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CIRE2131

POWER AMPLIFIERS A power amplifier is a large signal amplifier that produces a replica of the input signal at its output. In some cases this output signal maybe inverted.

CLASSES OF AMPLIFIERS Amplifiers are categorized into different classes based on their performance. One of the major classification factors is Conduction time also called conduction angle or duty cycle. All relate the time ON to the time OFF. We will use θc to denote conduction angle. Power handling ability is related to conduction angle. This produces the basic categories of: Class A → θc = 360 ° (handles the lowest power level) θc = 180 ° (handles the medium power level) θc < 180 ° (handles the highest power level)

Class B, AB → Class C →

CLASS “A” AMPLIFIER (7-1, Page 323) • • • Are biased so that the entire input waveform is amplified without clipping. The DC bias and input signal level are set so the output signal is undistorted, unclipped. Mid-point bias is used to permit maximum output voltage swing. Maximum power occurs at Vout MAX . Operates at 100% duty cycle. i.e. the transistor is “ON” all the time and is dissipating energy in the form of heat. Is the least efficient of all the classes of amplifiers; typically < 10% efficiency with 25% the theoretical best. Also called LINEAR amplifier
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Durham College

CIRE2131

Power Efficiency of a Class “A” Amplifier Power efficiency =

Powerout Pout PLOAD P = = = L =η Powerin Pin PDC PDC

Where PDC = power supplied by the DC power source % Efficiency =

η

* 100%

Maximum Power Transfer • Happens when RL = Zout (ZL = Zout) Power Analysis • We need to evaluate the following in order to perform a power analysis for a class “A” amplifier: - PD = power in the device - PL = power in the load - PDC or PS = power of the DC source or supply •

η

= power efficiency

In order to determine these “powers” we need to know/calculate the following: - DCLL - DC Q-point (ICQ & VCEQ) - ACLL - Vout MAX

Class “A” Power Consideration PD = power in the device

PD = VCEQ * I CQ
This is a DC parameter and it doesn’t change if the AC input is present

PL = power in the load

PL =
PL =

VRL2
2

RMS

RL
VRL pp 8RL

Where VRL is the RMS value of the voltage, recall V RMS = 0.707 *

V pp 2

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CIRE2131

PDC or PS = power from the DC supply

PDC = PS = VCC * I CC ; where ICC = IBIAS+ ICQ
Class “A” Power Example Given the following circuit, let’s determine the power values for this class A amplifier

ANALYSIS: a) DCLL (DC load Line)

I CSAT =

VCC 10V = = 1.7 mA RC + RE1 + RE 2 5kΩ + 500Ω + 500Ω

VCEOFF = VCC = 10V
b) DC

VB =

R2 10 KΩ   * VCC = 10V *   = 1.7Volts R1 + R2  50 KΩ + 10 KΩ 

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CIRE2131

VE = V B − VBE = 1.7V − 0.7V = 1Volt IC = VE 1V = = 1mA RE1 + RE 2 500Ω + 500Ω

VRC = I C * RC = 1mA * 5 KΩ = 5V VCE = VCC − VRC − VE = 10V − 1V − 5V = 4V
c) ACLL (AC Load Line)

rC = RC // R L = 5 KΩ // 50 KΩ = 4.5 KΩ re' = 25mV = 25Ω 1mA

rE = re' + R E1 ≈ R E1 = 500Ω VCEQ rC + rE 4V = 1.8mA 4.5 KΩ + 500Ω

iCSAT = I CQ +

= 1mA +

VCE POff =V CEQ + I C * (rC + rE ) = 4V + 1mA(4.5 KΩ + 500Ω) = 9Volts

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CIRE2131

Vout MAX = 2(VCEQ ⇒ SAT ) = 2(4V − 1V ) = 6V
Use Vout MAX = 6V PP

or

2(VCEQ ⇒ CUTOFF ) = 2(9V − 4V ) = 10V

This is the end of the preliminary work, now the actual power calculations start. d) Power Calculations

PD = VCEQ * I C = 4V * 1mA = 4mW
2 2 VRL pp (6V p − p ) PL = = = 90 µW = 0.09mW 8RL 8 * 50 KΩ

PDC = VCC * I CQ ≈ 10V * 1mA = 10mW
The reason for this is that

I BIAS =

VCC 10V = = 0.17 mA R1 + R2 (50 + 10) KΩ

PDC = VCC * ( I CQ + I BIAS ) = 10V * (1mA + 0.17 mA) = 11.7 mW

η= or PL 0.09mW * 100% = * 100% = 0.9% PDC 10mW

η=

PL 0.09mW * 100% = * 100% = 0.77% PDC 11.7 mW

Change value of RL for maximum power and we will have:

RL = Z OUT = RC = 5 KΩ

V pp (6V p − p ) PL = RL = = 0.9mW 8RL 8 * 5 KΩ P 0.9mW * 100% = 9% η = L *100% = PDC 10mW
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