Motor burn out protection circuit

This is  a motor protection circuit. It will protect the AC motor burn out  because  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 of load at once. Until the voltage is restored to its original condition. So to get back into the circuit again. Ideally, blog diagram in Figure 1.

block diagram
Figure 1 the block diagram

The working of motor protection circuit
As Figure 2 is a schematic of this project. There are an op-amp-IC1 as input voltage comparator, it will compare voltage between the reference voltage of ZD1- 3.6 volts 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.

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

The pin 3 will connected to capacitor C3 to reduce the fluctuating voltage to smooth. It may cause IC1, the relay-RY1 works with 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 function
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 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 a 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 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 starting with input voltage AC 215 V, but relay will fast cut off the load when input fall to AC 198V by can adjust potentiometer -VR1.

Parts you will need

Resistors ¼W +-5%
R1, R3, R8_______1K
R6, R7___________4.7K
VR1- potentiometer__4.7K
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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Stephen Richard

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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