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I have made a full wave rectifier circuit to obtain 5VDC, 9VDC and -9VDC. To prevent the ripple, i have selected C1 and C2 as 4uF which were shown below. I will use a 12V, 6W transformer to get AC input signals. Do you have a suggestion related to values of these capacitors to avoid the ripple or should i choose bigger capacitances like 100uF? Boxes are 7809, 7909 and 7805 regulators. Thank you so much.

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    \$\begingroup\$ What exactly are the white boxes? What is the maximum output current required at 5V, 9V and -9V? \$\endgroup\$ Aug 2, 2017 at 7:16
  • \$\begingroup\$ I have edited the question. Upper one is 7809, below side is 7909, box at the right side is 7805 regulators. Approximately, 100-200 mA is enough, i don't need so much current. \$\endgroup\$
    – layout789
    Aug 2, 2017 at 7:47
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    \$\begingroup\$ Put the actual regulators and correct AC frequency into your simulation, apply loads that draw 200mA and look for ripple on the outputs. \$\endgroup\$ Aug 2, 2017 at 8:35

2 Answers 2

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A bigger value is preferable, 4uF is when you have a decent DC into the regulator, 100 - 1000uF in parallel with a 100nF for higher frequency decoupling. a 100nF is needed on the output side as well.

Ref.: http://www.electronicshub.org/understanding-7805-ic-voltage-regulator/

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Precise anwer comes from examining the charge stored in a capacitor

$$ Q = C * V $$

Take the derivative of that, with respect to small steps in time, and we see

$$ dQ/dT = C * dV/dT + V * dC/dT $$

Now assume the capacitor is constant (this is your filter capacitor), and replace dQ/dT with "I" (current) to find

$$ I = C * dV/dT $$

Rearrange this, to have dV (ripple voltage) on the left, thus

$$ dV = I * dT / C $$

Now let dT be 8 milliseconds (1/120 Hertz), I be one amp, and C be 10,000uF

You have dV (ripple) = 1 amp * 0.008 / 10,000uF = 0.008 / 0.01

or 0.8 volts ripple. At 1 amp, 10,000uF.

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