Class a and Ab Amplifiers

Adamson UniversityCollege of Engineering

Electronics and Communications Engineering Department

Class B and AB Amplifiers

Engr. Bernadeth B. ZariInstructor

ECAD Class B and Class AB Amplifiers 1

CLASS B AND CLASS AB PUSH-PULL AMPLIFIERS

When an amplifier is biased at cut-off so that it operates in the linear region for 180° of the input cycle and is in cut-off for 180°, its is classified as Class B amplifier.

Class AB amplifiers are biased to conduct for slightly more than 180°.

The primary advantage of a class B or class AB amplifier over a class A amplifier is that either one is more efficient than a class A amplifier. You can get more output power for a given amount of input power.

A disadvantage of class B or class AB amplifier is that it is more difficult to implement the circuit in order to get a linear reproduction of the input waveform.

Class B Operation

Class B operation is provided when the dc bias leaves the transistor biased just off, the transistor turning on when the ac signal is applied. This is essentially no bias, and the transistor conducts current for only one-half of the signal cycle.

The Q-Point Is at Cut-off

The class B amplifier is biased at the cutoff point so that ICQ

= 0 and VCEQ = VCE (cutoff). It is brought out of cutoff and operates


in its linear region when the input signal drives the transistor into conduction.

Fig. 9 – 7 Common collector class B amplifier

Class B Push-Pull Operation

The circuit in Figure 9-7 only conducts for the positive of the cycle. To amplify the entire cycle, it is necessary to add a second class B amplifier that operates the negative half of the cycle. The combination of two class B amplifiers working together is called push-pull operation.

Fig. 9 – 8 Block representation of push – pull operation


Input (DC) Power

The power supplied to the load by an amplifier is drawn form the power supply (or power supplies) that provide the input or dc power. The amount of this input power can be calculated using

where Idc is the average or dc current drawn from the power supplies. In class B operation, the current drawn from a single power supply has the form of a full-wave rectified signal, whereas that drawn from two power supplies has the form a half-wave rectified signal from each supply. In either case, the value of the average current drawn can be expressed as

Where I(p) is the peak value of the output current waveform. Using equation in the power input equation results in

Output (AC) Power

The power to delivered to the load (usually referred to as a resistance RL) can be calculated using any one of a number of equations. If one is using an rms meter to measure the voltage across the load, the output power can be calculated as

If one is using an oscilloscope, the measured peak or peak-to-peak output voltage can be used:


The larger the rms or peak output voltage, the larger is the power delivered to the load.

Efficiency

The efficiency of the class B amplifier can be calculated using the basic equation

Using equations in the efficiency equation above results in

[Using I(p)=VL(p)/RL]. Equation shows that the larger the peak voltage, the higher is the circuit efficiency, up to a maximum value when VL (p) =VCC, this maximum efficiency then being


Power Dissipated by Output Transistors

The power dissipated (as heat) by the output power transistors is the difference between the input power delivered by the supplies and the output power delivered to the load,

Where P2Q is the power dissipated by the two output power transistor. The dissipated power handled by each transistor is then

Example: For a class B amplifier providing a 20-V peak signal to a 16-Ω load (speaker) and a power supply of Vcc= 30V, determine the input power, output power, and circuit efficiency.Solution: A 20-V peak signal across a 16-Ω load provides a peak load current of

The dc value of the current drawn from the power supply is then

And the input power delivered by the power supply voltage is

The output power delivered to the load is

For a resulting efficiency of


Maximum Power Considerations

For class b operation, the maximum output power is delivered to the load when VL(p)=VCC:

The corresponding peak ac current I(p)is then

So that the maximum value of average current from the power supply is

Using this current to calculate the maximum value of input power results in

The maximum circuit efficiency for class B operation is then

When the input signal results in less than the maximum output signal swing, the circuit efficiency is less than 78.5%.


For class B operation, the maximum power dissipated by the output transistors does not occur at the maximum power input or output condition. The maximum power dissipated by the two output transistor occur when the output voltage across the load is

For a maximum transistor power dissipated of

Example: For a class B amplifier using a supply of VCC= 30V and driving a load of 16Ω, determine the maximum input power, output power, and transistor dissipation.

