Induction Motor Protection Circuit and working

This is  a motor protection circuit. It will protect the AC motor burn out. Because of under voltage input. The principle is simple, it will check for voltage at all times. Normal voltage is 220V AC.

If the voltage falls below a preset. Relay will not work and the LED is a warning. Can be set to cut no-off in the voltage during the 168-227V.

As AC main in normal level, the relay will be connect to apply AC line to motor. And AC-line fall down lower than 10% of normal voltage the relay will cut off the load at once.

Until the voltage is restored to its original condition. So to get back into the circuit again. Ideally, block diagram in Figure 1.

block diagram of motor protection circuit

Figure 1 the block diagram

The working of the motor protection circuit

Figure 2 is a schematic of this project. There is an op-amp-IC1 as an input voltage comparator.

It will compare voltage between the reference voltage of ZD1- 3.6 volts the Zener diode , and input voltage from pin 3 of IC1 this voltage is divided by resistor-R4 and VR1.

We will set this voltage slightly lower than the reference voltage.

So this will apply a relay pulls in and connects AC electrical current to the motor or AC load.

Schematic of Motor burn out and under voltage protection

Figure 2 The schematic of Motor burn out and under-voltage protection

Pin 3 will be connected to capacitor C3 to reduce the fluctuating voltage to smooth. It may cause IC1, the relay-RY1 works with the error. This will not be good for this circuit.

We will also connect the resistor-R5 across pin 6 and pin 3 of IC1, to set circuit in Schmitt trigger form.

This is a special feature, it will cut off the load at once, when the AC main is under voltage into setting value.

If the main have ripple voltage,this will not works because the motor may be damaged.

Restarting with increasing AC voltage about 215 volts (as setting). So relay pulls in as Graph of time function in Figure 3.

Graph of time on motor protection

Figure 3

Reasons to Use IC1-CA3130 BecauseWhen conditions do not work, to make the output voltage is near 0 volts really. As a result transistor Q1 can not work absolutely.

But the common comparator ICs in not work state. The output voltage is 0.6 volts which may make the transistor can work.

The disadvantage of this IC is not able to tolerate the higher voltage of 16 volts. Thus in this circuit, we use voltage from the 15.8 volts power supply, then flow through the divider circuit of R1 and R2 cause have an 8.7 volts supply of IC1 only.

The transistor Q2 will drive LED to alert when under-voltage until electric devices cannot work.

But in normal Q1 will ON cause the voltage across Q2 is low so most current will flow to ground. Therefore, it is virtually no voltage to bias the Q2 As a result LED goes out.


Before make the PCB as Figure 4 as copper PCB layout. Then soldering all parts on PCB as Figure 5, to begin with soldering low devices first.

For example resistors, diodes then put higher parts down until connecting a transformer and more parts. Do not hurry, especially for pins to make sure the correct position. Then wiring the different positions.

Actual size of single sided copper PCB layout

Figure 4 Actual size of single sided copper PCB layout.

the components layout

Figure 5 the components layout.

This motor protection circuit will be starting with input voltage AC 215 V. But the relay will fast cut off the load when the input falls to AC 198V by can adjust potentiometer -VR1.

Parts you will need

Resistors 0.25W +-5%
R1, R3, R8: 1K
R2: 1.2K
R4: 10K
R5: 47K
R6, R7: 4.7K
VR1: 4.7K Potentiometer

C1: 470uF 25V, Electrolytic
C2: 10uF 16V, Ceramic
C3: 100uF 16V, Electrolytic
C4: 150pF 50V, Ceramic

D1-D5___1N4001__1A 50V Diode
ZD1_____3.6V/0.5W Zener
Q1, Q2_____BC548___0.4A 40V NPN transistors
RY1___Relay _____ 12V__1C
Box, PCB, Wires with plug 220V, etc.

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This Post Has One Comment

  1. Am doing this project in my final year this time…..but can you please help me why this transistors are connected two…..and how the circuit is going to be installed in the motor so as to protect it?…please and please!!

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