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enter image description here

What is the function of capacitor C in this diagram?

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    \$\begingroup\$ "This question does not show any research effort". I don't think it's possible to read about a bridge rectifier without the capacitor being mentioned. \$\endgroup\$
    – pipe
    Feb 23 at 14:58

2 Answers 2

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Here's the circuit without C:

schematic

simulate this circuit – Schematic created using CircuitLab

There's a source of 16V RMS AC voltage being rectified by D1 to D4. The voltage across load \$R_L\$ looks like this:

enter image description here

It's been full-wave rectified by the diodes, so it's always positive, but it's pulsing rather violently, 100 times per second. That's no good for any circuit that requires smooth DC to operate (which is almost every single circuit in existence), so we add capacitor C to "smooth" it out:

schematic

simulate this circuit

The capacitor becomes a reservoir of energy, which gets charged during the "peaks" above, and is used to provide current to the load in between those peaks. Now the output voltage across \$R_L\$ looks like this:

enter image description here

It still has ripples in it, but it's much smoother and steadier than before. That's why we call C a "smoothing capacitor". Its purpose is to "smooth" out the pulses of rectified AC.

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Can we "stop time" ?

The processes in this circuit occur too quickly for our human thinking and are repeated periodically like in a movie. In fact, only two frames of this "movie" are enough to understand what is happening in the circuit; the rest are redundant. It would be great if we could somehow "stop time" at these two moments and study the schematic through the CircuitLab Live DC Simulation.

How to do it

We can do it if we replace the capacitor with a "rechargeable battery" charged to the same voltage as the capacitor at this instant. Then we can see the meter readings or hover the mouse over the circuit to see more quantities. We can change the voltage across this simulated capacitor to shift the observed moment. To eliminate the conflict between the voltage sources, it is necessary to set some small internal resistance (e.g., 1 Ω) of the battery.

This trick can be used to explore other capacitor circuits too, e.g. AC amplifiers, voltage multipliers, etc.

Exploring the circuit

We can imagine the diode bridge as a diode switch that briefly turns on the input voltage to the capacitor and the load when Vin > Vc, and turns it off when Vin < Vc. So for the purpose of intuitive understanding, we can replace the AC input voltage source with a constant DC voltage source whose voltage is equal to the input voltage at that moment.

Vin > Vc; switch is ON

When the input voltage exceeds the voltage across the "capacitor", the diode switch D is ON and the input voltage is applied to both the capacitor and load. So the input source is doing two jobs simultaneously - charging the capacitor and powering the load. They are both loads so the current (green arrows) enters both.

schematic

simulate this circuit – Schematic created using CircuitLab

Vin < Vc; switch is OFF

When the input voltage is lower than the voltage across the "capacitor", the diode switch D is OFF. There is no input voltage applied and the capacitor begins to play this part - it is powering the load. The current circulates between the capacitor and the load.

schematic

simulate this circuit

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  • \$\begingroup\$ Would the downwaters be so kind as to explain what they don't like about my "stopping time" trick? Or should we be repeating the same textbook explanations over and over again? \$\endgroup\$ Feb 24 at 10:29

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