# How Choke Input Filter works

In this educational electronics tutorial, it is explained what choke input filter is, what it does, where it is used. Also it is explained how to calculate the inductor and capacitor values of an choke input filter.

The choke input filter is an L section filter that consist of a series inductor and parallel capacitor. The choke input filter is used to remove ripples of signal from a rectifier output to the load. Thus the choke filter is sits between the rectifier and the load. The rectifier can be half wave rectifier, full wave rectifier or a bridge rectifier. Below is choke input filter diagram with bridge rectifier.

Here the input is a sine wave of 120V peak and 60Hz which passes through the bridge rectifier. The output from the bridge rectifier is shown below.

When this signal passes through the choke input filter we get the following signal at the output of the filter.

The input signal into and the output signal from the choke input filter is shown below.

The following graph shows the the ac input signal, the transformer output, the rectified output and the choke input filter output on the same graph for comparison purpose.

Below circuit diagram shows the placement of the probe.

Now we show how to calculate the the dc output voltage at the rectifier output. See previous tutorial Output DC voltage and Frequency of Half-wave, Full-wave and Bridge Rectifier for other types of rectifiers.

The input signal is 60Hz, 120V rms sine wave. So the primary peak voltage is,

$$V_{p,peak} = \sqrt{2} V_{p,rms} = \sqrt{2} \times 120V = 169.7V$$

The transformer has turn ratio of N=10 and we know that,

$$N= \frac{V_{p,peak}}{V_{s,peak}} = \frac{Primary \space Peak \space Voltage}{Secondary \space Peak \space Voltage}$$

and so the secondary peak voltage is,

$$V_{s,peak} = \frac{V_{p,peak}}{N}=\frac{169.7V}{10}=16.97V$$

This secondary peak voltage goes into two paths with two diodes in each path during each half cycle. This was explained above. Therefore each path gets voltage which is half of the above calculated secondary peak voltage $$V_{s,peak} = 16.97V$$. Therefore, the voltage at each path is,

$$V_{sp(in)} = \frac{V_{s,peak}}{2} = \frac{16.97V}{2}=8.48V$$

Now each path has voltage drop across two diodes and each diode has 0.7V drop hence the peak voltage at the end of the two diodes is,

$$V_{sp(out)} = V_{sp(in)} - 0.7V - 0.7V = 8.48V - 1.4V=7.08V$$

The DC value of bridge rectifier is then,

$$V_{dc} = \frac{2 V_{sp(out)}}{\pi}=\frac{2 \times 7.08V}{3.14}=4.5V$$

Thus the dc value at the bridge rectifier output is 4.5V. This signal waveform is shown in the following graph.

Once we have the bridge rectifier output the next is to calculate the capacitor and the inductor values.

The rule of choosing inductor and capacitor value is to make the capacitive reactance($$X_C$$) much lower than the load resistance($$R_L$$) and the inductance reactance($$X_L$$). That is,

$$X_C << R_L$$ and $$X_C >> X_L$$

Let choose load resistance of $$R_L=1K \Omega$$ and so let choose capacitance reactance $$X_C=100 \Omega$$. Then we can calculate the capacitor value,

$$X_C= \frac{1}{2 \pi f C}$$

The frequency at the bridge rectifier is twice then the input frequency so,

$$f_{out}=2 f_{in}= 2 \times 60Hz = 120Hz$$

or, $$C= \frac{1}{2 \pi f X_C} = \frac{1}{2 \times 3.14 \times 120Hz \times 100 \Omega} = 13.26uF$$

Similarly let the inductance reactance $$X_L=1 K \Omega$$ because $$X_C=100 \Omega$$, then,

$$X_L= 2 \pi f L$$ and therefore,

$$L= \frac{X_L}{2 \pi f } = \frac{1K \Omega}{2 \times 3.14 \times 120Hz} = 1.32H$$

The choke input filter ac output voltage can be calculated as follows.

$$V_{out} \approxeq V_{in} \frac{X_C}{X_L} = 7.08V \frac{100 \Omega}{1k \Omega} = 0.7V$$

That the frequency at the output of the choke input filter is 120Hz is shown below.

Watch the following video of how the choke input filter works with the bridge rectifier.

So here we have illustrated with calculation example of how the choke input filter works. Choke filters are used after rectifier in a power supply electronics to remove the ac signal ripples. This is achieved because of the inductor. The inductor is opposes ac signal changes through it and thereby allows dc signal to pass but prevents signal to pass through it.

The disadvantage of choke input filter is that it is bulky because of the inductor required for the power supply unit in electronics circuits. The power line frequency is 60Hz or 50Hz and for this large inductor is required such as the 1.32H inductor used in the above example. Larger inductor will have larger dc resistance and therefore larger voltage drop which can cause design issues. Today electronics circuits are geared towards low voltage, light electronics parts and so the choke input filter is not used as much as it was used in earlier days. These days the capacitor input filter is used for filtering instead of choke input filter or the inductor input filter.