Class A BJT Amplifiers
Discrete Semiconductor Devices and Circuits
Don’t just sit there! Build something!! 
Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of sample problems and checking their answers against those provided by the textbook or the instructor. While this is good, there is a much better way.
You will learn much more by actually building and analyzing real circuits, letting your test equipment provide the “answers” instead of a book or another person. For successful circuitbuilding exercises, follow these steps:
 Carefully measure and record all component values prior to circuit construction, choosing resistor values high enough to make damage to any active components unlikely.
 Draw the schematic diagram for the circuit to be analyzed.
 Carefully build this circuit on a breadboard or other convenient medium.
 Check the accuracy of the circuit’s construction, following each wire to each connection point, and verifying these elements onebyone on the diagram.
 Mathematically analyze the circuit, solving for all voltage and current values.
 Carefully measure all voltages and currents, to verify the accuracy of your analysis.
 If there are any substantial errors (greater than a few percent), carefully check your circuit’s construction against the diagram, then carefully recalculate the values and remeasure.
When students are first learning about semiconductor devices, and are most likely to damage them by making improper connections in their circuits, I recommend they experiment with large, highwattage components (1N4001 rectifying diodes, TO220 or TO3 case power transistors, etc.), and using drycell battery power sources rather than a benchtop power supply. This decreases the likelihood of component damage.
As usual, avoid very high and very low resistor values, to avoid measurement errors caused by meter “loading” (on the high end) and to avoid transistor burnout (on the low end). I recommend resistors between 1 kΩ and 100 kΩ.
One way you can save time and reduce the possibility of error is to begin with a very simple circuit and incrementally add components to increase its complexity after each analysis, rather than building a whole new circuit for each practice problem. Another timesaving technique is to reuse the same components in a variety of different circuit configurations. This way, you won’t have to measure any component’s value more than once.
Complete the table of output voltages for several given values of input voltage in this commoncollector amplifier circuit. Assume that the transistor is a standard silicon NPN unit, with a nominal baseemitter junction forward voltage of 0.7 volts:

V_{in} V_{out}
0.0 V
0.5 V
1.0 V
1.5 V
5.0 V
7.8 V
Based on the values you calculate, explain why the commoncollector circuit configuration is often referred to as an emitter follower.
Complete the table of output voltages, output currents, and input currents for several given values of input voltage in this commoncollector amplifier circuit. Assume that the transistor is a standard silicon NPN unit, with a nominal baseemitter junction forward voltage of 0.7 volts:

V_{in} V_{out} I_{in} I_{out}
0.0 V
0.4 V
1.2 V
3.4 V
7.1 V
10.8 V
Calculate the voltage and current gains of this circuit from the numerical values in the table:


The following schematic diagram shows a simple commoncollector transistor amplifier circuit:

Explain why the AC voltage gain (A_{V(AC)}) of such an amplifier is approximately 1, using any or all of these general “rules” of transistor behavior:
 I_{E} = I_{C} + I_{B}
 I_{E} ≈ I_{C}
 V_{BE} ≈ 0.7 volts
 β = [(I_{C})/(I_{B})]
Remember that AC voltage gain is defined as [(∆V_{out})/(∆V_{in})].
Calculate the approximate voltage gain (A_{V}) for the following commoncollector amplifier circuit, expressing it as a ratio and as a decibel value. Also calculate the quiescent DC voltage measured across the load resistor (V_{load(DC)}). Assume a silicon transistor:

 A_{V} (as a ratio) ≈
 A_{V} (in decibels) ≈
 V_{load(DC)} ≈
Predict how all transistor currents (I_{B}, I_{C}, and I_{E}) and the output voltage signal will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

 Capacitor C_{in} fails open:
 Solder bridge (short) past resistor R_{1}:
 Resistor R_{1} fails open:
 Resistor R_{E} fails open:
For each of these conditions, explain why the resulting effects will occur.
Complete the table of voltages and currents for several given values of input voltage in this commonemitter amplifier circuit. Assume that the transistor is a standard silicon NPN unit, with a nominal baseemitter junction forward voltage of 0.7 volts. For the last row of the table, give qualitative answers (increase, decrease, or same) representing what each of the quantities will do given an increasing base voltage (V_{B}):

V_{B} V_{E} I_{C} V_{RC} V_{CE} V_{C} (V_{out})
0.0 V
0.5 V
1.0 V
1.5 V
2.0 V
2.5 V
3.0 V
increase
Calculate the voltage gain of this circuit from the numerical values in the table:

The following schematic diagram shows a simple commonemitter transistor amplifier circuit:

