What are Relays?

Relays are electromechanical devices that has an electromagnet to operate a pair of movable contacts from an open condition to a closed condition. The advantage of relays is that it takes a relatively small amount of power to operate the relay coil, but the relay itself can be used to control AC circuits and devices like motors, heaters, lamps which themselves can draw a lot more electrical power.

How do they work?

All relays contain a sensing unit, the electric coil, which is powered by AC or DC current. When the applied current or voltage exceeds a threshold value, the coil activates the armature, which operates either to close the open contacts or to open the closed contacts. When a power is supplied to the coil, it generates a magnetic force that actuates the switch mechanism. The magnetic force is, in effect, relaying the action from one circuit to another. The first circuit is called the control circuit; the second is called the load circuit.

What is it’s need?

Relay is an output device (actuator) which come in various shapes, sizes and designs, and is very useful in electronic circuits. They are useful when we need to control(ON and OFF) a high power device using a electronic circuits.

The design and types of relay switching circuits is huge, but many small electronic projects use transistors and MOSFETs as their main switching device as the transistor can provide fast DC switching (ON-OFF) control of the relay coil from a variety of input sources so here is a small collection of some of the more common ways of switching relays.

Step 1: Pin Configuration of Relay

Basically it depends on manufacturer but in most of the case it is somewhat like shown in figure

here NC--> normally connected (it is in contact with common when relay coil is de-energized )

NO--> normally open (it comes in contact with common when relay coil get energized)

WARNING***Kindly see the datasheet of manufacturer before using a relay for it's pin configuration

Step 2: Relay Switch Circuit Using NPN Transistor

A relay switch circuit can be designed using NPN transistor. In case of NPN transistor as shown in fig when V1 is 0v or turn off then base current ( ib ) will not flow due to which transistor remain in cut-off region and hence ic will not flow and relay coil remain de-energized but when base current ( ib ) is provided sufficiently the maximum amount of collector current starts flowing which energized the relay coil and hence NO and Com of relay will get attached.

So if a large enough positive current is now driven into the Base to saturate the NPN transistor, the current flowing from Base to Emitter (B to E) controls the larger relay coil current flowing through the transistor from the Collector to Emitter.

How to design the above Circuit?

How your design will be sustainable..

  1. It must be suitable for use with TTL circuit (means circuit behave well if base voltage V1 is in range 3.4-5V i.e. TTL logic high)
  2. The 2N2222A transistor can be replace by any other 2N2222 transistor and it should not affect the functioning of the circuit even if it’s ᵝ (current gain) is different.
  3. You can also consider working temperature range for your circuit.

So let’s go through different step to design a better Relay driving circuit.

Step 3: Let's Design the Relay Driver Circuit

Step-1: Finding the minimum current required for energizing the relay coil

So we are now going to find the threshold current to drive a relay properly. As my relay works on 5V DC so to find the coil current which is basically ic ( collector current ) we need to measure the resistance of Relay coil.

ic = 5V/ Rcoil

but if I say there will be one more voltage drop in the path of this current that is VCE (collector Emitter junction drop) which is 300mV-1V(datasheet). If we consider worst case we should give 1V extra to the Relay coil circuit so that sufficient amount of voltage is provided to the Relay coil. Than we have

ic = (6V-1V)/ Rcoil

In my case Rcoil is 71.9Ω so ic=69.541mA

Step-2: Selection of Transistor

  • · VCE(max) must be at least grater than 2 times the Relay operating DC voltage.
  • · IC(max) must be at least grater than 2 times the current required by Relay coil.
  • · Switching frequency or speed must be grater than your requirement.
  • · SOA( safe operating area) of that transistor must match your environmental conditions.
  • · It should be cheap but do not compromise above requirements.

For my project I found 2N2222 perfect (datasheet data)

| Symbol | Parameter | Min | Max |

| VCE | Collector- Emitter voltage | - | 30V |

| Ic | Collector current (DC) | - | 800mA |

| hFE | DC current gain (ᵝ) | 35 | 300 |

| fT | Transition Frequency | 250MHz | - |

| Tamb | Operating ambient temprature | -65 ˚C | +150 ˚C |

Step-3: Calculating the value of Base current (ib)

To calculate the value of base current we must know the relation between Base current and Collector current i.e ᵝ = iC/iB (current gain).

