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## Q. 30.5

What is the dc load voltage and ripple in Fig. 30-13? ## Verified Solution

The rms secondary voltage is

$V_2=\frac{120 V}{5}=24 V$

The peak secondary voltage is

$V_P=\frac{24 V}{0.707}=34 V$

Assuming an ideal diode and small ripple, the dc load voltage is

$V_L = 34 V$

To calculate the ripple, we first need to get the dc load current

$I_L=\frac{V_L}{R_L} =\frac{34 V}{5 k\Omega }=6.8 mA$

Now we can use Formula (30-9)$V_R=\frac{I}{fC}$ to get

$V_R=\frac{6.8 mA}{(60 Hz)(100 \mu F)}=1.13 Vpp \approx 1.1 Vp-p$

We rounded the ripple to two significant digits because it is an approximation and cannot be accurately measured with an oscilloscope with greater precision.

Here is how to improve the answer slightly: There is about 0.7 V across a silicon diode when it is conducting. Therefore, the peak voltage across the load will be closer to 33.3 V than to 34 V. The ripple also lowers the dc voltage slightly. So the actual dc load voltage will be closer to 33 V than to 34 V. But these are minor deviations. Ideal answers are usually adequate for troubleshooting and preliminary analysis.

A final point about the circuit: The plus sign on the filter capacitor indicates a polarized capacitor, one whose plus side must be connected to the positive rectifier output. In Fig. 30-14, the plus sign on the capacitor case is correctly connected to the positive output voltage. You must look carefully at the capacitor case when you are building or troubleshooting a circuit to find out whether it is polarized or not.

Power supplies often use polarized electrolytic capacitors because this type can provide high values of capacitance in small packages. As discussed in earlier courses, electrolytic capacitors must be connected with the correct polarity to produce the oxide film. If an electrolytic capacitor is connected in opposite polarity, it becomes hot and may explode. 