Explain why the voltage gain (A_{V}) of such an amplifier is approximately [(R_{C})/(R_{E})], using any or all of these general “rules” of transistor behavior:
 I_{E} = I_{C} + I_{B}
 I_{E} ≈ I_{C}
 V_{BE} ≈ 0.7 volts
 β = [(I_{C})/(I_{B})]
Remember that (AC) voltage gain is defined as [(∆V_{out})/(∆V_{in})]. Hint: this question might be easier to answer if you first consider how to explain the unitygain of a commoncollector amplifier circuit (simply eliminate R_{C}, replacing it with a direct connection to −V, and consider V_{E} to be the output voltage).
Determine what would happen to the voltage gain of a commonemitter transistor amplifier circuit if the following resistance values were changed (consider one change at a time):

 Resistance R_{C} increased; A_{V} . . .
 Resistance R_{E} increased; A_{V} . . .
 Resistance R_{bias1} increased; A_{V} . . .
 Resistance R_{bias2} increased; A_{V} . . .
A student attempts to calculate the voltage gain of the following commonemitter amplifier circuit, and arrives at an incalculable value (dividebyzero error):

According to a simple formula for approximating the voltage gain of this type of amplifier, it would indeed seem as though this circuit would have infinite voltage gain with zero emitter resistance. However, even with no emitter resistor installed in such a circuit, the transistor itself contains a small amount of resistance intrinsic to the semiconductor material, commonly symbolized as r′_{e}:

The problem is, this resistance value r′_{e} is far from stable. Determine some of the factors influencing the value of the transistor’s intrinsic emitter resistance, and explain why a circuit such as the one first shown in this question would be very unstable (possibly resulting in the selfdestruction of the transistor!).
Calculate the approximate voltage gain (A_{V}) for the following commonemitter amplifier circuit, and also calculate the quiescent DC voltages measured at the three terminals of the transistor with respect to ground (V_{B}, V_{E}, and V_{C}). Assume a silicon transistor:

 A_{V} ≈
 V_{B} ≈
 V_{E} ≈
 V_{C} ≈
Calculate the approximate voltage gain (A_{V}) for the following commonemitter amplifier circuit, and also calculate the quiescent DC voltages measured at the three terminals of the transistor with respect to ground (V_{B}, V_{E}, and V_{C}). Assume a silicon transistor:

 A_{V} ≈
 V_{B} ≈
 V_{E} ≈
 V_{C} ≈
Calculate the approximate voltage gain (A_{V}) for the following commonemitter amplifier circuit, and also calculate the quiescent DC voltages measured at the three terminals of the transistor with respect to ground (V_{B}, V_{E}, and V_{C}). Assume a silicon transistor:

 A_{V} ≈
 V_{B} ≈
 V_{E} ≈
 V_{C} ≈
Calculate the approximate voltage gain (A_{V}) for the following bypassed commonemitter amplifier circuit, assuming a quiescent (DC) emitter current value of 750 μA. Also calculate the quiescent DC voltage measured at the transistor’s collector terminal with respect to ground (V_{C}). Assume a silicon transistor:

 A_{V} ≈
 V_{C} ≈
Calculate the approximate voltage gain (A_{V}) for the following commonemitter amplifier circuit, expressing it both as a ratio and as a figure in decibels. Also calculate the quiescent DC voltages measured at the three terminals of the transistor with respect to ground (V_{B}, V_{E}, and V_{C}). Assume a silicon transistor:

 A_{V} (as a ratio) ≈
 A_{V} (in decibels) ≈
 V_{B} ≈
 V_{E} ≈
 V_{C} ≈
Predict how all transistor currents (I_{B}, I_{C}, and I_{E}) and the output voltage signal will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

 Capacitor C_{in} fails open:
 Solder bridge (short) past resistor R_{1}:
 Resistor R_{1} fails open:
 Resistor R_{C} fails open:
 Resistor R_{E} fails open:
 Capacitor C_{bypass} fails shorted:
For each of these conditions, explain why the resulting effects will occur.
Complete the table of voltages and currents for several given values of input voltage in this commonbase amplifier circuit. Assume that the transistor is a standard silicon NPN unit, with a nominal basebase junction forward voltage of 0.7 volts:

V_{E} V_{B} I_{B} I_{C} V_{RC} V_{C}
0.0 V
0.5 V
0.8 V
1.0 V
1.1 V
1.2 V
1.3 V
Calculate the voltage gain of this circuit from the numerical values in the table:

