This is a simple means of calculating the required size of the input filter capacitor in a basic power supply, or calculating the peak-to-peak ripple voltage in an existing supply. It works by assuming that the capacitor supplies current to the load approximately 70% of the cycle—the remaining 30% is supplied directly by the rectified voltage and during this period the capacitor is charged as well.

**Graphic**

**Source or native rfflow file **

I was surprised how quickly this graphic went together using Rfflow drawing software. I think this is the first time I was able to create realistic looking sine waves—of course, I have lots of experience with Rfflow.

**Peak-to-peak ripple voltage**

Peak-to peak ripple voltage is a most useful parameter. By knowing this, the minimum head voltage of a voltage regulator may be easily determined. For instance, the LM7812 series regulator requires a minimum of 2.5V across it to function within specifications. That means the valley of the peak-to-peak ripple voltage may not drop below 14.5V.

**Math**

Learn these basic relationships—I use them at least weekly…

Q = C * V, C = capacitance in farads, Q = charge in coulombs, V = voltage potential

Q = I * T, I = current flow in amps, Q = charge in coulombs, T = time in seconds

F = 1 /T, F = frequency in hZ, T = period in seconds

Therefore,

I * T = C * V, or C = (I * T) /V

The delta “Δ” symbol indicates a change, such as a change in voltage or a change in time

*C = I * ΔT / ΔV*

In this case, ΔT = 1 /F e.g. the time period of a 50hZ waveform = 1 /50hZ = 0.02seconds

Therefore,

C = I / (ΔV * F)

Now including the 70% factor we get the final relationship:

C = 0.7 * I /(ΔV * F)

C = capacitance in farads, I = current in amps, ΔV = peak-to-peak ripple voltage, F = ripple freq in hZ

Note that ripple frequency in a full-wave rectifier is double line frequency. For half-wave rectification, the ripple frequency is the line frequency.

Solving for ΔV

*ΔV = 0.7 * I /(C * F)*

**The 70% factor**

This is an assumption—in cases where the capacitor is oversize, it will be perhaps 80%, but this is close enough for government work and not very critical.

**Putting it to work, a practical example**

Assume that we want to make a 9V, 500mA power supply using the LM7809 voltage regulator device, 12V transformer, bridge rectifier and filter capacitor. Line frequency is 50hZ. How large should we make the filter capacitor?

From the spec sheet, we learn that the dropout voltage of the LM7805 is 2.5V. Therefore, the valley of the peak-to-peak ripple should be 9V + 2.5V = 11.5V. Assuming that the average rectified DC voltage is 12VDC, the minimum negative peak of the ripple voltage = 12V – 11.5V = 0.5V. The peak-to-peak ripple voltage is double the peak or 1.0V. Plugging it into the formula:

*C = 0.7 * I /(ΔV * F)* = 0.7 * 0.5A /(1Vp-p * 100hZ) = 0.0035farads or *3500µf*

**Selecting the capacitor**

Since 3500uf is not a standard value, simply scale up to 3900uf. The maximum value is limited by size and pocketbook—it is not unreasonable to go with 4700uf or 10,000uf—ripple voltage will be proportionately reduced. The DC voltage rating is generally selected to be double the applied voltage or 25V in this case. 35, 50 or 63V would also be acceptable voltage ratings. For experimentation, a 15V capacitor would be OK.

**In practice**

Should you make such a power supply, do not be at all surprised if the DV voltage you measure is much higher than expected. The reasons are numerous. First, you may not be loading the output to 0.5A and/or the transformer may be rated for higher current. In such cases, the capacitor will tend to peak charge to the peak of the AC waveform. The resulting voltage may be 15 or 16V or so. It is important to know this because the LM7909 regulator must now dissipate perhaps double the expected power. In other words, it will require a larger heat sink to run at an acceptable temperature.

**For the future**

Novel voltage doubler circuits

**Undocumented words and idioms** (for our ESL friends)

close enough for government work—idiom, there is a right way and a wrong way to accomplish a given task…and then there is the government way…or the army way…

limited by the pocketbook—idiom, limited by available funds

C = 0.7 * I /(ΔV * F) = 0.7 * 0.5A /(1Vp-p * 100hZ) = 0.0035farads or 3500µf

Sir Please Explain how did you got 100Hz Here, if the frq. is 50Hz from Suply.

Please Replay

The ripple frequency is double the line frequency–check out the waveforms in the graphics above and you will see how this works.

Sir can u explain me

C=0.7*0.5A/(1Vp-p*100hz)

What is 1Vp-p?

What is 0.7?

I hope u sir u clear my dought

1Vp-p is the AC peak to peak ripple voltage that appears across the capacitor. This adds and subtracts from the DC voltage level.

The factor “0.7″ is the assumption that the capacitor supplies the load for 70% of each half-cycle, and the rectifier output supplies the load for the remaining 30% while charging the capacitor at the same time.

Thank u sir for anser

1Vp-p is the AC peak to peak ripple voltage that appears across the capacitor. This adds and subtracts from the DC voltage level.

Can u tell me how to find outripple factor With formula

From the spec sheet, we learn that the dropout voltage of the LM7805 is 2.5V. Therefore, the valley of the peak-to-peak ripple should be 9V + 2.5V = 11.5V. Assuming that the average rectified DC voltage is 12VDC, the minimum negative peak of the ripple voltage = 12V – 11.5V = 0.5V.

could you plz expain this with figure??

In this case, the ripple voltage component is 0.5V peak, or 1V peak to peak. This provides the delta V that we can plug into the capacitor calculation formula. Confusing, but simple at the same time.

you have explained so simply there is no ripple in mind.