# How does a capactior reduce voltage ripples

The image below shows a very common use case of these capacitors in a full bridge rectifier.

Here is what I think:

The AC source acts as an independent voltage source, ie, it produces a fixed time-varying potential difference across its terminals irrespective of other components in the circuit. Since the voltage across the AC source is independent, the voltage source across the rectifier is also independent.

Now, according to Kirchoff's laws, the potential difference across the terminals of the rectifier must be equal to the poterntial difference across the capacitor. Applying the same law in the loop containing the rectifier and the load, the voltage across the rectifier is equal to the voltage across the load.

So according to me, the voltage across the load is governed only by the voltage produced by the rectifier and the capacitor does not do anything. Can anyone tell me where am I wrong?

• What is V+ when the load is not present? Remember that current does not flow back through the rectifier, so the capacitor will not be discharged. Commented Mar 4, 2016 at 3:29
• I'm quite sure that this is a duplicate of several other posts. Anywise, for an answer, draw the output waveform of your bridge without a cap, just with a load....."bunny-hop anyone?" - The capacitor stores some excess charge at the peak of each 'ripple,' then discharges into the load while the AC input crosses zero. This smooths the output waveform considerably. Commented Mar 4, 2016 at 3:31
• @RobhercKV5ROB But what about the arguments I posted above? Are they not valid? Commented Mar 4, 2016 at 3:33
• @Akshit I think you're forgetting the action of the diodes. The AC may act as a voltage source, but if there's a higher voltage in the cap, then that voltage can't back-feed into the AC source through the diodes, so the rectified, smoothed voltage never drops to 0 during AC waveform zero crossing events. Commented Mar 4, 2016 at 3:42

The AC source acts as an independent voltage source, ie, it produces a fixed time-varying potential difference across its terminals irrespective of other components in the circuit. Since the voltage across the AC source is independent, the voltage source across the rectifier is also independent.

Correct.

Now, according to Kirchoff's laws, the potential difference across the terminals of the rectifier must be equal to the potential difference across the capacitor.

Not correct. There can be a voltage across the diodes if they're off (not conducting). To see this more clearly, imagine removing the load and leaving only the capacitor and the diode rectifier. As the AC voltage rises, it charges the capacitor through the diodes. But when the AC voltage starts to fall again, it can't pull charge out of the capacitor -- the diodes only conduct one way! (That's the whole point of a rectifier.) The capacitor is still charged to the max AC voltage and stays that way forever. The AC source never supplies any more current.

Back to the real world. With a load, the capacitor drains over time. At the peak of the AC half-cycle, the AC voltage becomes greater than the capacitor voltage. The diodes turn on and the AC source charges the capacitor back to its maximum value. This is shown at the bottom of your picture.See the dashed purple line in your picture labeled "Waveform without capacitor"? That's the input voltage from the AC source. The black line labeled "Waveform with capacitor" shows the capacitor being charged up at the peak of the half-cycle, then draining slowly due to the load once the diodes turn off.

A higher capacitance means the capacitor (output) voltage drains more slowly:

$$\frac {dV_{Capacitor}} {dt} = \frac {I_{Load}} {C}$$

UPDATE: You asked whether the voltage across the rectifier is equal to the voltage across the load. The answer is yes, but only when the diodes are on. In that case, one pair of diodes acts (almost) like a short circuit, and you get this:

simulate this circuit – Schematic created using CircuitLab

When the diodes are off $(V_{Cap} > V_{AC})$, they act like open circuits, and you get this:

simulate this circuit

Diodes are nonlinear circuit elements. Their behavior changes based on the voltages and currents in the rest of the circuit!

For KVL, you need to look at it like this (simplified a bit):

simulate this circuit

One possible loop equation is:

$$V_{AC} = V_{D1+D2} + V_{Load}$$

Or:

$$V_{AC} = V_{D1+D2} + V_{Cap}$$

since the capacitor and load are in parallel. Now when the diodes are off, the cap/load is completely disconnected from the AC source ("floating"), so figuring out the diode voltage gets a little weird, especially since we're using ideal diodes. Fortunately, you don't have to worry about it. When the diodes are off, you can use KVL on the capacitor/load loop:

$$V_{Cap} = V_{Load}$$

(Of course, real diodes have a forward voltage drop when they conduct, but I'm leaving that out here for simplicity. You can put it in the $V_{D1+D2}$ term in the KVL equations above.)

• Is the next statement, which I wrote, correct:"(using KVL,) the voltage across the rectifier is equal to the voltage across the load"? Commented Mar 4, 2016 at 4:39
• Only when the diodes are on. I updated my answer with a longer explanation. Commented Mar 4, 2016 at 5:51

The AC source acts as an independent voltage source,

Yes, but the diodes can only pass current in one direction. If the input voltage to the rectifier is lower than the output voltage, then the ac source plus rectifier will no longer act like an independent voltage source, it will act like an open circuit.

During this part of the cycle is when the load will draw current from the capacitor instead of from the ac source. If the capacitor value is too low, the current drawn by the load can drop the capacitor voltage below the source voltage provided by the source+rectifier, leading to the source acting like a source again, and producing the waveform labeled "waveform without capacitor" in your diagram.

If the capacitor is chosen well, then the capacitor can supply the current to the load when the ac input voltage is below the output voltage, and you get the waveform labelled "waveform with capacitor" in your diagram.