Commonemitter, commoncollector, and commonbase amplifier circuits are sometimes referred to as groundedemitter, groundedcollector, and groundedbase, respectively, because these configurations may actually be built with those respective terminals connected straight to ground.
Although this may not be very practical for ease of biasing, it can be done. Draw the rest of the circuit necessary to provide classA operation for each of these (partial) transistor circuits. Be sure to show where the DC power source, signal input, and signal output connect:

Define what a commoncollector transistor amplifier circuit is. What distinguishes this amplifier configuration from the other singleBJT amplifier configurations, namely commonemitter and commonbase?
Also, describe the typical gains (voltage and current) of this amplifier configuration, and whether it is inverting or noninverting.
Define what a commonemitter transistor amplifier circuit is. What distinguishes this amplifier configuration from the other singleBJT amplifier configurations, namely commoncollector and commonbase?
Also, describe the typical gains (voltage and current) of this amplifier configuration, and whether it is inverting or noninverting.
Define what a commonbase transistor amplifier circuit is. What distinguishes this amplifier configuration from the other singleBJT amplifier configurations, namely commoncollector and commonemitter?
Also, describe the typical gains (voltage and current) of this amplifier configuration, and whether it is inverting or noninverting.
Predict how all transistor currents (I_{B}, I_{C}, and I_{E}) and the output voltage signal will be affected as a result of the following faults. Consider each fault independently (i.e. one at a time, no multiple faults):

 Capacitor C_{out} fails open:
 Solder bridge (short) past resistor R_{1}:
 Resistor R_{1} fails open:
 Resistor R_{C} fails open:
 Resistor R_{E} fails open:
 Capacitor C_{bypass} fails shorted:
For each of these conditions, explain why the resulting effects will occur.
Temperature changes are well known to affect transistor operation. For instance, if we were to apply a constant voltage between the base and emitter of a transistor and increase its temperature over time, the collector current would increase:

First, describe why the collector current changes, if the input voltage is held constant. Then, determine the relative degree of output voltage change (∆V_{out}) resulting from this thermal effect in the following two amplifier circuits:

What is different in the responses of these two circuits to temperature changes? Why does one circuit respond so much differently than the other?
If both these amplifier circuits had AC signal inputs, and were biased for Class A operation, what effect would an increase in temperature have on each of them? State your answer in terms of AC voltage gain and Qpoint.

One major different between a commonemitter amplifier configuration and a commoncollector amplifier configuration is a principle called negative feedback, where changes in output voltage “feed back” to influence the amplifier’s input signal, which in turn influences the output voltage again. Commoncollector amplifier circuits have large amounts of negative feedback inherent to their design.
The absence or presence of negative feedback in an amplifier circuit has profound effects on voltage gain (A_{V}). Compare the relative voltage gains of the following amplifiers:

At first, the low voltage gain of the commoncollector amplifier may appear to be a disadvantage of that circuit design. However, there is one major benefit relevant to the commoncollector amplifier’s voltage gain, being a direct result of negative feedback. What is this advantage?
One way to reap the advantages of negative feedback enjoyed in commoncollector amplifier circuits, in a commonemitter amplifier, is to add components that intentionally “feed back” some of the output signal to the transistor’s input in a degenerative fashion:

Explain what the term negative (or degenerative) means with reference to feedback, and explain how each of these techniques works to produce this type of feedback. Also, explain one disadvantage of applying negative feedback to a commonemitter amplifier circuit.
Amplifier distortion occurs when its gain varies as a function of the instantaneous signal amplitude. That is, some parts of the signal waveform become amplified more than others, and this results in the waveform taking on a slightly different shape.
All active devices, bipolar junction transistors included, are nonlinear to some extent. This term means that their gain varies throughout their operating ranges. During the 1920’s, an electrical engineer named Harold Black was pondering this problem in the design of telephone system amplifiers. His solution came to him in a flash of insight one day, as he was commuting from work on a ferry boat. Explain what his solution to this problem was.
A popular method of “reclaiming” some of the lost voltage gain resulting from the addition of an emitter resistor (R_{E}) to a commonemitter amplifier circuit is to connect a “bypass” capacitor in parallel with that resistor:

Explain why this technique works to increase the circuit’s AC voltage gain, without leading to the problems associated with directly grounding the emitter.
Some commonemitter amplifier circuits use partial bypassing of emitter resistance, with the bypass capacitor connected in parallel with only one of two series resistors:

Explain the purpose of this arrangement. How does this differ in performance from the simple oneresistor emitter feedback design, or a groundedemitter amplifier with no emitter resistor at all?
A student builds this commonemitter amplifier so they he may amplify the audio signals from a microphone to power a speaker:

Unfortunately, the results are considerably less than expected: although some sound does come out of the speaker, it is not enough to be considered a success. Another student inspects the design and cryptically mumbles something about “poor impedance matching,” leaving the first student somewhat confused.
Explain what impedance matching means in this context, where the mismatch might be in this circuit, and what might be done to correct it.
Each of the following faults will cause this audio amplifier circuit to stop working. Determine what diagnostic voltage measurement(s) would positively identify each one of the faults.

