Capacitor Power Supply circuit

One of the major problems that is to be solved in an electronic circuit design is the production of low voltage DC power supply from Mains to power the circuit. The conventional method is the use of a step-down transformer to reduce the 230 V AC to a desired level of low voltage AC. The most simple, space saving and low cost method is the use of a Voltage Dropping Capacitor in series with the phase line.

1.  IC2 is connected as a differential amplifier and compares the signals at its two inputs.
2. Referring to the circuit diagram: the input comprises a mains switch, fuse, transformer, bridge rectifier and smoothing capacitor (C2).
3. The difference between the inputs is the voltage drop across 'current’ sensor R4. This IC feeds the current sensing input (pin 2) of the L200.
4. P1 in the feedback loop of the 741 is used to vary the output current of the circuit. IC1 must be mounted on a suitable heat sink as it dissipates nearly all the power of the circuit.
5. The reference level output from I pin 4 of IC1 goes to the voltage divider made up of R5 and P2 (this pot sets the value of the output voltage).
6. The power supply can quite easily be built into a case and a voltmeter and ammeter mounted on the front panel. ln view of the accuracy of the circuit these should ideally be digital meters, but virtually any type will do.
7. If you compare the expense and the rating of this power supply you will get a surprise, because the output voltage and current are fully adjustable between O. . . 18 V and 0 . . . 1.8 A respectively and costs have still been kept very reasonable.
8. Diode D5 and capacitor C1 produce a negative auxiliary voltage, which is stabilized by zener diode D6 and capacitor C4.
9. All this is necessary to enable the output voltage to be adjusted down to zero volts. During the construction of this part of the circuit bear in mind that the positive lead of electrolytic capacitor C4 is connected to earth! Regulation is provided by IC1 and IC2. Capacitor C3 suppresses any residual transients at the input of lC1 and it should therefore be connected as closely as possible to IC1 similarly C4 and IC2).
10. The negative voltage provides the negative supply for the two ICs.