Solution: The maximum output power is

The maximum input power drawn from the voltage supply is

The circuit efficiency is then

As expected. The maximum power dissipated by each transistor is


Under maximum conditions a pair of transistors each handling 5.7 W at most can deliver 28.125W to a 16Ω load while drawing 35.81W from the supply.

The maximum efficiency of a class B amplifier can also be expressed as follows:

So that

Example: Calculate the efficiency of a class B amplifier for a supply voltage of VCC = 24V with peak output voltages of:

a. VL(p)= 22V.b. VL(p)= 6V.

Solution: Using equations gives


a.

b.

Notice that a voltage near the maximum [22 V in part (a)] results in an efficiency near the maximum, where as a small voltage swing [6 V in part (b)] still provides an efficiency near 20%. Similar power supply and signal swings would have resulted in much poorer efficiency in class A amplifier.

CLASS B AMPLIFIER CIRCUIT

There are two approaches for using push-pull amplifiers to reproduce the entire waveform. The first approach uses transformer coupling. The second uses two complementary symmetry transistors; these are a matching pair of npn / pnp BJT’s or a matching pair of n-channel/p-channel FET’s.

Transformer Coupling

Transformer coupling is illustrated in Figure 9-9. The input transformer has a center-tapped secondary that is connected to ground, producing phase inversion of one side with respect to the other.

The input transformer thus converts the input signal to two out-of-phase signals for the transistors. Notice that both transistors are npn types.


Q1 will conduct on the positive part of the cycle and Q2 will conduct on the negative part.

Fig. 9 – 9 Transformer coupled push-pull amplifier, Q1 conducts during the positive half-cycle; Q2 conducts during the negative half cycle. The two halves are combined by the output transformer

Complimentary Symmetry Transistors

Figure 9-10 shows one of the most popular types of push-pull class B amplifiers using two emitter-followers and both positive and negative power supplies. This is a complementary amplifier because one emitter-follower uses an npn transistor and the other a pnp, which conduct on opposite alternations of the input cycle.


(a) During a positive half cycle

(b) During a negative half cycle

Fig.9 – 10 push-pull ac operation

Crossover Distortion

When the dc base voltage is zero, both transistors are off and the input signal voltage must exceed VBE before a transistor conducts. Because of this, there is a time interval between the positive and negative alternations of the input when neither


transistor is conducting, as shown in Figure 9-11. The resulting distortion in the output waveform is called crossover distortion.

Biasing the Push-Pull Amplifier for Class AB Operation

To overcome crossover distortion, the biasing is adjusted to just overcome VBE of the transistors; this result in a modified form of operation called class AB.

In class AB operation, the push-pull stages are biased into slight conduction, even when no input signal is present. This can be done with a voltage divider and diode arrangement, as shown in Figure 9-12.

Fig. 9 – 12 biasing the push pull amplifier to eliminate crossover distortion

When the diode characteristics of D1 and D2 are closely matched to the characteristics of the transistor base-emitter junctions, the current in the diodes and the current in the


transistors are the same; this is called a current mirror. This current mirror produces the desired class AB operation and eliminates crossover distortion.

Assuming that both transistors are identical, the drop across D1 equals the VBE of Q2. Since they are matched, the diode current will be the same as ICQ. The diode current and ICQ can be found by applying Ohm’s Law to either R1 or R2 as follows:

AC Operation

Consider the ac load line for Q1 of the class AB amplifier in Figure 9-12. It’s Q-point slightly above cutoff. The ac cutoff voltage for a two-supply operation is at VCC with an ICQ as given earlier. The ac saturation current for a two-supply operation with a push-pull amplifier is:

Example: Determine the ideal maximum out voltage and current for the circuit.


Solution: The ideal maximum peak output voltage is Vout(peak) VCEQ = VCC = 20 V

The ideal maximum peak current is Iout(peak) ICsat = RL = 20 V/16 Ω = 1.25 A

Single-Supply Push-Pull Amplifier

Push-pull amplifiers using complimentary symmetry transistors can be operated from a single voltage source as shown in Figure 9-16. The circuit operation is the same as that described previously, except the bias is set to force the output emitter voltage to be VCC/2 instead of zero volts used with two supplies.


