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Explain working of Class B power amplifier. Derive expression for efficiency. OR Explain power amplifier in brief.

Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits

Marks: 10M

Year: Dec 2016, May 2016

1 Answer
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Please refer below question for working of Class B power amplifier: Working of Class B PA.

Explain working of Class B push-pull power amplifier.What is cross over distortion?OR Explain Class B power amplifier and methods to remove cross over distortion.OR Short note on cross over distortion. (Till Waveform of Class B PA.)

Derivation of expression for efficiency of Class B Power Amplifier:

D.C. load can be obtained from below equation:

$Vcc - Icq - Rcoil - Vce = 0$

where Rcoil is D.C. resistance of primary $Rcoil \approx 0$

Hence $Vce = Vcc$ is a d.c. load line parallel to y-axis.

enter image description here

when ac input is applied, maximum output voltage equals to Vcc can be obtained. i.e. $Vop \approx Vcc $

$iop = \frac{Vop}{R_L'} = \frac{Vcc}{R_L'}$

ac load line can be drawn from two points first point on X-axis with $Vce = Vcc = Vop$ and second point on Y-axis with $i_C = iop = \frac{Vcc}{R_L'}$ as shown below:

enter image description here

$Poac = Vorms \times iorms$

$Poac = \frac{Vop}{\sqrt2} \times \frac{iop}{\sqrt2} = \frac {Vop \times iop}{2} = Vcc \times \frac{Vcc}{R_L'} \times \frac{1}{2}$ = $\frac{Vcc^2}{R_L'}$

$Poac = \frac{Vcc^2}{R_L'}$

$Pindc = Vcc \times Idc$

$Idc$ can be obtained from waveform of $i_c$. Nature of current i is same as full wave rectifier whose d.c component i.e. $Idc$ $= \frac{2I_m}{\pi}$

For Class B efficiency is > 50% but less than <78.5%

$Pindc = Vcc \times \frac{2I_m}{\pi} where I_m = iop = \frac{Vcc}{R_L'}$

$Pindc = Vcc \times \frac{Vcc}{R_L'} \times \frac{2}{\pi}$

$Pindc = \frac{2Vcc^2}{\pi R_L'}$

Efficiency =$ \eta = \frac{Poac}{Pindc} \times 100 = \frac{Vcc^2}{2R_L'} \times \frac{\pi R_L'}{2Vcc^2} \times 100 = \frac{\pi}{4}$ $\times 100$

$\eta = 78.5\%$

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