Half-Wave Single-Sided Supply

A typical application for diodes is in the conversion of an AC signal into a constant DC signal which can then be used to power electrical circuits. To make such a circuit we'll use the 24V transformers provided. These transformers take the wall signal of approx 120Vrms and drop it down to 24Vrms at a frequency of 60Hz. By attaching a rectifier we can transform the AC signal so that it only passes load current in one direction. Finally, if we filter the signal out of the rectifier, we can obtain a fairly constant DC signal that can be used as a power source for electronic circuits. Power circuits similar to these are found in virtually all electronic equipment, including computers, stereos and televisions.

To understand how the filter works, consider the circuit shown in Fig. 1.3.

  

Figure 1.3: DC Power Supply without Load

As the signal at V_o increases, the capacitor charges up virtually instantaneously since there is very little resistance in the circuit. After the waveform reaches its peak value and starts to decrease, the voltage on the capacitor is higher that the voltage going into the rectifier. Thus, the diodes will be reverse biased, and no current will flow out of capacitor so it cannot discharge. Thus, the capacitor will be pinned at the peak voltage of the AC signal. Now, if we attach a load resistor R_L across the capacitor, as shown in Fig. 1.4, the capacitor will discharge through R_L with time constant $\tau =R_LC$.

  

Figure 1.4: DC Power Supply with Load R_L

If we make C very large so that R_LC is much larger than the period of the AC signal we're rectifying, then the capacitor will not have very much time to discharge, and the output voltage will be almost constant with only a small AC ripple. To approximately determine the amount of AC ripple, we can start from the definition of capacitance

$$V=\frac{Q}{C}$$ (3)

Differentiating both sides with respect to time gives

$$\frac{dV}{dt}=\frac{1}{C}\frac{dQ}{dt}=\frac{I}{C}$$ (4)

Multiplying out by the small time interval $\Delta t$ yields to first order:

$$\Delta V = \frac{I}{C}\Delta t$$ (5)

If we take $\Delta t$ to be approximately the period of a cycle of signal we are rectifying, then $\Delta t \approx \frac{1}{f}$, and the ripple voltage is proportional to the current through the load and inversely proportional to the frequency of the source voltage

$$\Delta V = \frac{I}{fC} = AC~ripple$$ (6)

Finally, if the current is used to drive a load R_L, then the ripple voltage can be approximated by

$$\Delta V = \frac{V_{avg}}{R_LfC} = AC~ripple$$ (7)

where V_avg is the time average voltage delivered to the load. Clearly, from a practical point of view, the more current that is required by the load, the greater the ripple and the less constant the DC voltage and the less ideal the DC power supply.