What is Class AB Amplifier : Working & Its Applications

An electronic device that can increase the current, voltage, or power of the given input signal is known as an amplifier. According to the circuit application, the amplifiers are divided into voltage and power amplifiers. The important factors of small-signal and large-signal (power) amplifiers are linearity, amplification, efficiency, the maximum power that a circuit can handle, impedance matching, and capacity of handling high voltage and current levels. The variation of the output signal over positive and negative cycles of the given input signal depends on the class of operation or mode of operation called amplifier classes. There are four types of amplifier classes, they are Class A, Class B, Class AB, and Class C. This article gives a brief description of the class AB amplifier, circuit diagram, working, applications, advantages, and disadvantages.


What is a Class AB Amplifier?

An amplifier that is used to overcome the cross-over distortion in the class B amplifier is known as the class AB amplifier. The combination of class A amplifier and class B amplifier gives the class AB amplifier. It is designed to overcome and eliminate the disadvantage of low efficiency in class A amplifiers and distortion in class B amplifiers and utilizes the advantages of both the amplifier classes. The circuit diagram of the class AB amplifier is shown in the figure below.

Class AB Amplifier Circuit Diagram
Class AB Amplifier Circuit Diagram

Construction/Design of Class AB amplifier

The best method to design an amplifier with high efficiency and low distortion is the class AB amplifier. It is the combination of class B configuration and class A configuration, which results in the class AB amplifier circuit. Hence the output stage of this amplifier circuit is combined with the advantages of class A and class B amplifiers and also reduces the issues of distortion and low efficiency.

Here, the two transistors T1 and T2 are considered for the circuit design. The transistor T1 is NPN type and the transistor T2 is the PNP type. The two forward-biased diodes D1 and D2 are connected in series to control the variations of VBE (emitter-base voltage) due to the temperature change as shown in the circuit diagram. The resistor R1 is connected in series with the D1 and the resistor R2 is connected in series with the D2.

Working

This type of amplifier circuit operates even for small power outputs. Because the class A amplifier operates for small current outputs and the class B amplifier operates for high current outputs. This is achieved by pre-biasing of the 2 transistors in the output stage of the amplifier circuit. Based on the pre-biasing and output current, each transistor conducts between 180° and 360° with respect to time. Therefore, the output stage of the amplifier works as a class AB amplifier.

Consider the T1 is the NPN transistor and T2 is the PNP transistor for this amplifier configuration, to obtain the complete output signal with the combination of positive and negative half cycles of the given input signal. In this configuration, crossover distortion can be eliminated due to the conduction of the input signal simultaneously by both T1 and T2 transistors in the transition period.

Therefore in the absence of an input AC signal, the transistor can conduct. By applying the small bias voltage using the diodes D1 and D2, the operating point is observed to be above the cut-off region. Here the amplitude and phase of the output signal are the same. Therefore the class AB amplifiers are also known as linear amplifiers.

Biasing Methods

There are different biasing methods for class ab amplifiers to create voltage intervals based on the simultaneous conduction of complementary transistors used in this configuration. The possible biasing methods with advantages and disadvantages are discussed below.

Voltage Biasing

It is a direct, independent, and easy method done with the voltage supply of a DC generator or batteries. But it is not implemented for real circuits due to the packaging, and high cost.

Voltage Divider Network

Here the voltage drop across the bases of the transistors will be 1.2V to 1.4V (across the R1+R2 total resistance). This value is related to 2VBE, where VBE is the transistor’s threshold voltage. The transistors T1 and T2 go above the cut-off region due to this voltage drop and therefore result in simultaneous conduction of a small portion of the input signal.

This method is not widely used and it is applicable only for specific push-pull amplifier configurations with certain resistor values and complementary transistors.

Resistor Biasing

The difference Between voltage biasing and resistor biasing is, the adjustable resistance (or) potentiometer is kept in between the two networks. The main advantage of this method is, the biasing of the complementary transistors (even with different electrical properties) is achieved with a controlling of resistance.

The inappropriate working of this configuration can result in temperature change due to the biasing above the cut-off region. Therefore another method called diode biasing is preferred.