We must consider current gain ᵝ as minimum as possible so that our Relay driving circuit can perform well when ᵝ > ᵝmin. I am taking ᵝmin=35 for 2N2222 (from datasheet).

Hence ib = ic / ᵝmin

On putting values of collector current & current gain , ib = 1.986mA.

Step 4: Designing on Process

Step-4: Choosing a Sustainable value for RB

If we look at the below circuit and applying KVL to the loop (Note: We are designing a Relay driving circuit which can be use with TTL circuits so I am considering Base Voltage 3.4V because it is the minimum voltage which a TTL circuit can give when in digital High condition) we have

Here VB is base voltage , VBE is (base emitter voltage drop which is nearly 0.7 as it is PN junction) and iB is 1.986mA.

RB= (3.4-0.7)/1.986 = 1.359 KΩ

Step-5: Breaking heavy current to enter in the transistor

Since an inductor (the relay coil) cannot change it's current instantly, the flyback diode provides a path for the current when the coil is switched off. Otherwise, a voltage spike will occur causing arcing on switch contacts or possibly destroying switching transistors.

** see figure of 3 inductors

Now the complete circuit we have designed is (see the figure)

Step 5: Observation of Delays...

Observation of Delays

1. Output delay: It is the delay which is due to mechanical delay of relay switch. We can measure it using an Digital Storage Oscilloscope(DSO).

  • · Apply a 3.4V pulse with 50% duty-cycle at the Base (on place of V1)
  • · Connect your 1st channel of DSO across the V1.
  • · Connect it’s 2nd channel across the LED connected in output side.
  • · Observe and measure the delay between the both the rising and falling edge of both input and output .

2. Energizing delay of Coil: The coil when transistor get ON takes some time to get energized i.e. ic takes time to reach it’s maximum value. To observe this follow these steps

  • · Connect a 1Ω resistor in series with coil to collector of transistor (as shown in figure)
  • · Apply a 3.4V pulse with 50% duty-cycle at the Base (on place of V1)
  • · Connect the oscilloscope probe across that 1Ω resistor.
  • · Observe the voltage waveform

Because current across 1Ω resistor

i= V/R => i = V

Hence we can say it’s a current waveform

The slope in the waveform shows that their is some delay of ic to reach it's maximum value.

In my experiment the delay is 3.25mSec

Image1: Current waveform (ic) showing the energizing delay when flowing in the Relay coil

3. De-energizing delay of Coil:

Now we know that coil is getting discharge through diode. If you want to observe this delay also then follow these instructions

  • · Connect a 1Ω resistor in series with coil but in parallel with diode as shown in figure
  • · Apply a 3.4V pulse with 50% duty-cycle at the Base (on place of V1)
  • · Connect the oscilloscope probe across that 1Ω resistor.
  • · Observe the current waveform\
  • . The slope at the falling side shows that even after the input ( base voltage) changes from (logic 1 to logic 0) id (dischargeing current of coil flows for a short time.

Image2: Current waveform (ic) showing the energizing and de-energizing delay when flowing in the Relay coil


That's it now you can design your own relay driving circuit.

Step 6: Bibliography

Experiment Performed at---> Techno India NJR Institute Of Technology, Udaipur, India

Performed under guidance of Proff. Pradeep C.

Instuments used:

  • Agilent DSO-X 2014A 4-channel Digital Storage Oscilloscope
  • Tektronix AFG3022B function generator
  • fluke Multimeter

<p>The utility of relays has always fascinated me, I even remember seeing a circuit that, once energized, would throttle back the relay holding current to an absolute minimum in a power conserving mode- brilliant too I thought. ☺</p>
ha ha :-)<br>

About This Instructable



Bio: Student of Electronics & Communication Engineering @ Techno India NJR Institute of Technology,Udaipur, Rajasthan, India
More by Ritvikdave:How to Use Multisim? Build Your first IOT using Arduino without additional modules Design a Sustainable Relay Driving Circuit using BJT 
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