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Power amplifier designing

Solved problems

Class B amplifier

Do not write text included in square bracket

Power transmitted to load, PL' = PL/nt
[nt = efficiency of transformer]

[If nt is not given assume nT = 90% or 0.9]

Q = (Pq max)/PL') = 2

[calculate Pq max]

Select transistor with Pd > 2 * Pq max

[Usually select ECN 149]

[calculate Vre]

Vceq = Vcc - Vre

[calculate Vceq]

Vce peak = Vceq - Vce sat

[calculate Vce peak]

Ic peak = (2 * PL')/Vce peak

[calculate Ic peak]

Icq = Ic peak + Icmin

Assume Ic min = 0

[calculate Icq]

[calculate Re]

Pre = sqr(Vre)/Re

[Select power rating of Re > (2 * Pre). Specify power rating as multiple
of 0.25 W]

Ce = 1/(2 * pi * FL * RL)

[If FL is not given assume FL = 50 Hz]

Since Ce is very high we leave Re unbypassed

s = (1 + hfe max)/(1 + ((hfe max * Re )/(Rb + Re))

Find Rb[Do not standardise]

Vr2 = Vbe + (Icq * Re)

Vr1 = Vcc - Vr2

Assume Vbe = 0.6V [for Si, 0.3 for Ge if not specified]

R1/R2 = Vr1/Vr2 .............(A)

[Get R1 in terms of R2 & substitute in Rb]

Rb = R1 parallel R2 = (R1 * R2)/(R1 + R2)

Find R2

Select lower standard value to make circuit indepent of beta

Substitute in (A) to find R1

Select higher standard value so that circuit draws minimum current
from supply

[calculate RL']

RL' = (sqr(N1/N2)) * RL

[calculate (N1/N2)]

Select audio frequency transformer with turns ratio 1:(N1/N2)

[Draw the circuit diagram with calculated values]

PL' FL = (Vce peak * Ic peak)/2

Pi dc = (Vcc * Icq) + (Vcc ^ 2)/(R1 + R2)

Half load efficiency, __n HL = (PL' HL)/(Pi dc)__

PL' HL = PL' FL / 2

Max power dissipation = power dissipation at no signal = Pd max of transistor

[If nt is not given assume nT = 90% or 0.9]

Q = (Pq max)/PL') = 1/5

[calculate Pq max]

Select transistor with Pd > 2 * Pq max

[Usually select ECN 149]

(Vceo/2) <= Vcc <= Vceo

[select Vcc = 25 V for ECN 149]

Vre = Vcc/10

[calculate Vre]

Vceq = Vcc - Vre

[calculate Vceq]

Vce peak = Vceq - Vce sat

[calculate Vce peak]

Ic peak = (2 * PL')/Vce peak

[calculate Ic peak]

Icq = Ic peak + Icmin

Assume Ic min = 0

[calculate Icq]

Idc full wave = (2 * Idc peak)/pi

Idc half wave = Idc peak/pi

[Calculate Re. Select nearest std value]

Pre = sqr(Vre)/Re

[Select power rating of Re > (2 * Pre). Specify power rating as multiple
of 0.25 W]

Ce = 1/(2 * pi * FL * RL)

[If FL is not given assume FL = 50 Hz. Select higher std value]

s = (1 + hfe max)/(1 + ((hfe max * Re )/(Rb + Re))

Find Rb[Do not standardise]

Vr2 = Vbe + (Idc half wave * Re)

Vr1 = Vcc - Vr2

Assume Vbe = 0.6V [for Si]

R1/R2 = Vr1/Vr2 .............(A)

[Get R1 in terms of R2 & substitute in Rb]

Rb = R1 parallel R2 = (R1 * R2)/(R1 + R2)

Find R2

Select lower standard value to make circuit indepent of beta

Substitute in (A) to find R1

Select higher standard value so that circuit draws minimum current
from supply

[calculate RL']

RL' = (sqr(N1/N2)) * RL

[calculate (N1/N2)]

Power rating of primary > PL'

Select audio frequency transformer with turns ratio 1:(N1/N2)

[Draw the circuit diagram with calculated values]

PL' FL = (Vce peak * Ic peak)/2

Pi dc FL = (Vcc * Idc FW) + (Vcc ^ 2)/(R1 + R2)

__Half load efficiency, n HL = (PL' HL)/(Pi dc HW)__

Pi dc HW = (Vcc * Idc HW) + (Vcc ^ 2)/(R1 + R2)

PL' HL = PL' FL / 2

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Standard values