how does the filter capacitor work in a rectifier circuit?

Question

My question concerns the following schematic

PICTURE NUMBER 1

If the Capacitor and the LOAD are in parallel, this means that there is more than one path for the current to flow, therefore, the unfiltered direct current from the rectifier will split between the two branches and part of it will reach the LOAD. My question is how is this prevented? Does really the unfiltered current split, or first it builds up the capacitor until it's fully charged, than it further processes towards the LOAD?

PICTURE NUMBER 2

Here's what I do understand and what I don't. First, the rectifier's operation process is fairy straightforward, it transforms the INPUT AC into rippled DC at the OUTPUT.

PICTURE NUMBER 3

I want to simplify the first schematic on the top.The purpose of the Filter Capacitor is to remove the ripples from the DC as much as possible in a manner that it will appear smooth on the LOAD, but on the schematic is shown that the unfiltered DC passes the LOAD because of the parallel circuit connection. Please explain to me what I got wrong about all this. Thanks in advance.

Answer 1

We commonly call these capacitors "reservoir" capacitors. They act as a reserve of power during periods of no voltage in much the same way that a water reservoir can supply water during dry spells.

Continuing with the water analogy, we notice that during the dry spells the water level will fall as water is consumed and this will result in falling pressure on the water mains. Similarly if current is drawn from the capacitor reservoir as in your "With Load" graph you will see the voltage droop between cycles.

When the water supply is restored pressure builds up in the line. Since the reservoir needs to be refilled water will flow into it. Meanwhile the load is demanding water and, since water can't flow in and out of the reservoir simultaneously, it should be clear that the mains pressure feeds both the reservoir and the load.

Note that during the "With Load" period of your graph that the supply voltage exceeds the reservoir voltage for about 1/3 of the time. That means that it has to supply the average current in 1/3 of the time so the pulsed current will be about three times that of the load current.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A semiconductor version of the rectifier. Note that R2 and D3 are there to enable visualisation of the un-smoothed full-wave rectified waveform on the simulator.

• The bottom trace shows current into the capacitor as positive.
• (1) Note that there is a 10 A current pulse on the first half-cycle to charge the reservoir capacitor.
• (3) Note that 80% of the time that C is discharging at -0.4 A.
• (2) Note that subsequent pulses are much shorter - about 20% of the cycle and peaking at almost 2.5 A.

Figure 2. Simulator results. (Click to enlarge.)

Answer 2

I believe you are new to electronics still trying to grasp with basic concepts and its flow of current.

If the Capacitor and the LOAD are in parallel, this means that there is more than one path for the current to flow, therefore, the unfiltered direct current from the rectifier will split between the two branches and part of it will reach the LOAD

It all depends on how you want to look at it and understand it. Technically yes there are 2 paths and the unfiltered current flows through capacitor while the constant current reaches to LOAD.

My question is how is this prevented? Does really the unfiltered current split, or first, it builds up the capacitor until it's fully charged than it further processes towards the LOAD?

So here you are mostly right on the second part. Basically capacitor doesn't allow a sudden change in voltage. So your capacitor is acting as a temporary bank. So when the rectifier output pulse reaches lower value the capacitor starts providing that excess voltage and at your LOAD you will see constant voltage (approximately constant).

It is very much acting like what Transistor has explained in his answer.

Answer 3

There is no un-filtered current coming from the rectifier - the rectifier is either conducting or it isn't. If it is conducting then the majority of the current it is providing flows into the capacitor to charge or recharge it. If this wasn't the case, then the capacitor is too small and you likely have a poor design that will create too much ripple voltage under normal load conditions.

So, if the capacitor has had energy taken from it by the load, its terminal voltage will have reduced a little bit AND the trick is to ensure that the amount the voltage has sagged-to isn't going to cause the load problems i.e. the "droop" isn't too much.

Then along comes the next peak in the AC waveform and the diode conducts to refill the capacitor with energy - engineers normally assume that the diode current that flows during this re-charging event is mainly going into the capacitor. As I said earlier, if this is not >75% true then you probably need to increase the value of the capacitor so that it is true.

Answer 4

The unfiltered current splits. How much goes where is determined by the instantaneous voltage coming from the rectifier, the instantaneous voltage on the capacitor, and the current requirement of the load.

As shown in the right-side drawing, the output voltage (the voltage on the capacitor) increases whenever it is less than the input waveform. this is the time when the input is both charging up the cap and driving the load.

When the input is less than the cap voltage, the rectifier is reverse biased and no current comes from the transformer. During this time the capacitor alone is powering the load, and hence its voltage decreases.