Electronic – How do capacitors help in smoothing potential difference from a full wave bridge rectifier

With a 800 Volts DC supply, and a 200 kΩ resistor in series with the capacitor, the maximum current the charging will draw is 4 mA.

How to calculate this:

• Assume an ideal capacitor i.e. Zero Equivalent Series Resistance. Real capacitors and their leads can only increase this ESR.
• At start of charging the current draw is maximum, as potential difference is maximum
• At this time, I = V / R, V = 800 Volts and R = 200 kΩ
• Thus I_max = 4 mA
• As the capacitor charges up, this current will reduce further as per the normal RC curve.

As this is lower than the 20 mA supply limit, the supply-limited maximum current will not come into play at all.

EDIT: Note that the time constant of the resistor and capacitor is 47 seconds, which means that it will take that long to reach 63% of its final voltage. It will take 3× that long (2:21) to reach 95% and 5× that long (3:55) to reach 99%. You can reduce those numbers by a factor of 5 by reducing the resistor to 40KΩ, which would just touch the source limit of 20mA at the beginning of the charge cycle. If you simply charge the capacitor with a constant current of 20mA, it will take 9.4 seconds.

When the relay coil is open circuited whilst current is flowing, without some "protection" device a high voltage spark can be generated that can destroy transistors. A capacitor across the coil can suppress this spike. The energy contained in the relay coil's magnetic field flows into the capacitor charging it up but, some of that energy is converted to heat in the resistance of the coil windings.

At some point the coil has pushed out all the available energy and the capacitor has a voltage across its terminals due to it "collecting" the energy and the cycle reverses with the capacitor (because it's in parallel with the relay coil) now forcing current to flow into the coil. But some of that energy transferred is again given off by heat in the coil resistance so a little less current is built-up within the coil.

At some point later the capacitor terminal voltage is zero meaning it has transferred all its energy into the current in the coil and that current starts the process of recharging the capacitor in the reverse direction.

The cycle repeats with a little less energy each time due to heat losses in the resistance of the coil and eventually (usually after a few milli seconds) it's done and there's no appreciable energy left in inductor or capacitor. The maximum voltage that the circuit reached would be (by design) no greater than twice the original supply voltage and probably a lot less - it all depends on how big the suppression cap is.

So, you apply power to the relay in the normal way and you are also supplying some energy to the cap but because it is dc, the cap soon charges up and doesn't draw current for most of the time the relay is energized.

Similarly, a diode doesn't draw any current when the power is applied to the relay because it is reverse biased but, when the relay opens, the current in the coil has to go somewhere (or make a big spark) and it suddenly finds it is connected across a forward biased diode. The energy circulates around coil and diode dissipating power from the diode and the relay coil and never attaining a voltage greater than supply voltage + 0.7V.

The two components protect the relay driver circuit but they are totally different parts and act very differently in this circuit.