Large Signal Amplifier Design Ryan Child 1. Background Large Signal Amplifiers belong to a class of amplifiers that are used for applications where high.

1

Large Signal Amplifier Design

Ryan Child

2

Background

Large Signal Amplifiers belong to a class of amplifiers that are used for applications where high current switching is needed, such as when driving:

• Electromechanical transducers (i.e. loudspeakers)• Motors of rotating machines (i.e. centrifuges)

When to use Small Signal and when to use Large Signal Amplifiers?

• Small Signal Amplifiers provide voltage gain but little current gain and are incapable of switching large amounts of current.

• Small Signal Amplifiers may be used in the earlier stages of amplification if the input resistance of the next stage is very large.

• Large Signal Amplifiers are typically used in the later stages of amplification and directly drive a load, which may require large amounts of current.

3

Requirements

Typical applications of the large signal amplifier lead to three primary objectives for any design:

1. Maximize output signal power (allows for driving larger loads)

2. Minimize distortion (output waveform should deviate from input waveform as little as possible)

3. Minimize quiescent power dissipation (keeps operation costs low and lengthens the life of the circuit)

4

Requirements and Specifications

This amplifier was designed to meet the following requirements, for an input signal of 50mV RMS and a load impedance of 16Ω:

Parameter Requirement SpecPower Supply ±24V ±24VMax Output Signal Power 10 W 13.5WBandwidth >10 kHz 15.8 kHzInput Z >1 kΩ 10 kΩ

5

Top-level Block Diagram

2 Stages:• Input stage

– Differential Amplifier provides voltage gain

• Output stage– Phase Splitter splits positive voltage near rail into positive and negative

swing around 0V, eliminates crossover distortion– Push-Pull amplifier provides large current gain and delivers power to

the load

6

Principles of Operation – Push-Pull Amplifier

1. Initially the input is 0V and Q1 and Q2 are both slightly on. The output is also at 0V.

2. When the input is maximum, VBE of Q1 is heavily forward biased, and VBE of Q2 is reverse biased. Q1 “pushes” current into the load, bringing it close to the input voltage.

3. When the input is minimum, VBE of Q2 is heavily forward biased, and VBE of Q1 is reverse biased. Q2 “pulls” current from the load, bringing it close to the input voltage.

4. The process repeats, pushing and pulling current to and from the load.

Vin

VBE of Q1

VBE of Q2

Diodes eliminate crossover distortion, by keeping Q1 and Q2 on even when Vin ≈ 0.

7

Principles of Operation – Signal Splitter

1. Initially, input is near the positive supply rail (≈ 21V)

2. As the input falls, VBE of Q10 increases, turning Q10 on. Excess current goes into the base of Q5, turning it on as well. VCE of Q10 decreases, raising the input of the push-pull circuit.

3. When the input is minimum, Q10 and Q5 are fully on, and the input to the push-pull circuit is at a maximum.

4. As input rises above the initial value, VBE of Q10 is lowered, and its emitter current is reduced. The collector current of Q9 must remain constant, so it pulls more current from Q6. VCE of Q10 increases, lowering the input of the push-pull circuit.

5. When input is maximum, Q10 and Q5 are nearly cut off, and the input to the push-pull circuit is at a minimum.

6. Repeat

Vin

VBE of Q10

VCE of Q10

R13, R14 and Q7 mimic the diodes to eliminate crossover distortion, but allow a “tunable diode drop” if a potentiometer is used.

∆ϕ=180˚

8

Principles of Operation – Differential Amplifier1. Voltage at base of Q2 is ≈ 0. The sum of the collector

currents through Q3 and Q2 is set by the current mirror and source Q8 and Q1. Initially, the input (base of Q3) is also at 0V. Therefore, VBE and collector currents of both Q2 and Q3 are approximately equal.

2. As the input rises, the voltage at the emitter of Q3 rises with it, cutting off Q2. This “steers” the current through Q3. The increased current through R3 pushes the output voltage down.

3. As the input falls, Q3 becomes cutoff because its emitter voltage is fixed by Q2. The collector current of Q3 decreases as a result, pulling the output voltage up.

4. This process repeats resulting in a gain and a DC shift of the input signal at the output. The gain is proportional to the load resistor (R3) and the current through it.

Vin

Vout

• Base of Q3 is “inverting” because an increase with respect to the base of Q2 causes the output to fall

• Base of Q2 is “non-inverting” because an increase with respect to the base of Q3 causes the output voltage to rise

– +

• Because the input is connected to the inverting input, phase is offset by 180˚

∆ϕ=180˚

Large Signal Amplifier – Complete Circuit

Gain Analysis of Input Stage (Differential Amplifier)

vB2 ≈ 0 V

Gain is highly nonlinear over input range

Larg

e Si

gnal

ana

lysi

s

When the output is maximum, we can simplify the circuit by noting that the lower two BJTs are off.

