Cover Image
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

10.4.2. 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.

CLOSER LOOK TO THE SMOOTHING FUNCTION OF A SMOOTHING CIRCUIT

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


figure

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


figure

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:


figure

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


figure

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%