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Context:

enter image description here Pg-485 of this pdf

Now we shall discuss the role of capacitor in filtering. When the voltage across the capacitor is rising, it gets charged. If there is no external load, it remains charged to the peak voltage of the rectified output. When there is a load, it gets discharged through the load and the voltage across it begins to fall. In the next half-cycle of rectified output it again gets charged to the peak value (Fig. 14.20). The rate of fall of the voltage across the capacitor depends upon the inverse product of capacitor C and the effective resistance RL used in the circuit and is called the time constant. To make the time constant large value of C should be large. So capacitor input filters use large capacitors. The output voltage obtained by using a capacitor input filter is nearer to the peak voltage of the rectified voltage. This type of filter is most widely used in power supplies.

So, as said in the above paragraph, the capacitor after being raised to the peak voltage, discharges to load as rectified voltage supply starts to decrease. The point I am being confused is, if the rectifier's voltage output voltage drops below the capacitor, then wouldn't the capacitor push back current into the rectifier as well? What effect would this have on the rectifier?

Edit: Thank you everyone for the help!

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Rectifiers act as one-way or "check" valves. Current only flows in the direction of the diode arrow.

enter image description here

Figure 1. The diode's check-valve analogy. Image source: What is an LED?.

If you look at the check-valve in the figure above, it should be clear that the spring normally keeps the ball in position and prevents back-flow. When “forward-biased” the ball shut-off can be moved against the spring but it will take some initial pressure to move the ball. This results in a pressure drop across the valve: the pressure downstream will be less than the inlet pressure.

In a similar manner the PN junction causes a voltage drop. For silicon it is about 0.7 V. Since there is a PN junction in the base-emitter of your transistor you can expect a 0.7 V drop across it when forward biased.

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  • \$\begingroup\$ I choose to accept this answer over the others due to the nice analogy it gave ^^ \$\endgroup\$
    – Buraian
    Mar 25 at 8:52
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No, because the rectifier is made of four diodes. It is also called a bridge rectifier or full-wave rectifier.

You may understand this better with this image:

enter image description here

D1 to D4 is the bridge rectifier.

The current cannot flow back, because it is blocked by the diodes.

Diodes are devices that let current flow only in one direction.

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No, the diodes will not let current to flow in reverse back to AC side, as the diodes only let the AC side to pass current to charge the capacitor. That is why the capacitor discharges only to the load and stays mostly charged.

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