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Economical On/Off Power Switch Schematic Circuit Diagram

Many appliances these days are switched on and off with the simple push of a ‘soft’ on/off button. In the ‘off’ position the appliance is merely in the sleep state and continues to use a small amount of energy — and that is just ‘not done’ nowadays. This circuit not only retains the feature of switching on and off with one simple pushbutton but also reduces the power consumption in the off-state to zero.

Economical On Off Power Switch Schematic Circuit Diagram

 

When pushbutton S1 is pressed, the circuit receives its power supply voltage via the capacitive voltage divider containing C1. The rectified voltage across C2 energizes the relay RE1.B via R3. LED D2 lights up. One set of the relay contacts is connected in parallel with S1 so that the relay will continue to be energized when S1 is released. The remainder of the circuit has no effect in turn on. C3 ensures that T2 blocks and capacitor C4 are not charged yet. Both conditions ensure that T1 does not have any base current and is therefore off. The relay can now close and the AC power line voltage across K1 is switched through to K2.

After power on, C4 will charge slowly. After about 0.25 sec the voltage is high enough to turn T3 on via zener diode D4. There is now a voltage at the emitter of T3. If S1 is now pressed then T1 will receive base current via T3 and the second contact of S1. T1 conducts and shorts the voltage across RE1.B, which de-energizes the relay. At the same time, T2 ensures that the circuit latches: T1 provides base current, via R6, for T2. This will conduct and provide base current for T1 via R7. So T1 will continue to conduct, even after S1 is released. C2 is discharged via R3. In this way the power supply voltage for latch T1-T2 will eventually disappear, so it will unlatch. Timing capacitor C4 is also discharged, via D5, so that the circuit is now ready for the next start. The entire circuit is completely disconnected from the mains, the current consumption is literally zero!

The value of capacitor C1 mainly depends on the relay that is used. As an example, we are using an Omron MY4-24VDC [1]. The relay is a 24V-type which is happy with a coil current of 40 mA and has contacts that allow for a load of up to 5 A. At 24 V across the relay there is a current of about 10 mA through LED D2. The total current when switching on is therefore about 50 mA. The value of capacitor C1 is roughly determined as follows:

XC1 = UC1/IC1 = (115 V – 24 V)/50 mA = 1.8 kΩ C1 = 1/2πfXC1 = 1 / (2 x 3.14 x 50 x 4120) = 1.47 uF (1.5 uF)

It is absolutely essential that this capacitor be suitable for operation at AC grid voltages and is preferably a Class X2 type, for example one from the MKP 336 2 X2 series made by Vishay [2]. The capacitor actually limits the total current that can flow through the circuit. When T1 conducts, C1 limits the current through T1 to about 50 mA. The magnitude of this current also gives an indication of the apparent power that the circuit draws:

PS = U x I = 115 V x 50 mA = 5.75 VA.

The actual real power of the circuit is smaller than this value since the cos ϕ of the circuit is certainly smaller than 1. Resistor R2 discharges capacitor C1 after switching off. This also has to be a type rated for 115 VAC (for example the MBE/SMA 0414 series [3]). Switch S1 to needs to be appropriate for AC power line operation. It is possible to replace R2 with two ‘ordinary’ resistors of 470 kΩ in series. Resistor R1 limits the switch on current through S1 when capacitor C1 is discharged.

[1] www.ia.omron.com/data_pdf/ data_sheet/ my_dsheet_gwj111-e1-03.pdf

[2] www.vishay.com/docs/28120/ mkp3362.pdf

[3] www.vishay.com/docs/28767/28767.pdf

 

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