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close this bookRadio and Electronics (DED Philippinen, 66 p.)
close this folder10. BLOCKS OF RADIOS / -1- / POWER SUPPLIES
close this folder10.4. SMOOTHING AND FILTER CIRCUITS
View the document10.4.1. THE RESERVOIR CAPACITOR
View the document10.4.2. FILTER CIRCUITS


To obtain a still steadier voltage we need a circuit which will allow the passage of dc current and avoid the flow of ac current through the load.

fig. 96

To achieve this aim there are two possible methods:

A- To drop the ac-voltage component at a part of the circuit - this means at a component connected in series to the load (see in fig. 96a)

fig. 96a

B- To bypass the ac-component of the current via a component connected in parallel to the load

Method A requires something with a low resistance for dc and a high one for ac-components. Fitting therefore is obviously an inductor.

Method B requires something with a high resistance for dc and a low one for ac components. Fitting therefore is obviously a capacitance.

The best smoothing can be achieved, if both methods are used in the same smoothing circuit. Therefore in power supplies for very high quality you will find so called LC-smoothing circuits containing inductors and capacitors.

But inductors (often called “chockes”) are bulky, heavy and expensive. Therefore they are in most of the cases replaced by a resistor and form then together with the capacitor so called RC-smoothing circuits.

But resistors do not have an increased impedance for ac-components (compared with their resistance for dc).

fig. 97

This fact makes the RC-smoothing circuits less effective than the LC ones - But in practice you will mostly find the RC ones.


1. Supposed the receiver represented here by the load resistor R draws at 10 Volts a current of 30 mA. That means, the load resistor represents


2. Supposed resistor is 100 Ohms. Then the voltage drop across this resister will be:

VR1=R1 × IL = R1 × I = 30 mA × 100 W = 3V


Thus the supply voltage in this case must be = Vs = VR1 + VL = 3 V + 10 V = 13 V

3. Supposed the ripple-voltage across the reservoir capacitor of the powersupply is 2 Volts and assuming, a full-wave rectification.

That means a ripplefrequency of 100Hz. Supposed the reservoir capacitor has a capacity of 500 µF then its reactance is:


The ripple voltage can be calculated according to the voltage division at Csmooth and Rsmooth


Thus the ripple voltage has been reduced from 2V at the input to 0.064 V at the output of the smoothing circuit.

This means, it has been reduced for 97%