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Class A NPN Common-Base and Cascode Amplifiers

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Introduction

Common-emitter and emitter-follower amplifiers are the most widely used single-transistor amplifiers. The common-base configuration, illustrated below in its basic NPN form, is used less frequently as a stand-alone voltage amplifier stage, mostly because it has a low input resistance, but it is often combined with a common-emitter stage to form a cascode amplifier. In a common-base voltage amplifier, the input voltage is applied to the emitter and the output voltage is taken from the collector, and the input and output voltages are in-phase. The in-phase relationship can be understood by observing that when the signal voltage applied to the emitter increases, the base-emitter voltage v_BE decreases, causing the emitter/collector current to decrease, decreasing the voltage drop across the collector resistor, R_C, thereby increasing the collector voltage. Because of its low input resistance, the common-base amplifier is sometimes used as a current-in/voltage-out amplifier. Its operation in the cascode configuration is something like this because it takes the collector current from the common-emitter stage and produces an output voltage at its collector.

As with all linear Class A BJT amplifiers, the transistor must operate in the forward active region in the common-base amplifier. This means that the base-emitter junction must be forward biased, the collector-base junction must be reverse biased, and operation must be prevented from entering the saturation region. These bias conditions can be seen in the schematic shown in the “Procedure” section for the common-base stage as well as the CE stage in the cascode amplifier. The base is often biased using a resistive voltage divider, voltage regulator, or available power supply, and an emitter resistor, R_E is used to establish the emitter current. The base should always be bypassed with a capacitor that produces a low impedance AC ground at all signal frequencies. If the amplifier is DC-coupled, the voltage source applied to the base must also provide a low DC resistance. The collector circuit is similar to what is used in the common-emitter amplifier, and in its simplest form consists of a resistor, R_C connected between the collector and the power supply.

The input of a common-base amplifier looks into the emitter, which is the same as looking into the output resistance of an emitter-follower amplifier. As we saw in the “Class A NPN Emitter-Follower Amplifier” lab, the emitter-follower output resistance is low, which is desirable for a voltage amplifier output, but is not desirable for a voltage amplifier input. With a low-impedance voltage source providing the base bias, the common-base input resistance is simply r_e, which is equal to the ratio of the thermal voltage V_T to the collector bias current I_C, or R_IN = V_T/I_C ≈ 26 mV/I_C at room temperature. The gain of a common-base amplifier can be calculated using detailed circuit analysis or approximated by inspection as was done with the CE amplifier. Using the same approach as was done with the CE amplifier, the gain of the common-base amplifier with a low-impedance base driving circuit is A_V = R_C/r_e. Besides the low input resistance issue, the gain of the common-base amplifier depends on r_e, which is small (producing high gain, which may not be desirable), temperature dependent, and nonlinear. When combined with a CE stage to form a cascode amplifier, we will see that the r_e of the common-base amplifier becomes the effective collector resistance of the CE stage, and gets cancelled out of the cascode amplifier gain equation when the emitter resistor is 100% capacitively bypassed. The greatest advantage of the cascode amplifier, however, is that it reduces the Miller effect, which causes the bandwidth of a CE stage to decrease as its gain increases. With the r_e of the common-base stage as the effective collector resistance of the CE stage and the r_e of the CE stage as the emitter resistance when the emitter bias-setting resistor is fully bypassed, the gain of the CE stage is -r_e/r_e ≈ -1 when the two transistors are well-matched. The voltage gain is provided by the collector current flowing through the collector resistor in the common-base stage, and the common-base amplifier bandwidth is not diminished by the Miller effect. The important benefit realized by the cascode configuration is that high gains can be achieved without significant bandwidth loss due to the Miller effect, but it requires two transistors. As with CE amplifiers, an emitter-follower stage is often added on the output of the common-base stage to drive low impedance loads.

In this lab, we will not build a standalone common-base amplifier, but will instead study the use of the common-base amplifier in a cascode amplifier.

Objective

To understand the operating principles of common-base and cascode amplifiers. To design, build, and test a cascode amplifier using two 2N3904 NPN transistor stages, with an input resistance of at least 1 KΩ and a voltage gain of a little less than 10. To understand how to set up the proper bias conditions for a cascode amplifier and verify that the bias voltages in the circuit are close to their designed values. Following completion of this lab you should be able to explain the basic operation of common-base and cascode amplifiers, and be able to calculate the voltage gain for each of these amplifiers.

