Products Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 30.7

What is the dc load voltage and ripple in Fig. 30-15? Compare the answers with those in the two preceding examples. ## Verified Solution

Since the transformer is 5 : 1 step down as in the preceding example, the peak secondary voltage is still 34 V. Assuming an ideal diode and small ripple, the dc load voltage is

$V_L =34 V$

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

Now, Formula (30-9)$V_R=\frac{I}{fC}$ gives

$V_R=\frac{6.8 mA}{(120 Hz)(100 \mu F)}=0.566 Vp-p\thickapprox 0.57 Vp-p$

Because of the 1.4 V across two conducting diodes and the ripple, the actual dc load voltage will be closer to 32 V than to 34 V.

We have calculated the dc load voltage and ripple for the three different rectifiers. Here are the results:

Half-wave: 34 V and 1.13 V

Full wave: 17 V and 0.288 V

Bridge: 34 V and 0.566 V

For a given transformer, the bridge rectifier is better than the half-wave rectifier because it has less ripple, and it’s better than the full-wave rectifier because it produces twice as much output voltage. Of the three, the bridge rectifier has emerged as the most popular.