Refrigerators tend to draw appreciable amounts of current each time their compressor switch ON, and this could happen many times per day. A soft start circuit to the compressor motor could probably tackle this issue and help save electricity. The idea was requested by Mr. Naeem Khan.

The capacitor in a capacitor start motor has nothing to do with the speed of the motor. The capacitor is there just to energize the field coil of the motor for helping the main winding to start the rotation, after which it's cut-of from the system.

In any case, the soft start circuit presented here is irrelevant to the type of AC motor used, it should hopefully work for all types of motor.

Referring to the figure we see an arrangement where the refrigerator is wired in series with a rectifier diode which has an SCR connected in parallel.

The operation is rather simple.

As soon as the internal relay of the refrigerator clicks ON, diode D1 provides a half wave AC to the refrigerator forcing a slow soft start to the motor, the SCR is unable to conduct immediately due to the the presence of the capacitor at its gate.

Therefore at the start, the refrigerator is able to get only a half wave AC through the rectifier diode, until the capacitor across the SCR gate/cathode charges up and fires the SCR.

During this period the half wave AC allows only around 50% of initial voltage to the refrigerator, providing a soft start to the motor, until within seconds the SCRs fires and restores full available power to the motor.

Once the SCR is fired it takes on the other half of the AC so that the refrigerator motor is able to gain its full rated torque.

R1 = 47K 1watt

D1 = 6 amp diode

D2 = 1N4007

Z1 = 50V 1 watt zener

C1 = 10uF/400V

Since initially the series diode converts the AC input into a half wave DC, it's important to know the average DC applied at the particular instant. It may be calculated using the formula:

Vdc av = Vp/Ï€

where Ï€ = 3.1416, and Vp = peak half wave value

The Ï€ value may be solved and the above formula may be further expressed as:

Vdc av = 0.318 Vp

The peak voltage may be calculated by using the following formula:

peak volts = RMS volts x 1.414

therefore we get:

Vp = Vrms x 1.414

For a 220V RMS, the above formula could be solved as:

Vp = 220 x 1.414 = 311.08V

For accuracy we can also include the 0.7V drop produced by the diode in our calculation:

Vdc av = (VP - 0.7)/Ï€

Solving the above equation with Vp = 311.08, we get:

Vdc av = (311.08 - 0.7)/Ï€ = 98.84V

If the refrigerator motor coil resistance is known the above DC average voltage could be used for calculating the initial soft-start power consumed by the motor, through the following formula:

P = I2R, where P stands for power,

I = current (amps) and R = resistance of the motor coil

I (amps) could be found by applying Ohms law:

IDC = VDC /R,

where R = resistance of the motor coil, and VDC = 98.84V obtained from the previous calculations. where Ï€ = 3.1416.

**Technical Specifications**

*I need your help regarding to control the starting torque(soft start) of the Refrigerator Compressor for energy saving purpose. All these compressor are capacitor start type. If you have any other idea to control these capacitor start compressor RPM then let me know.**Awaiting your reply soon.***The Design**

The capacitor in a capacitor start motor has nothing to do with the speed of the motor. The capacitor is there just to energize the field coil of the motor for helping the main winding to start the rotation, after which it's cut-of from the system.

In any case, the soft start circuit presented here is irrelevant to the type of AC motor used, it should hopefully work for all types of motor.

Referring to the figure we see an arrangement where the refrigerator is wired in series with a rectifier diode which has an SCR connected in parallel.

The operation is rather simple.

### How the Circuit Operates

As soon as the internal relay of the refrigerator clicks ON, diode D1 provides a half wave AC to the refrigerator forcing a slow soft start to the motor, the SCR is unable to conduct immediately due to the the presence of the capacitor at its gate.

Therefore at the start, the refrigerator is able to get only a half wave AC through the rectifier diode, until the capacitor across the SCR gate/cathode charges up and fires the SCR.

During this period the half wave AC allows only around 50% of initial voltage to the refrigerator, providing a soft start to the motor, until within seconds the SCRs fires and restores full available power to the motor.

Once the SCR is fired it takes on the other half of the AC so that the refrigerator motor is able to gain its full rated torque.

### Circuit Diagram

#### Parts List

R1 = 47K 1watt

D1 = 6 amp diode

D2 = 1N4007

Z1 = 50V 1 watt zener

C1 = 10uF/400V

**Switch ON Starting Power Calculation **

Since initially the series diode converts the AC input into a half wave DC, it's important to know the average DC applied at the particular instant. It may be calculated using the formula:

Vdc av = Vp/Ï€

where Ï€ = 3.1416, and Vp = peak half wave value

The Ï€ value may be solved and the above formula may be further expressed as:

Vdc av = 0.318 Vp

The peak voltage may be calculated by using the following formula:

peak volts = RMS volts x 1.414

therefore we get:

Vp = Vrms x 1.414

For a 220V RMS, the above formula could be solved as:

Vp = 220 x 1.414 = 311.08V

For accuracy we can also include the 0.7V drop produced by the diode in our calculation:

Vdc av = (VP - 0.7)/Ï€

Solving the above equation with Vp = 311.08, we get:

Vdc av = (311.08 - 0.7)/Ï€ = 98.84V

If the refrigerator motor coil resistance is known the above DC average voltage could be used for calculating the initial soft-start power consumed by the motor, through the following formula:

P = I2R, where P stands for power,

I = current (amps) and R = resistance of the motor coil

I (amps) could be found by applying Ohms law:

IDC = VDC /R,

where R = resistance of the motor coil, and VDC = 98.84V obtained from the previous calculations. where Ï€ = 3.1416.

**Warning: The circuit has not been tested or verified practically, and the effects are unknown. Initially try the circuit by using a 200watt bulb. The bulb should brighten up slowly in comparison to when it's connected directly to mains.****Also the entire circuit is directly linked with mains and is therefore extremely dangerous while plugged in and without an enclosure.****Need Help? Please send your queries through Comments for quick replies!**

## Comments

Would you please provide the specs of the SCR?

We can try BT151, I think it would work for the above application, its RMS current handling capacity is only 4 amps but since the initial surge is sinked by the diode, the scr would stay aloof from it.

Have you tested this circuit? I think capacitor start induction run motor is used in the compressors, (if you use a lamp, it will glow) such motors do start or run in half or full wave DC ( only universal motors, used in mixie, power tools etc; can start/ run in AC as well as DC

Another doubt, when we use a diode in series with a load, it gets half of the supply voltage. So the power consumption reduces to 1/4 ie. 25% not 50%, if voltage reduces to 1/3, the power reduces to 1/9 so

Please read the warning message at the bottom, I have clearly mentioned my doubts there.

I have given the example of a bulb just to make sure nothing goes in smoke while using it with any other form of load.

In the motor soft start circuit for the refrigerator, you did not specify the type of the thyristor to be used. Are we free to use any type of thyristor.

Regards

Bernard Tendengu

best regards

https://homemade-circuits.com/adding-soft-start-to-water-pump-motors/

for identifying the zener voltage you can connect the respective zener across a DC voltage source through a 10K resistor, and measure the potential right across the zener diode pins, this will give you the required data regarding the device.

As you can see there's hardly any component in the design and I have explained its working clearly which can be easily understood by anybody.

If you are having difficulty understanding it You can put this link to any electronic forum and ask their response about this and see if you can get an appropriate reply.