Materials and Apparatus

Data sheet handout for the 2N3904 NPN transistor
Computer running PixelPulse software
Analog Devices ADALM1000 (M1K)
Solderless breadboard and jumper wires from the ADALP2000 Analog Parts Kit
(2) 2N3904 NPN transistor from the ADALP2000 Analog Parts Kit
(1) 47 Ω resistor from the ADALP2000 Analog Parts Kit
(2) 100 Ω resistor from the ADALP2000 Analog Parts Kit
(2) 470 Ω resistor from the ADALP2000 Analog Parts Kit
(1) 2.2 KΩ resistor from the ADALP2000 Analog Parts Kit
(1) 6.8 KΩ resistor from the ADALP2000 Analog Parts Kit
(1) 47 μF capacitor from the ADALP2000 Analog Parts Kit
(2) 47 μF capacitor from the ADALP2000 Analog Parts Kit
(1) 220 μF capacitor from the ADALP2000 Analog Parts Kit

Procedure

Construct the following circuit on the solderless breadboard
Refer to the illustration below for one way to install the components in the solderless breadboard. Note that this circuit has been added to the CE amplifier breadboard from the “Class A NPN Common-Emitter Amplifier” lab.
Run PixelPulse and plug in the M1K using the supplied USB cable
Update M1K firmware, if necessary
Set up the M1K to source voltage/measure current on Channel A and measure voltage on Channel B
Set up Channel A source waveform for a 100 Hz “Sine” output that swings between 2.0 V and of 4.0 V
Observe the output voltage into the 1 KΩ load resistor on Channel B and verify that it is swinging nominally +/- 1 V about the 2.5 V baseline and that it is in-phase with the input signal
Observe the voltage at the emitter on Channel B and verify that it is swinging nominally +/- V about a 1.7 V bias voltage and that it is also in-phase with the input signal. Note that these voltages may vary somewhat due to resistor tolerances
Note any visible distortions in the output signals
Remove the input signal and measure the DC bias voltages at the base, emitter, and collector, and verify that these are at their designed levels, allowing for resistor tolerances
Calculate the voltage gain, power gain, and efficiency of this amplifier; compare the power dissipation of this circuit with that of the CE amplifier in the “Class A NPN Common-Emitter Amplifier” lab
Set up Channel A source waveform for a 100 Hz “Sine” output that swings between 2.3 V and of 2.7 V
Remove the AC-coupling capacitor and bias resistors from the amplifier input and connect it to the output of the CE amplifier from the “Class A NPN Common-Emitter Amplifier” lab as shown in the schematic. Note the change in the emitter-follower bias point
Refer to the illustration below for one way to interconnect the two amplifiers on the solderless breadboard
Verify that the voltage across the 47 Ω load resistor is swinging +/- 1V about the 2.5 V bias voltage
Calculate the rms power dissipation in the load resistor
Estimate the voltage swing that would be present across the 47 Ω load resistor if it were driven by the CE stage alone, without the emitter-follower buffer in palce
Reduce the swing of the input voltage to 2.45 V to 2.55 V
Replace the 47 Ω load resistor with a 10 Ω resistor
Verify that the output voltage swings 500 mV_P-P about the 2.5 V bias voltage
Calculate the load current into the 10 Ω load resistor

Theory

The emitter-follower amplifier presents a high input resistance to signals applied to its base and provides a low resistance effective voltage source at its output. These characteristics make the emitter-follower amplifier well suited for use as a voltage buffer amplifier. Buffer amplifiers are used to buffer heavy loads from signal sources that have a high output resistance. A good buffer has a high input resistance so as not to significantly load the output resistance of the source that is being buffered and a low resistance voltage source output that is capable of driving heavy loads with minimal loading loss.

Our amplifier needs to drive a 47 Ω load with a +/-1 V sine wave, which requires approximately +/-21 mA. We will make the base bias point at about mid-supply. Starting with a mid-supply base bias voltage, we can estimate the emitter voltage to be 2.5 V - 0.7 V = 1.8 V. We have a 68 Ω resistor in the kit, and if we use this for our R_E, we get an emitter current of 1.8 V/68 Ω ≈ 26.5 mA. This exceeds our minimum current of 21 mA, so we will start with this. If we want to make a first-round estimate of the voltage drop incurred due to base current, remembering that i_E ≈ i_C, we use β = 200 to estimate the base current as i_B ≈ 26.5 mA/200 ≈ 133 μA. The base bias current is on the high side because we are using a large emitter current. The voltage drop due to the base current can be estimated to be (133 μA)(2.2 KΩ||2.2 KΩ) ≈ 0.15 V. This reduction will in turn reduce the emitter voltage, which will reduce the emitter current, which will reduce the base current and its associated voltage drop, so a reasonable estimate for base bias voltage reduction due to base bias current would be about 0.1 V, so we will use 2.4 V for the base voltage. Now, a more accurate emitter bias voltage can be established as 2.4 V - 0.7 V = 1.7 V. The emitter bias current can be calculated as 1.7 V/68 Ω = 25 mA.