Diode Biasing

The diodes can generate 0.7V of constant voltage, above a certain current value. This property is utilized to provide a 1.4V constant potential between the bases of T1 and T2 transistors. The advantage of this method is the self-adjustment of the voltage drop across the diodes D1 and D2 at any temperature change.

The increase in temperature can decrease the threshold voltage of diodes and decrease in biasing of T1 and T2 transistors, resulting in the limited thermal runaway.

Operating Characteristics

The class AB amplifier provides linearity with high efficiency and no cross-over distortion. In this configuration, the biasing methods are chosen based on the operating point, which includes the intermediate conduction angle of the class A and B amplifiers.

The position of this operating point is based on the degree of linearity and efficiency. It works as a class A amplifier if its operating point is near the class A amplifier operating point. In this case, it provides high linearity with less efficiency.

The operating characteristics of class AB amplifier configuration are shown in the figure below. The conduction angle of the output signal of this amplifier is greater than 180° and less than 360°. The location of the Q-point is noted above the cut-off region. Therefore in the absence of an input AC signal, the transistor can conduct.

By applying the small bias voltage using the diodes D1 and D2, the operating point is observed to be above the cut-off region. Here the amplitude and phase of the output signal are the same. Therefore the class AB amplifiers are also known as linear amplifiers.

Operating characteristics
Operating characteristics

Consider the T1 is the NPN transistor and T2 is the PNP transistor for this amplifier configuration, to obtain the complete output signal with the combination of positive and negative half cycles of the given input signal. In this configuration, crossover distortion can be eliminated due to the conduction of input signal simultaneously by both T1 and T2 transistors

The efficiency of class AB amplifier is given as,

Ƞ = π/4. Vac/V supply

Where Vac is the AC fluctuation of the output signal. The maximum efficiency depends on the maximum value of Vac and the location of the operating point.

If the biasing of class AB amplifier is limited to the cut-off point, Vacmax = V supply, then Ƞmax = π/4 = 78.5%

If the biasing of class AB amplifier is limited to the class A amplifier operating point, Vacmax = Vsupply/2, then Ƞmax= π/8=39.3%

If the operating point is close to the class B amplifier (conduction angle of the signal is between 180° to 270°), then Ƞmax ranges between 58.9% to 78.5%

To eliminate the cross-over distortion virtually, a small quiescent bias is applied on transistors T1 and T2. If the T1 and T2 are matched transistors, the emitter-base junction of each is biased with VBB/2, and when input Vi=0, then results in Vo=0. Therefore the equation for quiescent collector currents are given as,

Icn = Icp = Is. e^(VBB/2.Vt)

When Vi increases, the base voltage of T1 increases, and V0 increases. Here the transistor T2 functions as the emitter-follower and supplies current to RL. Therefore the equation for output voltage is given as,

V0 = Vi +VBB/2 – VBEn.

The collector current of T1 is given as

Icn = Il + Icp. (Where base currents are neglected)

Since Icn should supply the load current. As VBEn increases, VBEp decreases because of constant VBB. The VBEn decrease results in an Icp decrease.

When Vi is negative, the base voltage of T2 is decreased and results in a decrease in Vo. Here T2 functions as an emitter-follower and sinks the load current.

With the increase in Icp, VBEp increases causing a decrease in Icn and VBEn.

Class AB Amplifier using MOSFET

The class AB amplifier can be designed by using MOSFET to generate an output power of 100W to drive the 8 ohms load. This type of power amplifier is designed due to its high efficiency, less harmonic, and Cross-over distortion. The circuit diagram of a class AB amplifier using MOSFET is shown in the figure below.

Class AB Amplifier Using MOSFET
Class AB Amplifier Using MOSFET

The required components to construct the circuit are given below.

  • PNP transistors -BC556 (Q1, Q2)
  • NPN transistors – MJE340 (Q3, Q4)
  • PNP transistors – MJE350 (Q5, Q6)
  • N- channel E-MOSFET – IRF530 (Q7)
  • P- channel E-MOSFET – IRF9530 (Q8)
  • Supply voltage +/- 50 Volts (V1, V2)
  • 4kohms -(R1, R4)
  • 100ohms -(R2)
  • 50kohms -(R3)
  • 1kohms -(R5)
  • 50kohms -(R6)
  • 10kohms -(R7)
  • 100ohms -(R8, R9)
  • 470ohms -(R10)
  • 100ohms -(R11)
  • 3kohms -(R12)
  • 0.33ohms -(R14, R15)
  • 10microFarads -(C1)
  • 18pF -(C2, C3)
  • 100nF -(C4)

The circuit works based on the multi-stage amplification of power principle, which consists of drivers, pre-amplifiers, and power amplification using MOSFET. The process of pre-amplification is done using a differential amplifier, driver stage with the current load, and power amplification using MOSFET with class AB amplifier.

