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I'm planing to use this circuit for my op amp test but I don't understand capacitor selection

Why C1 and C2 selected 2200 uF but C3 and C4 selected 1000 uF? (C3 is 1000 uF picture is wrong)

I tried to use 220 uF and 100 uF and voltage levels were same. I think C3 and C4 for stabilizing the output but what's the purpose of C1 and C2?

Also if capacitor selection only effects on output current, is there equation for it?

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  • \$\begingroup\$ If I know correctly (I am an amateur), C1 and C2 must not be electrolytic caps. Infact you don't need capacitor before bridge rectifiers. A crude rule is 1000uF per 1 amp of output current for C3 and C4. \$\endgroup\$ – sribasu Sep 14 at 7:52
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Why C1 and C2 selected 2200 uF but C3 and C4 selected 1000uF ? (C3 is 1000 uF picture is wrong)

C1 and C2 are in series with the negative supply. The impedance is given as \$ Z_C = \frac {1}{2 \pi fC} \$ and for 2200 μF at 50 Hz that's just under 1.5 Ω each. (See Capacitor impedance calculator.) If you tried to pull 1 A from the negative supply you would drop 3 V across the capacitors so your supply would not be symmetrical.

C3 and C4 are holding the voltage up between half-cycles of the supply. For modest levels of ripple the ripple will be given by \$ V_r = \frac {I_{load}}{2 f C} \$. See Electronics notes for more.

I tried to use 220uF and 100uF and voltage levels were same I think C3 and C4 for stabilizing the output but what's the purpose of C1 and C2 ?

C1 and C2 create an isolated supply for the negative rail since you haven't got a centre-tapped transformer.

Also if capacitor selection only effects on output current , is there equation for it ?

Given.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A much simpler supply.

You can make a much simpler supply as shown in Figure 1. This won't suffer the voltage drop in the negative rail that your circuit does and will give symmetrical voltages if the currents drawn are symmetrical.

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    \$\begingroup\$ A much simpler supply is usually called a Delon bridge. \$\endgroup\$ – Janka Sep 13 at 21:55
  • \$\begingroup\$ In figure 1. Virtual ground has direct contact with ac , isn't that will cause problem ? Also +V and -V is peak or ? \$\endgroup\$ – Mordecai Sep 13 at 21:55
  • \$\begingroup\$ @Janka: Never heard of it in over forty years of reading. \$\endgroup\$ – Transistor Sep 13 at 21:56
  • \$\begingroup\$ @Mordecai: No. Ground is just a reference point. Here we've chosen the bottom of the transformer as ground because you wanted a symmetric supply. V+ and V- will be DC supplies with ripple that can be calculated on the linked Electronic Notes page. \$\endgroup\$ – Transistor Sep 13 at 21:58
  • \$\begingroup\$ @Janka your comment really helped because now I understand this circuit is just a modified voltage multiplier circuit \$\endgroup\$ – Mordecai Sep 13 at 21:59
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Top ac input goes +ve, C2 charges via D3 & D6. When top ac input goes -ve C2 dumps its charge into C3 which is charging via D1 and the 0V supply.
When the top ac input goes -ve, the Bottom ac input goes +ve charging C1 via D4 & D6.The bottom ac input then returns -ve forcing C1 to dump its charge into C3 which is charging via D2 and the 0V supply.

So each of C1 and C2 is alternately charged and then, in turn, dump their charge into C3.

It's a charge pump.

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