LM 317: Regulator

The trouble with zener diode driven power supply is that one needs a zener of the same voltage as the desired voltage output. We can overcome this using a voltage regulator such as LM 317

Figure 9.21: A power supply using voltage regulator

This regulator maintains a constant voltage $V_ref$of 1.25 volts across two of its terminals. So, connecting it in the above configuration and neglecting the $I_adj$, we have when $V_in>V_0$, $(V_0R_1)/(R_1+R_2)=1.25$. For $V_0=15$ V, if we fix $R_1$ to be $1K\Omega$, we get $R_2$ to be $11.0K\Omega$.

The next thing we need is the value of $C$ required. Suppose the load is such that $I_L=200$ mA. The current thorugh $R_1$ and $R_2$ is $\frac{15}{12}=1.25$mA. Therefore, $I=201.25$mA. For no ripple, the capacitor should be able to supply this current without the voltage dropping below 15 V. Now, taking the worst possible instant of time, we have,

$$\frac{V_p-15}{1/2f} = \frac{i}{C}$$

$$V_p = \frac{i}{2fC}+15$$

$$\mathrm{For C}=10\mu F$$

$$V_p = 216\mathrm{volts}$$

On the other hand, for a 1mF capacitor, $V_p=201.25/100+15 = 17.0$ volts.