The main benefit of using MOSFET over BJT(Bipolar Junction transistor) is its simple drive circuit, less susceptibility to thermal stability, and high input impedance. In the pre-amplification process, the amplified signal noise-free is produced by the pre-amplifier with the help of a 2-stage differential amplifier circuit. The 1st stage of the pre-amplifier is in differential mode using an emitter coupled amplifier with PNP transistors.

The 2nd stage is a differential amplifier circuit with an active load, that helps in enhancing the voltage gain. To ensure the output current remains constant regarding the changes in the input voltage signals, the current mirror circuit is utilized. The amplified input signal is fed to the class AB power amplifier stage to produce an output signal with high power.

The class AB amplifier using MOSFET is referred to as a Hi-fi amplifier circuit, which is suitable for various applications such as keyboard amplifiers, general-purpose amplifiers, guitar amplifiers, and subwoofer amplifiers. It produces less distortion of about 0.1%, damping factor>200, 1.2V of input sensitivity with a bandwidth range of 4Hz-4KHz.

Class AB Amplifier Power Calculation

The total power of a Class AB amplifier is calculated from the input power and the output power.

Input Power

It refers to the amount of power supplied to the load by the class AB amplifier from the power supply. It is also known as DC power.

It is calculated from, Pi(dc) = Vcc. Idc

The amount of current drawn by each transistor is similar to the current of a full-wave signal. Therefore,

Idc = 2/π.I(peak)

Hence the formula total input power is given as,

Pi(dc) = Vcc. 2/π. I(peak)

The amount of current drawn by each transistor for a half-wave signal is the same as the average value of the half signal.

Icc = Ic(sat) / π

Pi(dc) = Ic(sat) Vcc/π

Output Power or AC Power

The maximum output power at the load is calculated from,

Pout = Iout(rms) x Vout(rms)

Output current Iout(rms) = 0.707Iout(peak) = 0.707Ic(sat)

The output voltage Vout(rms) = 0.707Vout(peak) = 0.707VCEQ

Therefore, the total output power is calculated from,

Pout = 0.5Ic(sat) x VCEQ

Substitute VCEQ = VCC/2. Then the maximum output power is,

Pout = 0.25Ic(sat) x VCC

Where VCEQ= maximum peak output voltage of both single and dual power supply of a Class AB amplifier.

Ic(sat) = peak output current of a class AB amplifier

OR

Pout=Vl^2/Rl=Vl^2(peak) /2Rl = Vl^2(p-p) /8Rl

Class AB Amplifier Efficiency

The efficiency of class AB amplifiers is high when compared to the efficiency of class B and class A amplifiers. The biasing of the class AB amplifier eliminates the cross over distortion
The efficiency is defined as the ratio of AC output power to DC input power. It is denoted by ‘Ƞ’

Ƞ = Pout/Pdc

Substitute Pout = 0.25.Ic(sat) Vcc

Pdc = Ic(sat) / π

Therefore,

The maximum efficiency of class AB amplifier will be,

Ƞ = Pout/Pdc

Ƞ= 0.25.Ic(sat) Vcc/ [(Ic(sat) Vcc) /π]

Ƞ= 0.25π

Ƞ= 78.5%

Class AB Power Amplifier Solved Problems

Example 1: Consider the following circuit diagram of Class AB power amplifier diode biasing with load resistance of 8 ohms, output power to load of 5 W, peak output voltage not greater than 80% of VCC, the minimum value of Id (<5mA). Assume, for T1 and T2, Isq = 10^-13, β = 75 and For diodes D1 and D2, Isd = 3X10^-14 A.

Determine the following.

A) Ibias (Bias current) and VCC

B) Quiescent collector current

C) Icn (collector current of NPN transistor)and Icp (collector current of PNP transistor) for the positive peak value of output voltage.

