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.
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
3. The load current depends on the load resistor for pre-selected output load 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.
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.
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