# How Capacitor Input Filter works with Half Wave Rectifier

Capacitor input filter are used to filter the rectifier output in power supply circuit to get DC signal from AC signal. Another type of power supply filter that could be used is the choke input filter but they are bulky for line power operation. The capacitor input filter is alternative to the choke input filter. Here the capacitor input filter works with half wave rectifier is explained. Also explained is how to calculate the capacitor value of the filter and the load resistor value.

### Capacitor input filter with Half Wave Rectifier

The following shows capacitor input filter(C1) after the half wave rectifier.

The input to the transformer is an ac signal of 120Vrms(peak of 169.7V) and frequency of 60Hz. The transformer has turn ratio of N=10 and so the secondary rms voltage is,

$$V_{rms,s} = \frac{V_{rms,s}}{N}$$

so, $$V_{rms,s} = \frac{120V}{10}=12V$$

And therefore the secondary peak voltage is,

$$V_{peak,s} = \sqrt{2}V_{rms,s} = \sqrt{2} \times 12V=16.97V$$

This voltage passes into the diode and the output signal peak out of the diode is,

$$V_{peak,s(out)} = V_{peak,s(in)} - V_D = 16.97V - 0.7V = 16.27V$$

where the $$V_D = 0.7V$$ is the voltage drop across the diode(D1).

This peak voltage is also the load voltage, $$V_L$$,

$$V_L = V_{peak,s(out)} = 16.27V$$

Then we can calculate the DC load current,

$$I_L = \frac{V_L}{R_L} = \frac{16.27V}{10k \Omega} = 1.62mA$$

Now we can calculate the capacitor input filter ripple voltage, which is peak to peak voltage. The ripple formula is,

$$V_r = \frac{I_L}{f C}$$

where, $$I_L=1.62mA$$ is the dc load current, f=60Hz the frequency of the signal and $$C = C1 = 10\mu F$$ is the capacitor input filter capacitance.

So, $$V_r = \frac{1.62mA}{60Hz \times 10\mu F} = 2.7V$$

So the ripple voltage at the load resistor is 2.7V peak to peak. The following graph shows the secondary peak voltage $$V_{peak,s(in)} = 16.97V$$ before the diode and the peak voltage $$V_L = V_{peak,s(out)} = 16.27V$$ and the ripple voltage of 2.7V.

How to choose the capacitor value for capacitor input filter?

One might ask how to choose the value for the capacitor when using capacitor input filter. For this first you must know what voltage and current you want at the load resistor.

1. What is the rms voltage you want at the load, say 12V

2. Calculate the peak voltage for the given rms voltage.

For 12V rms voltage at the output, the secondary peak voltage is 16.97V and we must subtract diode voltage(0.7V for silicon diode) to get the secondary peak voltage at the load which is $$V_L=16.27V$$.

3. Calculate the load resistor value $$R_L$$ for your required dc load current.

Say we need load current of $$I_L=1.62mA$$, then calculate the resistor value,

$$R_L = \frac{V_L}{I_L}=\frac{16.27V}{1.62mA}=10 k\Omega$$

The load resistor $$R_L$$ sets up the load current

4. Calculate the capacitor value using the ripple formula

$$C= \frac{I_L}{f V_r }$$

Suppose we would like the ripple voltage $$Vr$$ to be 2.7V then we can calculate the capacitor value as follows,

$$C= \frac{1.62mA}{60Hz \times 2.7V}=100\mu F$$

So in this way you can calculate the value of capacitor for the capacitor input filter and the load resistor value.

Some notes:

1. Increasing the capacitor value will reduce ripple voltage

2. Increasing the load resistor value will reduce ripple voltage

The following video shows simulation in proteus softwarae of how the capacitor input filter works with half wave rectifier.

As can be seen in the graph provided above, the output of the capacitor input filter still contains ripples. To get steady DC voltage a voltage regulator is added after the capacitor. This is illustrated in the next tutorial Power supply Design-Halfwave rectifier, capacitor filter, Zener voltage regulator.