The input resistance looking into the base of the transistor used in the emitter-follower amplifier R_i,base is the same as that of the CE amplifier

${R_{i,base}} = β{r_e} + (1 + β){R_E}$

Using β = 200 from the Introduction to Transistors lab, and substituting numbers from this lab, we have

${R_{i,base}} = 200({{26 mV}/{25 mA}}) + (1 + 200)({{68 Ω}||{47 Ω}}) ≈ 208 Ω + 11.5 KΩ ≈ 5.6 KΩ$

When we use the voltage divider to provide base bias, we need to include the resistors in the divider in parallel with the input resistance looking into the base, so the total input resistance R_iis

${R_i} = 2.2 KΩ||2.2 KΩ||5.6 KΩ ≈ 0.92 KΩ$

Note that this result falls slightly short of our design goal of having a minimum input resistance of 1 KΩ. This is primarily due to the 2.2 KΩ resistors that were used to set up the base bias voltage. We could increase the value of these resistors a small amount and raise the input resistance to at least 1 KΩ, but the next largest value in the kit is 4.7 KΩ, which would give us considerably more unnecessary voltage drop on the base bias voltage. The best solution, if all commercially available resistor values were available, would be to use the smallest available resistor value that meets the input resistance requirement.

For large β (≥ 100) the output resistance looking into the emitter of the emitter-follower amplifier R_o is calculated as

$R_o = r_e + {R_S}/(1 + β)$

where R_S is the total equivalent source resistance in the base circuit. In our circuit, R_S is equal to the parallel combination of the base bias resistors and the source resistance feeding the base. Note that when the emitter-follower amplifier is fed by the low resistance voltage source output of the M1K, R_S ≈ 0 and the output resistance is simply the incremental emitter resistance r_e.

$R_o = r_e = {26 mV}/{25 mA} ≈ 1 Ω$

When we place the emitter-follower on the output of the CE amplifier from the “Class A NPN Common-Emitter Amplifier” lab, we have nonzero R_S, and need to include it in out calculation of R_o

$R_o = r_e + {R_S}/(1 + β) = 1 Ω + ({470 Ω}||{2.2 KΩ}||{2.2 KΩ})/(1 + 200) ≈ 1 Ω + 1.6 Ω = 2.6 Ω$

We will see a small voltage divider loss when driving the 47 Ω load, and a more pronounced loss when driving the 10 Ω load. With the emitter-follower amplifier buffering the CE amplifier, we will have voltage divider factors of 47 Ω/49.6 Ω ≈ 0.95 for the 47 Ω load and 10 Ω/12.6 Ω ≈ 0.79 for the 10 Ω load.

We can calculate the efficiency of the emitter-follower amplifier studied in this lab.

For the emitter-follower amplifier by itself driving +/-1 V into the 47 Ω load resistor, the power into the load is:

$P_LOAD = (1/√2)({1 V_PEAK})(1/√2)({21.3 mA_PEAK}) = 10.6 mW_rms$

Note that the power into the load could also have been calculated using the [v_LOAD(rms)]²/R_L formula.

The quiescent power drawn from the supply is:

$P_SUPPLY = {V_CE(Q)}{I_C(Q)} = (5.0 V - 1.7 V)(25 mA) ≈ 82.5 mW$

The efficiency is therefore:

For the emitter-follower amplifier buffering the CE amplifier driving +/-1 V into the 47 Ω load resistor, the power into the load is again:

The total quiescent power drawn from the supply, using the results already calculated in the CE amplifier lab, is:

$P_SUPPLY = {P_SUPPLY}(CE Amp) + {V_CE(Q)}{I_C(Q)} = 11.7 mW + (5.0 V - 2.1 V)(31 mA) ≈ 102 mW$

The efficiency is therefore:

If we substitute the 10 Ω load for the 47 Ω load, we can only deliver about +/- 0.25 V to the load, and the power into the load is

$P_LOAD = (1/√2)({0.25 V_PEAK})(1/√2)({25 mA_PEAK}) = 3.1 mW_rms$

The efficiency is now only

It is important to note that the voltage divider losses between the emitter-follower output and the load were omitted.

We can calculate the power gain of the two cascaded amplifiers with the 47 Ω load using the results obtained in the “Class A NPN Common-Emitter Amplifier” lab. The voltage gain is essentially the same as it was for the CE amplifier because the voltage gain of the emitter-follower amplifier is very close to unity and the input resistance of the CE is unchanged. The power gain is therefore

${A_P} = {A_V}^2({R_i}/{R_L}) = (-5.1)^2({1.2 KΩ}/{47 Ω}) ≈ 664$

${A_P}(dB) = 10{log_10}(664) ≈ 28 dB$

From this result, we can see how adding the emitter-follower output stage increased the overall power gain by providing current gain to drive the heavier load.

Observations and Conclusions

Emitter-follower amplifiers provide significant current gain and unity voltage gain
The operation of an emitter-follower amplifier involves a form of negative feedback
Emitter-follower amplifiers have high input resistance and low output resistance
Emitter-follower amplifiers are often used to buffer heavy loads that require large output currents from sources that have high source resistances
Adding an emitter-follower stage to a CE amplifier can significantly increase the power gain of the overall amplifier

Wiki

Analog Devices Wiki

Table of Contents

Class A NPN Common-Base and Cascode Amplifiers

Introduction

Objective

Materials and Apparatus

Procedure

Theory

Observations and Conclusions

Analog Devices Wiki

Table of Contents

Class A NPN Common-Base and Cascode Amplifiers

Introduction

Objective

Materials and Apparatus

Procedure

Theory

Observations and Conclusions

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