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enter image description here How can I maximize the ouputs?

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  • \$\begingroup\$ To get the lowest ripple possible you should make the capacitor the largest possible. \$\endgroup\$ Commented Apr 30, 2018 at 1:20
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    \$\begingroup\$ The general rule for audio amplifiers is 2,000 uF per amp of current consumed. This is to make sure 60HZ/120HZ ripple is very low. You would need inductors in the Henries to filter 120 HZ ripple from a bridge rectifier. Their DC resistance helps more than their inductance. \$\endgroup\$
    – user105652
    Commented Apr 30, 2018 at 1:33
  • \$\begingroup\$ Your schematic generates a + supply and a - supply and a ground. Perhaps you should add another inductor to the - supply, similar to the + supply's L1. Or did you intend to have a unipolar DC supply? \$\endgroup\$
    – glen_geek
    Commented Apr 30, 2018 at 1:48
  • \$\begingroup\$ Yes! this is a bipolar DC supply. Im trying to get a Variable positive regulated power supply and a Variable negative regulated power supply using LM337 and LM317. PS- the secondary voltage of the transformer is about 17.76V rms. \$\endgroup\$
    – Shawty
    Commented Apr 30, 2018 at 2:06
  • \$\begingroup\$ Yes it but depends on the load R and % ripple. for 5% ripple make R/Xc(f) =20 . The inductor doesn't do very much as is with XL(f)/R and XL(f)/Xc(f) being probably very low . R=tbd = Vcc/Icc \$\endgroup\$ Commented Apr 30, 2018 at 2:23

2 Answers 2

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Consider this circuit.. you will see many volts "ripple" across R1.

schematic

simulate this circuit – Schematic created using CircuitLab

You really need two capacitors, one for the positive supply and one for the negative, each connected to ground. Peak-to-peak ripple @60Hz will be approximately 0.0083*Iout/C, so for a 1A output current you need 8.3mF or 8,300uF to get 1V p-p ripple.

Tony is correct, the 100uH and 1uF don't do much of value so I am ignoring them.

Also, it appears you have a mistake in your simulation- 120VAC mains is 170V peak, 340V peak-to-peak. The stated voltage is already RMS, no need to multiply it by \$\sqrt{2}\$.

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If C1 and C2 share the same via to GND, and if the depth/periphery ratio of that via is 1:1 indicating 0.0005 ohm of resistor if the via is plated to standard 1.4 mil thickness, then bigger values of Capacitance are merely better conductors of ripple from the top of C1 to the top of C2.

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