Fig. 9 – 13 Single ended push-pull amplifier

AMPLIFIER DISTORTION

A pure sinusoidal signal has a single frequency at which the voltage varies positive and negative by equal amounts. Any signal varying over less than 360° cycle is considered to have distortion. When distortion occurs, the output will not be an exact duplicate (except for magnitude) of the input signal.

Distortion can occur because the device characteristic is not linear, in which case non-linear amplitude distortion occurs. This can occur with all classes of amplifier operation. Distortion


can also occur because the circuit elements and devices respond to the input signal differently at various frequencies, this being frequency distortion. One technique for describing distorted but period waveforms uses Fourier analysis, a method that describes any periodic waveform in terms of its fundamental frequency component and frequency components at integer multiples – these components are called harmonic components or harmonics.

Example:

A signal that is originally 1000 Hz could result, after distortion, in a frequency component at 1000 Hz (1 kHz) and harmonic components at 2 kHz, 3 kHz, 4 kHz, and so on.

The original frequency of 1 kHz is called the fundamental frequency; those at integer multiples are the harmonics.

The 2 kHz component is therefore called a second harmonic.

The 3 kHz is the third harmonic, and so on. The fundamental frequency is not considered a harmonic.

Fourier analysis does not allow for fractional harmonic frequencies – only integer multiples of the fundamental.

HARMONIC DISTORTION

A signal is considered have harmonic distortion where there are harmonic frequency has an amplitude A1 and the nth frequency component has an amplitude An, a harmonic distortion can be defined as


The fundamental component is typically larger than any harmonic component.

Example: Calculate the harmonic distortion components for an output signal having fundamental amplitude of 2.5V, second harmonic amplitude of 0.25V, third harmonic amplitude of 0.1V, and fourth harmonic amplitude of 0.05V.

Solution: Using Eq. yields:

Total Harmonic Distortion

When an output signal has a number of individual harmonic distortion components, the signal can be seen to have a total harmonic distortion based on the individual elements as combined by the relationship of the following equation:

X 100%

Example: Calculate the total harmonic distortion for the amplitude components given in Example 12.13.


Solution: using the computed values of D2 = 0.10, D3 = 0.04 and D4 = 0.02 in eq., we obtain:

X 100%

Second Harmonic Distortion

Figure 12.20 shows a waveform to use for obtaining second harmonic distortion. A collector current waveform is shown with the quiescent, minimum, and maximum signal levels, and the time at which they occur is marked on the waveform. The signal shown indicates that some distortion is present. An equation that approximately describes the distorted signal waveform is:

The current waveform contains the original quiescent current ICQ

which occurs with zero input signal; an additional current Io due to the non zero average of the distorted signal; I2, at twice the fundamental frequency. Although other harmonics are also present, only the second is considered here. Equating the resulting current at a few points in the cycle to that shown on the current waveform provides the following three relations:

At point 1 (t = 0)


At point 2 (t = ),

At point 3 (t = ),

Solving the preceding three equations simultaneously gives the following results

Referring to equation, we can express the definition of second harmonic distortion as

Inserting the values of I1 and I2 determined above gives

In a similar manner, the second harmonic distortion can be expressed in terms of measured collector-emitter voltages:


Example: Calculate the second harmonic distortion if an output waveform displayed on an oscilloscope provides the following measurements:

a. VCEmin = 1V, VCEmax = 22V, VCEQ = 12V.b. VCEmin = 4V, VCEmax = 20V, VCEQ = 12V.

Solutions:

a.

b.

Power of a Signal Having Distortion

When distortion does occur, the output power calculated for the undistorted signal is no longer correct. When distortion is present, the output power delivered to the load resistor RC due to the fundamental component of distorted signal is

The total power due to all the harmonic components of the distorted signal can then be calculated using

The total power can also be expressed in terms of the total harmonic distortion,


Example: For a harmonic distortion reading of D2 = 0.1, D3 = 0.02, and D4 = 0.01, with I1 = 4A and RC = 8Ω, calculate the total harmonic distortion, fundamental power component, and total power.

Solution: The total harmonic distortion is

The fundamental power calculated using the equations is then

The total power calculated using the equation is then

(Note that the total power is due mainly to the fundamental component even with 10% second harmonic distortion.)


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