Power Supply

Many electronic applications such as radio sets, toys, walkmans etc. require a d.c. ower supply (usually 6V or 3V). One way to achive it is through the use of dry cells in series. But an economically more convenient solution would be the use of the a.c. supply to generate the desired d.c. output. Power supplies are used to achieve precisely this result.

Let us first state the problem at hand. We are given a 50 Hz, 230 V r.m.s (i.e. $230\times\sqrt{2}$ V peak ) a.c. supply and our objective is to design a circuit which would take this as input and give as output a constant d.c. voltage, say 6 V.

In order to be able to use our common circuit elements (which run on small voltages), we first reduce the amplitude of the input to say 6 V through the use of a transformer.

Figure 9.1: The First Step

Figure 9.2: A transformer

The output of an ideal transformer shown above is governed by the following equations:

$$v_{in}\times{}i_{in} = v_{out\times{}i_{out}}$$

$$\frac{v_{in}}{v_{out}} = \frac{N_1}{N_2}$$

where, both the input and the output are a.c. We denote the transformer in the ciruit as following.

Figure 9.3: A transformer

The next step is to use this small a.c. voltage to generate the required d.c. The simplest manner in which this task is accomplished is by the use of a diode in a circuit known as Half wave rectifier. The figure is shown below. The working of the circuit is as follows. The diode can conduct only ion one direction, i.e. only when the voltage applied to the circuit is such that the current flows in the forward direction. Otherwise, the diode simply blockes the current. Now, when the load is simply a resistor $R_L$, the equation for the output current should be proportional to the output voltage and therefore, the output volatage is zero for half of the cycle and equal to the input voltage for the other half of the cycle.

Figure 9.4: Half wave rectifier

Figure 9.5: Output of the half wave rectifier

Now, suppose that instead of the resistor, we have an inductor as the load. In this case, the output voltage at any time will not be proportional to the current at the time. Recall that for an inductor, $v_L=L\frac{di_L}{dt}$. Therefore, the output voltage can go negative in this case as long as the current is not negative. After all, diode prevents only negative currents and not negative voltages. So, as long as the diode conducts (due to non-zero forward current), the input voltage is transferred to the output. So, barring the transients which occur due to the initial values of the voltage and current through the inductor, eventually the diode will conduct at all times with the output voltage being equal to the input and the current being phase shifted from the input by 90 degrees with a d.c. component added to make it positive valued at all times. The figures below two different cases, one in which the $v_in$ is $6\sin{\omega{}t}$ and the other in which it is $6\cos{\omega{}t}$. The initial current through the inductor is assumed to be zero in both the cases.

Figure 9.6: The inductive load
Figure: Output of the half wave rectifier when the input is $\sin{\omega{}t}$
Figure: Output of the half wave rectifier when the input is $\cos{\omega{}t}$

In case the load is purely capacitive, once the capacitor is charged to its maximum value, not forther charging takes place. Also, as the current cannot be negative the discharge also doesn't take place. The output is therefore a pure d.c.

Figure 9.9: The capacitive load