# Power supply Design-Halfwave rectifier, capacitor filter, Zener voltage regulator

Here it is illustrated how to design DC power supply with half-wave rectifier, capacitor input filter and Zener diode based voltage regulator. This is a detailed power supply design calculation with Zener diode regulator. Almost all electronics devices operate on dc power supply which can range from 3.3V to 20V. So this can be useful in ones basic electronics projects.

The block diagram of a DC power supply is shown below.

In the above block diagram, the high voltage represents AC main line voltage which here we will assume to be a 220V rms and 50Hz. The step down transformer lowers the high voltage 220 to 14V RMS. For this a step down transformer with turn ratio of 15 is used. The output of the step down transformer is thus low voltage ac signal of rms voltage around 14V. Next is the rectifier which can be either a half wave, full wave or a bridge rectifier. In this tutorial the half wave rectifier will be used. The reservoir or smoothing filter converts the ac signal into unsmoothed dc signal or pulsating dc signal. It is a dc signal in the sense that it does not change polarity. The filter is also called capacitor input filter. The capacitor input smoothing filter converts the unsmoothed dc signal to smoothed dc signal with some ripple. The smoothed dc signal is then converted to regulated dc signal using the voltage regulator. The voltage regulator can be made of discrete components like Zener diode and transistors or a dedicated voltage regulator IC(Integrated Circuit) can be used. Here we will use a simple Zener diode based voltage regulator.

The following shows the circuit diagram of a complete DC power supply using half-wave rectifier, capacitor input filter and a 1N5349 Zener diode based voltage regulator.

The RMS voltage at the secondary winding of the step down transformer is,

$$V_{rms,s} = \frac{V_{rms,p}}{15}$$

that is, $$V_{rms,s} = \frac{220V}{15}=14.66V$$

And therefore the peak voltage at the secondary winding is,

$$V_{peak,s} = \sqrt{2}V_{rms,s} = \sqrt{2} \times 14.66V = 20.74V$$

This is the input to the 1N4001 diode which has a voltage drop of 0.7V. So the peak output from the diode is,

$$V_{peak,s,out} = V_{peak,s,in} -0.7V= 20.74V -0.7V = 20.04V$$

This is the voltage that appears at the output of the diode. The following shows the graph of input and output signal of the diode rectifier.

The ripple voltage is given by the following equation,

$$V_R=\frac{I_s}{fC}$$

where, $$I_s$$ is the series current that passes through the current limiting resistor R1, f is the ripple frequency which is 50Hz and C is the capacitor filter which is 10uF in this example.

The series current $$I_s$$ is sum of the Zener diode current and the load current.

$$I_s = I_z+I_L$$

The load current is given by,

$$I_L = \frac{V_L}{R_L}$$

The load voltage $$V_L$$ is the same as the zener voltage $$V_Z$$ which is 12V and the load resistor value is $$R_L=10K\Omega$$. So the load current is,

$$I_L = \frac{12V}{10K\Omega}=1.2mA$$

So the power supply design procedure are as follows.

Say output load voltage is 12V and output load current is 1.2mA

2. From the required output voltage choose Zener diode that has the required reverse breakdown voltage

Since we want the output load voltage is 12V the Zener diode 1N5349. This zener diode has maximum power of 5W.

3. Choose load resistor that has current of 1.2mA and load voltage is 12V

$$R_L = \frac{12V}{1.2mA}=10K\Omega$$

4. Choose stepdown transformer that produces secondary voltage greater than 12V rms.

5. Choose the current limiting resistor such that the zener diode is operating in breakdown region.

The zener diode will be in breakdown region if the Thevenin voltage is greater than the Zener voltage.

The Thevenin voltage is,

$$V_{TH} = \frac{R_L}{R_1+R_L}V_{peak,s,out}$$

or,  $$V_{TH} = \frac{10K\Omega}{1K\Omega+10K\Omega}20V = 18V$$

Since the Thevenin voltage(18V) is greater than the Zener voltage(12V) the Zener diode is in breakdown voltage.

So in this way one can design DC power supply using half-wave rectifier, capacitor input filter and 1N5349 Zener diode voltage regulator. Here we have used simple capacitor input filter but one can also use other types of filter like Choke Input Filter.