*This 20th article in the series of “Do It Yourself: Electronics”, explains the basic working of IC 555 and generating a square wave using it.*

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Playing with raw electronics (without any microcontroller), further boosted the confidence of Pugs to dive into non-microcontroller electronics. This time he decided to explore the ever popular IC 555, loosely also known as the timer IC.

## 555 Functionality

555 is basically an 8-pin IC, with pin 1 for GND, pin 8 for Vcc, and pin 3 for Vo – the output voltage, which goes either high (Vcc) or low (GND), based on the other pins.

Vo goes high if the trigger pin 2 senses voltage less than 1/3 of Vcc. Vo goes low if the threshold pin 6 senses voltage greater than 2/3 of Vcc.

Pin 5 can be used as a control voltage always fixed to 2/3 of Vcc. Putting reset pin 4 low any time makes Vo go immediately low. So, if not in use it is recommended to be tied to Vcc.

Discharge pin 7 becomes GND when pin 6 senses voltage greater than 2/3 of Vcc and becomes tristate (open) when pin 2 senses voltage less than 1/3 of Vcc. In other words, discharge pin 7 becomes GND when Vo goes low and becomes open when Vo goes high.

## Generating a Square Wave

Given this background, one of the common uses of the 555 IC is to generate a square wave of any particular frequency and duty cycle (on pin 3), by varying some analog voltage between GND and Vcc (on pins 2 and 6), more precisely between 1/3 Vcc and 2/3 Vcc, both inclusive. And this analog voltage is typically achieved by charging / discharging a capacitor through one or more resistors. Thus, the time constants given by τ = RC, R being the resistance, and C being the capacitance in the corresponding charging & discharging paths, controlling the corresponding on & off cycle of the square wave.

Let’s consider the following circuit with R1 as a variable resistance (pot) between 0-10KΩ, and R2 as fixed resistance of 4.7KΩ, and C as a 1μF capacitor.

Square Wave using 555 (Schematic)

In the on cycle (when Vo (pin 3) is high), pin 7 would be open. Pins 2 & 6 can be assumed tristate. Hence, then C is getting charged towards Vcc through R = R1 + R2.

In the off cycle (when Vo (pin 3) is low), pin 7 would be GND. Pins 2 & 6 can be assumed tristate. Hence, then C is getting discharged towards GND (pin 7) through R = R2.

Moreover, note that in the on cycle as soon as capacitor voltage reaches 2/3 Vcc, Vo (pin 3) becomes low, and pin 7 becomes GND, i.e. off cycle starts.

And, in the off cycle as soon as capacitor voltage drops to 1/3 Vcc, Vo (pin 3) becomes high, and pin 7 becomes tristate, i.e. on cycle starts.

And the above sequence keeps on repeating, thus giving a square wave on Vo (pin 3), with on time t_on controlled by charging through R1 + R2 and off time t_off controlled by discharging through R2.

From RC circuit analysis, we have that voltage *Vc* across a capacitor C, getting charged through resistance R, at time t is given by:

,

where Vs is supply voltage (Vcc in our case), Vi is the initial voltage on the capacitor.

So, *t_on* could be obtained from the fact that it starts with inital voltage Vi = 1/3 Vcc, and ends when *Vc* = 2/3 Vcc, being charged by Vs = Vcc through R = R1 + R2. That is,

,

which on simplifying gives:

… (1)

Similarly, from RC circuit analysis, we have that voltage *Vc* across a capacitor C, getting discharged through resistance R, at time t is given by:

,

where Vi is the initial voltage on the capacitor.

So, *t_off* could be obtained from the fact that it starts with initial voltage Vi = 2/3 Vcc, and ends when Vc = 1/3 Vcc, being discharged through R = R2. That is,

,

which on simplifying gives:

… (2)

## Live Demo

Pugs doesn’t get a punch unless he sees the theory working in practice. That’s where, he sets up the above circuitry on a breadboard as shown in the figure below:

WARNING: Do NOT put the pot to a value of zero, as that will short Vcc & GND, and may blow off the circuit. A safety workaround could be to put a fixed 1K resistor in series with the pot.

Square Wave using 555 (Breadboard)

The audio jack is being used for observing the waveforms on the home-made PC oscilloscope, as created in his previous PC Oscilloscope article.

Below are the three waveforms Pugs observed for the values of R1 being adjusted to 1.28KΩ, 4.2KΩ, 8.6KΩ:

R1 = 1.28K, R2 = 4.3K

R1 = 4.15K, R2 = 4.3K

R1 = 8.60K, R2 = 4.3K

From the waveforms, Pugs approximately have the following *t_on* & *t_off*:

R1 = 1.28KΩ => *t_on* = 3.8ms, *t_off* = 3.0ms

R1 = 4.15KΩ => *t_on* = 6.0ms, *t_off* = 3.0ms

R1 = 8.60KΩ => *t_on* = 9.0ms, *t_off* = 3.0ms

Now, as per equations (1) & (2), for C = 1μF, R2 = 4.7K, and the above three R1 values, we should have got the following:

R1 = 1.28KΩ => *t_on* = 4.1ms, *t_off* = 3.3ms

R1 = 4.15KΩ => *t_on* = 6.1ms, *t_off* = 3.3ms

R1 = 8.60KΩ => *t_on* = 9.2ms, *t_off* = 3.3ms

Pretty close, but the *t_off* not really satisfactory. That triggered Pugs to take out his multimeter and check the resistance of the fixed resistor R2, he used. Ow! that actually measured 4.3K. Recomputing using R2 = 4.3K, gave values amazingly close to the observed values.

## Summary

Thus by appropriately choosing the R1, R2, and C values one should be able to get a square wave of a desired frequency given by *1 / (t_on + t_off)* and duty cycle given by *t_on / (t_on + t_off)*. Obviously, the frequency would have a practical upper limit dictated by the 555 IC, though it is typically in MHz. What about duty cycle? Note that as per relations (1) & (2), *t_on* will be always greater than *t_off*. Thus, duty cycle would be always greater than 0.5.

So, what if we need duty cycle less than 0.5, or at least equal to 0.5, where *t_on* = *t_off*. This is what Pugs is working out on. Watch out for the next article.

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