Yurba07 14-12-2020, 16:18 Hello everyone, it took me a relay to turn off the fan in the bathroom.
Having rummaged through the Internet, I stopped at this scheme:
I want to warn you right away that this and similar circuits with a transformerless power supply do not have galvanic isolation from the network and you must definitely follow safety measures when setting up and using these devices !!!
In general, the scheme is working, but very voracious. Own consumption is outrageously huge! This is due to the rectifier on two diodes VD1 and VD2, which are located after the quenching capacitor C1. Since the capacitor can only pass alternating current, it turns out the following: diode VD2 passes the positive half-wave, charging the capacitor C3, and the negative half-wave passes the diode VD1, which actually discharges the capacitor, I could be wrong … As a result, the total current consumed from the network when the timer is running large enough that negates the principle of energy saving by automatically disconnecting the load after a certain time.
There are two options for reducing the current consumed by the circuit itself:
1 – The use of a banal low-power transformer power supply with a power of 1-2 W. This will give many times less consumption than with the indicated transformerless power supply.
2 – Put a full diode bridge in this circuit, which I did.
As a result, its own power consumption has decreased to 10-12 W. Yes, that's a lot.
But it suited me for two reasons:
1 – The connected fan consumes 150 W, against its background another 10 W of power for
7-8 minutes (you may need a different shutdown delay time!), Will not greatly affect total consumption.
2 – Small dimensions of a transformerless power supply unit.
Nothing interferes with instead of a transformerless power supply unit on the elements R1, C1, VD1, VD2, put a small transic + 4 diodes + electrolyte + linear stabilizer type L7812 + electrolyte .. Everything !!!!
The running time of the timer and, accordingly, the fan off delay depends on the values of the resistor R2, C3. The higher their value, the longer the turn-off delay will be. Well, and, accordingly, vice versa.
The principle of operation is as follows:
When the contacts of the S1 button are closed, the mains voltage of 220 V flows through the fuse F1 to the quenching capacitor C1, on which the excess voltage drops. Further, a reduced alternating voltage (about 13 V) is fed to the diode bridge from which the rectified voltage is filtered by the capacitor C2 and feeds the timer circuit. Stabilization at the PSU output is carried out by the VD2 Zener diode. At the moment of power supply, the capacitor C3 of the timing circuit of the timer is discharged and there is a low voltage at 2, 6 feet (logical 0). Accordingly, the timer instantly switches to the on state and a positive voltage (logical 1) appears on its 3 leg, the VT1 transistor opens and relay K1 is triggered, it blocks the contacts of the S1 button (self-supply). When the S1 button is released, power to the circuit and the fan will be supplied through the closed contacts of the K1 relay. Then the capacitor C3 starts charging and when the voltage on the 2, 6 legs of the timer reaches 2/3 of the supply voltage (in my case, about 8.5 V), the timer turns off. On the 3rd leg, the voltage will disappear (logical 0), the VT1 transistor will close, the relay will turn off and turn off the fan and itself with its contacts. As a result, there is no current consumption in standby mode!
The process is repeated the next time you press it.