Gain Analysis of Output Stage

In reality this will be lower due to nonlinearities in the output stage.

≈1 Because the resistance seen by

the differential pair into Q10 (base) is much lower than Q4 (base and collector), we can simplify the analysis by assuming that most of the current will come from Q10 for given change in input voltage.

Gain Analysis of Complete Amplifier

Ri = 47(β10 + 1) = 9.2kΩ

Ri ≈ 33.5kΩ(β4 + 1)=4.6MΩ

The total phase difference from Vin to Vout is 360˚, so the output is in phase with the input. If we feed a fraction of the output signal back to the non-inverting input of the differential amplifier, it will be subtracted from the input due to the common-mode rejection of the amplifier. The result is a lower overall gain and greater bandwidth.

Gain Analysis of Complete Amplifier with Feedback

The emitter resistors in the push-pull circuit serve to reduce the spread of internal emitter resistances and are typically ≈0.47Ω

Component Calculations

The power requirement gives us the currents required in the push-pull BJTs:

As a rule of thumb, the bias current for a network of BJTs should be 5 to 10x the base current. Therefore, we chose 50 mA for the bias current in the output stage.

To find the appropriate resistor values for the bias network, we need to determine a suitable DC level for the input to the output stage. We want to make sure we satisfy the power requirement, without driving the BJTs Q9 and Q10 of the signal splitter stage into saturation:

We want to select resistors for the current mirror such that Q9 supplies about 50 mA, and the base voltage of the current mirror reflects the DC voltage at the input.

Component Calculations Continued

To eliminate crossover distortion, we pick resistors for a 0.7 V drop and a current of approximately 50 mA:

For the differential amplifier, we choose a typical current source value of 1 mA. After calculating the quiescent current coming from the output stage, we can find a suitable value for the collector resistor R3.

Component Calculations Continued

R1, R9 and R10 were chosen to create a 1mA source for the differential pair:

=R10=543ΩR9=100ΩR1=3.9kΩ

To satisfy the requirement of an input resistance ≥ 1kΩ, we chose a 10kΩ resistor at the inverting input (R2).

Component Calculations Continued

Reusing the 10kΩ component to set the voltage of the non-inverting input, we pick 10kΩ for R4 and R7. The two resistors in parallel result in a resistance of 5kΩ at that input. To match the resistances of the two inputs, we put an additional 4.7kΩ resistor (R23) in series with parallel configuration of R4 and R7.

A large capacitor was chosen to couple the input waveform to the inverting input of the differential amplifier. 4.7µF was large enough to pass the lowest frequencies required.

Power BJTs (2N2955 and 2N3055) were used for the output stage. They were chosen such that their common-emitter current gains were large and did not differ much from transistor to transistor.

The feedback resistor was calculated to eliminate clipping for the desired input:

Static Power Dissipation

Total static power dissipation = 5.11 W

Push-Pull Circuit SPICE and Oscilloscope Plots

Oscilloscope measurement after bypassing the diodes. Crossover distortion is apparent in the output (probe 2)

Phase Splitter SPICE Plot

Differential Amp SPICE and Oscilloscope Plots

AC-Coupled Open Loop SPICE and Scope Plots

Direct-Coupled Open Loop SPICE and Scope Plots

Direct-Coupled Closed Loop SPICE and Scope Plots

Measured Output Stage Voltage Transfer Characteristics

Measured and Simulated Midband Gains of Various Stages

Measured Frequency Response of Direct-Coupled Amplifier

Measured Frequency Response of AC-Coupled Amplifier

Conclusions Large signal amplifiers capable of switching large amounts of current can be built

using BJTs and typically include multiple stages, such as a differential stage, a signal splitter and push-pull stage.

Unlike common-emitter, common-collector and common-base configurations, a differential amplifier can be directly coupled to an output stage with no loss in gain.

Currents in a differential pair must be balanced. If they are not, there might not be sufficient current to allow the output to swing fully in both directions.

Bias voltages on either side of a signal splitter and push-pull amplifier must be symmetrical about 0V to ensure an output voltage of 0V for an input of 0V, and to maximize available swing at the output.

For best results, the common-emitter current gains of the BJTs on either side of the signal splitter and push-pull amplifier should deviate as little as possible, no more than 5 to 10%.