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. 7.6

Finding the Values of L and C for a Series-Loaded Resonant Inverter to Yield Specific Output Power

A series resonance inverter in Figure 7.8a with series loaded delivers a load power of $P_L = 1 kW$ at resonance. The load resistance is $R =10 \Omega$. The resonant frequency is $f_0 = 20 kHz$. Determine (a) the dc input voltage $V_s$, (b) the quality factor $Q_s$ if it is required to reduce the load power to $250 W$ by frequency control so that $u = 0.8$, (c) the inductor L, and (d) the capacitor C. ## Verified Solution

a. Because at resonance $u = 1$ and $|G(j\omega )|_{max}=1$, the peak fundamental load voltage is $V_p=V_{i(pk)}=4V_s/\pi$.

$P_L=\frac{V_p^2}{2R} =\frac{4^2V_s^2}{2R\pi ^2}$       or              $1000=\frac{4^2V_s^2}{2\pi ^2\times 10}$

which gives $V_s = 110 V$.

b. To reduce the load power by $(1000/250 = ) 4$, the voltage gain must be reduced by 2 at $u = 0.8$. That is, from Eq. (7.35),

$|G(j\omega )|=\frac{1}{[1+Q_s^2(u-1/u)^2]^{1/2}}$                                      (7.35)

we get $1 + Q_s^2(u – 1/u)^2 = 2^2$, which gives $Q_s = 3.85$.

c. $Q_s$ is defined by

$Q_s=\frac{\omega _0L}{R}$

or

$3.85=\frac{2\pi \times 20 kHz\times L}{10}$

which gives $L = 306.37 \mu H$.

d. $f_0=1/2\pi \sqrt{LC}or 20 kHz -1/[2\pi \sqrt{306.37 \mu H\times C} ]$ which gives $C = 0.2067 μF$.