Solution:

A).Ibias (Bias current) and VCC

Vo(rms) = √(Pl.Rl = √(5×8)= 6.32V

Vo(peak) = √2Vo(rms)

= √2×6.32 = 8.9V

Assume Vcc = 12V

Vcc = Vo(peak) /0.8 = 11.8 V

To maintain the 5mA minimum current flow through the diodes D1 and D2,

Ibias = Ibn + Id = 14.7+5=19.7mA

Assume Ibias = 20mA

B). Quiescent Collector Current

At quiescent condition Id = 20mA, Vi= 0, and neglect Ibn

VBB = 2Vt ln(Id/Isd) = 2×0.0026 ln[(20X10^-3)/(3×10^-14) ] = 1.41V

Assume T1 and T2 are the transistors

VBEn = VBEp = VBB/2 = 0.70V

Therefore,

Icq = Isq. e^(VBB/2Vt)

= 10^-13.e^(0.70/0.0026)

Icq = 67mA

C) At the positive peak output voltage

Ien(max) = Il(peak) = 1.12mA

Ibn(max) = 14.7mA

Id = Ibias – Ibn(max)

= 20-14.7 = 5.3mA

Now,

VBB = 2×0.0026ln[(5.3×10^-3) / (3×10^-14) ] = 1.34V

Icn(max) = β/(1+β) = 75/(1+75) = 1.10A

VBEn = Vt. ln (Icn(max) /Icq)

= 0.0026 ln (1.10/10^-13) = 0.78V

VBEp = VBB-VBEn

= 1.34-0.78 = 0.56V

Icp = Isq. e^(VBEp/Vt)

= 10^-13 e^(0.56/0.0026)

= 0.28mA

Therefore, if the output voltage is at its positive peak value then,

Icn = 1.10 A and Icp = 0.28 mA

Example 2: Consider the above class AB amplifier circuit with Vcc = 25V and Rl = 4ohms. Calculate input power, output power, power handled by T1 and T2, the efficiency of input voltage of 12V(rms).

Solution: Given Vcc = 25V, Rl = 4ohms

Input voltage Vi(rms) = 12V

Input voltage Vi(peak) = √2Vi(rms)

= √2×12 = 17V

If the operating point is considered in the cut-off region then, the peak input voltage is equal to the load voltage at its peak

Vi(peak) = Vl(peak) = 17V

Load current Il(peak) = Vl(peak) /RL

= 17/4 = 4.25 Amps

Idc = 2/π Il(peak) = 2/π x 4.25 = 2.71 Amps

Input power Pi(dc) = Vcc/Idc

= 25/4.25= 67.75W

Output power Po(ac) = Vl^2(peak) /2Rl

= 17^2/(2X4) = 36.12W

Power dissipation of each transistor is,

P = Pt/2 =( Pi – Po) / 2 = (67.75-36.12) /2

= 15.8 W

Efficiency Ƞ% = Po/Pi x 100

= 36.12/67.75 x 100

= 53.3%

Advantages and Disadvantages

The advantages of a class ab amplifier include the following.

  • The main advantage of the class AB amplifier is the elimination of cross-over distortion.
  • Provides high efficiency and amplification of the signal than class A amplifier
  • It is a linear amplifier because the amplitude and the phase of the output signal are the same as the input
  • Preferred for radios, audio systems, and TV receivers.
  • Provides high efficiency and high-frequency response
  • Less harmonic distortion

The disadvantages of a class ab amplifier include the following.

  • Practically the efficiency of the class AB amplifier is less when compared to the efficiency of the class B amplifier
  • Construction of circuit is complex
  • Cost is high
  • Inappropriate biasing produces spikes in the output signal, resulting in crossover distortion.

Applications

The applications of class ab amplifier include the following.

  • Used in Audio Frequency (AF) amplifiers in TVs
  • Used in Public addressing systems
  • Used in TV receivers
  • Used in radio receivers
  • Used in CD players.
  • Used satellite and other wireless communication systems

Thus, this is all about an overview of Class AB amplifier – definition, circuit diagram, class AB amplifier using MOSFET, derivation of power and efficiency, applications, solved problems, advantages and disadvantages. Here is a question for you, “What is the difference between class A, class B, and class C amplifiers? “

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