# Working of this circuit?

Discussion in 'General Electronics Chat' started by hubble, Jul 16, 2015.

1. ### hubble Thread Starter Member

Jun 17, 2009
22
0
Hello. I am trying to figure out this circuit. I understand the rectifier part but am struggling to understand what are these components doing: R2, C2, R3, C3, D1/2/3/4 and how they are doing what they are doing? I mean, how to go about solving this particular part of the circuit?

2. ### Bordodynov Active Member

May 20, 2015
670
194
D1, D2, C2, C3-classic rectifier voltage doubler. It gets a negative voltage. R2 limits the current of the capacitor C2. R2 together with capacitor C3 filters the output voltage.
Resistor R3 and diodes d3 and d4 are the simplest voltage stabiliser, about -1.2 volts.
Use the Spice circuit analysis program.
Such as free, but very high quality programme LTspice.

3. ### AnalogKid Distinguished Member

Aug 1, 2013
4,685
1,297
This is a tricky one. The circuit is 1/2 normal power supply and 1/2 charge pump. The bridge rectifier and C1 form a normal positive voltage power supply, with R1 acting as a load on the main filter capacitor. There is not ground symbol, but we will establish the line running through the center of the circuit, connecting to the left side of the bridge, D1, R1, C3, C1, D3, and both meters as GND.

This charge pump uses a capacitor to create negative DC from positive AC. R2 is just a current limiter; we can leave it out for the moment. C2 acts as an AC coupling capacitor. It has the input power AC on it, going both above and below GND. But its - end can't go above GND because it is "caught' by D1. When the - end of C2 gets more than 0.7 V above GND, D1 conducts and clamps the - end of C2 and +0.7 V. The + end of C2 continues to be pulled up by the input AC, and this charges up C2. On the negative half cycle on the input AC, both ends of C2 go down. This pushes the - end of C2 below GND. The action is similar to the way the diodes in a bridge rectifier catch the ends of the secondary winding to create unipolar DC.

So now on each half-cycle the - end of C2 goes negative (below GND). The charge in C2 goes through D2 to charge up C3 just like in a normal power supply, except that a) these are negative voltages so the diodes appear backwards; and b) the energy is coming through a capacitor rather than an inductor (the transformer secondary winding) so there isn't a lot of current available. While it is common for a power transformer secondary to have an impedance of 1 ohm or so, for a capacitor to have an impedance of 1 ohm at 60 Hz it would need to be 2,700 uF. Your 47 uF cap has an impedance of 57 ohms, which reduces the available voltage and current to the rest of the circuit. So the voltage developed across C3 will not be as large as the voltage across C1. D3 and D4 are forward biased through R3 by the negative voltage on C3, and act as a 1.4 V zener diode.

The key to understanding this circuit is to keep a flexible attitude to things being *relatively* positive and negative with respect to other things. GND is negative with respect to the positive end of C1. That same GND is *positive* with respect to the negative end of C3.

ak

cmartinez likes this.
4. ### hubble Thread Starter Member

Jun 17, 2009
22
0
I am still not getting this. How does the negative cycle of an AC push the -ve plate of C2 below ground? Also, how can an electrolytic capacitor take negative cycle on the positive plate? Will it not harm the capacitor?

Does it mean that the -ve end of C2 goes negative on both the +ve and -ve halves of an AC signal after the first charging cycle?

5. ### cmartinez AAC Fanatic!

Jan 17, 2007
3,698
2,759
Very interesting circuit. AnalogKid has already explained it thoroughly, and I agree with Bordodynov, you should download LTspice and sim it. It's an extremely useful program for this sort of exercises. And it's also very easy to learn. Drawing your circuit in it should only take a few minutes. After that, you can probe it in whatever node you're curious about, and start adding or removing or changing component values to see what happens... that sort of playing around will improve your understanding significantly.

6. ### AnalogKid Distinguished Member

Aug 1, 2013
4,685
1,297
C2 is acting as a coupling capacitor. Think about a coupling capacitor between two audio amplifiers. Whatever AC voltage is on one side is on the other side. When one side goes up or down, the other side tracks it. This is completely independent of any DC charge on the capacitor. So it is very common for the positive side of a capacitor to move in the negative direction. This is not the same thing as the positive side of a capacitor going below the negative side. That could indeed cause problems for an electrolytic capacitor.

ak