# Choosing capacitance for recifier

So I'm trying to use a couple of rectifiers. Specifically, the "NTE5326" and the "36MB120A."

I've tried following videos on how to calculate for a smoothing-capacitor but to no avail.

If you use these rectifiers can you please give the capacitance value for each, and show how you got your value?

• It depends on f, Vp, Ip, ESR, Irms and your criteria for example: **Vpp ripple or Vmin for LDO dropout, Irms ripple current max, leakage time constant, voltage multiplier ratio ....etc. which one? and how much>? – Tony Stewart Sunnyskyguy EE75 Apr 17 at 12:14
• or your expected DC amps and volts for some input Vac and Hz or some power in watts ?? any criteria?? – Tony Stewart Sunnyskyguy EE75 Apr 17 at 12:21
• "I've tried following videos on how to calculate for a smoothing-capacitor" - well, there's your problem. – JRE Apr 17 at 12:39

Using Q = C * V, charge = capacitance times voltage, we take the derivatve with respect to time, to produce

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

Then assume C is constant, making the 2nd term become zero, and assume dQ/dT = I, and we have

Icurrent = dQ/dT = C * dV/dT

simulate this circuit – Schematic created using CircuitLab

Now you need 3 variables: I (one amp), T for 50Hertz (50 cycle per second, 0.02 seconds) and V the ripple voltage looking like a sawtooth slow fall and abrupt recharge time of 1 volt

And rearrange the I = C * dV/dT to become

C = I * dT/dV

C = 1 amp * 0.02seconds / 1volt = 0.02 Farad or 20,000 microFarads

If you use full wave rectifier at 60Hz, thus 120Hertz recharge events per second, the math is

C = 1 amp *( 1/120seconds ) / 1volt = 0.00833 Farads, or 8,333 microFarads

If 0.1 amp load, and your regulator accepts 2 volts ripple, then

C = 0.1 amp * (1/120seconds) / 2volt = 416 uF (470 is a standard value)

Those are just ordinary bridge rectifiers. The choice of filter capacitor has little to do with the rectifier, and everything to do with the load you need to drive.

The capacitor gets charged in short bursts at the peaks of the AC waveform (at twice the line frequency for a bridge rectifier), and it supplies the load current during the intervals in between. Therefore, you can model the capacitor voltage as a sawtooth waveform. This waveform is called the "ripple voltage". The peak-to-peak voltage of this waveform gets larger with increasing load current, and it gets smaller if you increase the capacitance.

So, the two things you need to know are the average load current, and the maximum ripple you can tolerate. The maximum voltage is determined by the transformer, and the minimum voltage is determined usually by whatever regulator you're using.

You can derive the formula from the basic equation for a capacitor:

$$Q = CV$$

Q is charge, C is capaitance, and V is voltage. Since charge is just current × time, you can write this as:

$$It = CV$$

Solving for C gives:

$$C = \frac{It}{V}$$

This tells us that for a given load current I, a charging period t (e.g., $$\\frac{1}{120}\$$ s @ 60 Hz) and a peak-to-peak ripple voltage V, you need a capacitor with the value C.

For example, suppose you have a 8 VAC (RMS) transformer and you want to produce 5 V @ 1 A using a 7805 regulator, which needs an input of at least 8 VDC.

The peak voltage on the capacitor will be 8 VAC × 1.414, minus two diode drops in the rectifier, or about 9.9 VDC. Since the regulator needs 8 VDC, we can only allow 1.9 V of ripple. Again, assuming 60 Hz, the charging period is 8.33 ms. The formula gives us

$$C = \frac{It}{V} = \frac{1\text{ A} \cdot 8.33\text{ ms}}{1.9\text{ V}} = 4.4\text{ mF}$$

You can use the full wave rectifier at 2x line frequency 2x60=120 Hz = 1/T then choose your ripple %Vpp and if you know load R and then you can compute C from product RC=τ then τ/T must have some higher ratio.

Rectifier ripple voltage formula

Then you can edit the numbers in a simulator and verify it.