Pin 3 ON and OFF will happen as long as power is supplied to the circuit. This causes the LED light to blink.

The speed that the LED light will blink is controlled by the interaction between the resistors (R1 and R2) and the capacitor (C1) in the circuit. Therefore, the rate that the light flashes can be controlled by changing any or all the values for R1, R2, and C1. Since changing values will affect the frequency that the LED will blink, electrical engineers are able to create a circuit with a specific rate / frequency using the 555 timer circuit. That is how long T1 (Time High / ON), and T2 (Time Low / OFF). The following formulas can be used to calculate the variables:

ParameterFormulaUnit
T1  (Time High)0.693 x (R1 + R2) X C1Seconds (s)
T2  (Time Low )0.693 x R2 x C1Seconds (s)
T    (Timer Period)0.693 x (R1 + 2 x R2) x C1Seconds (s)
F    (Frequency)1.44 / (R1 + 2 x R2) x C1Hetz (Hz)
DC (Duty Cycle )(T1 / T) x 100%Percentage (%)

NOTE: It might be difficult to produce an exact rate based on available resistor values; however, by using a variable resistor for R2 precise values can be dialled in. This allows the engineer to calibrate or even adjust the frequency as needed; therefore, new parameters can be set as needed after the fact. That is why this simple 555 timer circuit has been designed with a #103 variable resistor for R2 instead of a standard fixed value resistor.

Tips and tricks for customizing your 555 timer circuit: 

  • T (Period Time) and F (Frequency) are inversely proportional;
  • Increasing C1 will decrease the F (Frequency)
  • Increasing R1 will increase T1 (High Time) but will not alter T2 (Low Time) 
  • Increasing R2 will increase both T1 (High Time) and T2 (Low Time) 
  • Increasing R2 will decrease DC (Duty Cycle 

Example Calculations:

In the provided simple 555 circuit diagram the value of resistors R1 is 1K, the value for R2 can go up to10K, and the value of capacitor C1 is 100uf

R1 = 1,000 Ohms R2 = 10 ~ 10,000 Ohms C1 = .0001 Farads

T1 (Time High) is the amount of time during which the pulse stays high (5V) in the output wave. This can be calculated as:

T1= 0.693 × (R1+R2) × C1
= 0.693 × (1,000 +10,000) × 0.0001
= 0.693 x (11,000) x 0.0001 = 0.693 x 1.1 
= 0.762 seconds 
= 762 milliseconds 

T2 (Time Low) is the amount of time during which the pulse stays low(0v) in the output wave. It can be calculated as:

T2= 0.693 × R2 × C1
= 0.693 × 10,000 × 0.0001
= 0.693 x 1
= 0.693 seconds
= 693 milliseconds 

T (Time Period) is the sum of Time low and Time high and can be calculates as:

T= 0.693 × (R1 + 2 x R2) x C1
= 0.693 x (1,000 + 2 x 10,000) x 0.0001
= 0.693 x (1,000 + 20,000) x 0.0001
= 0.693 x (21,000) x 0.0001
= 0.693 x 2.1
= 1.455 seconds
Or
= T1 + T2     
= .762 + .693     
= 1.455 seconds

F (frequency) is just the inverse of time. The frequency can be calculated as:

F= 1.44 / (R1+2×R2) × C1
= 1.44 / (1,000 + 2 x 10,000) x 0.0001
= 1.44 / 2.1
= 0.686 Hertz
Or
= 1/T     
= 1 / 1.445     
= .692 Hertz

NOTE: T1 andT2 have been rounded in previous steps which may result in a slight discrepancy in the answers.

DC (Duty Cycle) is given in terms of percentage, if T1 is equal to T2 then the pulse has 50%. We can calculate the duty cycle as. 

DC= (T1 / T) x 100%
= (0.762 / 1.455) x 100%
= 52.3%

Practice Problems

  1. Determine T1 (Time High), T (Time Period), and F (frequency) if the setting on the variable resistor at location R2 is adjusted to the following values: 
  • 500 Ω
  • 2,500 Ω
  • 5,000 Ω
  • 7,500 Ω 
  1. Determine the value for R2 that is necessary for T (Period Time) to be equal to 1 second. Show all of your calculations.

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