Push pull class b amplifier

RaghaviSwaminathan2 772 views 21 slides Mar 13, 2022
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About This Presentation

This is a ppt presentation on PUSH PULL CLASS B AMPLIFIER.
The circuit of a push-pull class B power amplifier consists of two identical transistors T1 and T2 whose bases are connected to the secondary of the center-tapped input transformer Tr1. The emitters are shorted and the collectors are given t...


Slide Content

PUSH PULL CLASS B AMPLIFIER BY, RAGHAVI.S ELECTRONICS AND COMMUNICATION ENGINEERING

ANALYSIS OF CLASS B AMPLIFIER INPUT OUTPUT TO GET FULL CYCLE – A PAIR OF TRANSISTORS IS USED THE 2 TRANSISTORS ARE IDENTICAL IN CHARACTERISTICS ARE CALLED MATCHED TRANSISTORS

2 CATEGORIES DEPENDING UPON THE TYPES OF 2 TRANSISTORS PUSH PULL CLASS B When both transistors are n-p-n (or) p-n-p either 2. COMPLEMENTARY SYMMETRY CLASS B When the transistors form a complementary pair ( i.e )one n-p-n and other p-n-p

PUSH PULL CLASS B AMPLIFIER - CIRCUIT DRIVER TRANSFORMER OUTPUT TRANSFORMER

OPERATION FOR POSITIVE HALF CYCLE: Ib2=0 Ic2=0 I/P O/P

FOR NEGATIVE HALF CYCLE I/P O/P Ib1=0 Ic1=0

OUTPUT POSITIVE HALF CYCLE 180* OUT OF PHASE NEGATIVE HALF CYCLE +

IMPORTANT CONCEPT IN OUTPUT TRANSFORMER : THE NUMBER OF TURNS EACH HALF OF THE PRIMARY = N1 SECONDARY = N2 HENCE TOTAL NUMBER OF PRIMARY TURNS = 2N1 RATIO: 2N1:N2

DC OPERATION Q point is adjusted on the X – axis such that V CEQ = VCC I CEQ = 0 Coordinates of the Q point (VCC,0) NO DC BIAS VOLTAGE

DC POWER INPUT Each transistor output is in the form of HALF RECTIFIED WAVEFORM. Im => Peak value of the output current of each transistor Due to HALF RECTIFIED, the DC VALUE (OR) AVERAGE VALUE is Im / Ꮧ

AC OPERATION WHEN AC SIGNAL APPLIED POSITIVE HALF CYCLE: Q1 conducts Lower half of the output transformer does not carry any current Only N1 no of turns carry the current . NEGATIVE HALF CYCLE: Q2 conducts Upper half of the output transformer does not carry any current  Only N1 no of turns carry the current. X X REFELECTED LOAD ON THE PRIMARY, R’ L = R L /n 2 , where n=N2/N1

SLOPE OF THE AC LOAD LINE IS -1/R’ L THE SLOPE IN TERMS OF Vm AND Im AS, 1/R’ L = Im / Vm ( i.e ) R’ L = Vm / Im

AC POWER OUTPUT Vrms = R’ L Irms R’ L = Vm / Im Irms = Vrms / R’ L

EFFICIENCY MAXIMUM EFFICIENCY

IMPORTANT CONCEPT MAXIMUM POSSIBLE THEORETICAL EFFICIENCY IN CASE OF PUSH PULL CLASS B AMPLIFIER IS > THE TRANSFORMER COUPLED CLASS A AMPLIFIER. FOR PRACTICAL CIRCUITS IT IS UPTO 65 TO 75% UNDER MAXIMUM EFFICIENCY CONDITION: 78.5 Vm = VCC – V CE(MIN)

POWER DISSIPATION MAXIMUM POWER DISSIPATION differentiating w.r.t vm

P d (max) per transistor,

HARMONIC DISTORTION

ADVANTAGES The efficiency is much higher than the class A operation. When there is no input signal, the power dissipation is 0 The even harmonics get cancelled. This reduces the harmonic distortion. DISADVANTAGES Two centre tap transformers are necessary The transformer , make the circuit bulky and hence costlier. Frequency response is poor.

THANK YOU ALL
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