 Microphone coil fails open:
 Capacitor C_{1} fails shorted:
 Resistor R_{1} fails open:
 Resistor R_{2} fails open:
 Capacitor C_{3} fails open:
 Transformer T_{1} primary winding fails open:
Sometimes you will see amplifier circuits expressed as collections of impedances and dependent sources:

With this model, the amplifier appears as a load (Z_{in}) to whatever signal source its input is connected to, boosts that input voltage by the gain factor (A_{V}), then outputs the boosted signal through a series output impedance (Z_{out}) to whatever load is connected to the output terminals:

Explain why all these impedances (shown as resistors) are significant to us as we seek to apply amplifier circuits to practical applications. Which of these impedances do you suppose are typically easier for us to change, if they require changing at all?
Complete the table of output voltages, output currents, and input currents for several given values of input voltage in this commoncollector amplifier circuit. Assume that the transistor is a standard silicon NPN unit, with a nominal baseemitter junction forward voltage of 0.7 volts:

V_{in} V_{out} I_{in} I_{out}
0.8 V
1.5 V
3.0 V
4.5 V
6.0 V
7.5 V
Calculate the amount of impedance “seen” by the input voltage source V_{in}, given the following definition for impedance:

The voltage divider network employed to create a DC bias voltage for many transistor amplifier circuits has its own effect on amplifier input impedance. Without considering the presence of the transistor or the emitter resistance, calculate the impedance as “seen” from the input terminal resulting from the two resistors R_{1} and R_{2} in the following commoncollector amplifier circuit:

Remember, what you are doing here is actually determining the Thévenin/Norton equivalent resistance as seen from the input terminal by an AC signal. The input coupling capacitor reactance is generally small enough to be safely ignored.
Next, calculate the input impedance of the same circuit, this time considering the presence of the transistor and emitter resistor, assuming a current gain (β or h_{fe}) of 60, and the following formula for impedance at the base resulting from β and R_{E}:


Develop an equation from the steps you take in calculating this impedance value.
The voltage divider network employed to create a DC bias voltage for many transistor amplifier circuits has its own effect on amplifier input impedance. Without considering the presence of the transistor or the emitter resistance, calculate the impedance as “seen” from the input terminal resulting from the two resistors R_{1} and R_{2} in the following commoncollector amplifier circuit:

Remember, what you are doing here is actually determining the Thévenin/Norton equivalent resistance as seen from the input terminal by an AC signal (consider the coupling capacitor reactance to be negligibly small).
Next, calculate the input impedance of the same circuit, this time considering the presence of the transistor and emitter resistor, assuming a current gain (β or h_{fe}) of 42:

Develop an equation from the steps you take in calculating this impedance value.
Determining the output impedance of a commonemitter amplifier is impossible unless we know how to model the transistor in terms of components whose behavior is simple to express.

When in its active mode, a transistor operates like a current regulator. This is similar enough to the behavior of a current source that we may use a source to model the transistor’s behavior for the sake of this impedance determination:

Now, apply the same steps you would use in determining the Thévenin or Norton equivalent impedance to the output of this amplifier circuit, and this will yield the amplifier’s output impedance. Draw an equivalent circuit for the amplifier during this Thévenizing/Nortonizing process to show how the output impedance is determined.
Explain each of the mathematical approximations for this typical commoncollector amplifier circuit:




What does each term in each expression represent, and why do they relate to one another as shown?
In a commoncollector transistor amplifier circuit with voltage divider biasing, the input impedance (Z_{in}) is a function of load impedance, emitter resistance (R_{E}), and the two biasing resistances (R_{1} and R_{2}). Often, the biasing resistances are of sufficiently low value to swamp the input impedance of the transistor, so that R_{1} and R_{2} constitute the heaviest load for any input signals driving the amplifier.


This is a shame, because the only practical purpose served by R_{1} and R_{2} is to provide a stable bias voltage so the transistor always functions in class A mode. In order to provide a stable bias, these resistors have to be relatively low in value compared to the impedance seen at the base of the transistor (resulting from the load). Otherwise, changes in dynamic emitter resistance (r′_{e}) could result in significant bias shifts. So, the naturally high input impedance of the commoncollector transistor configuration is spoiled by the necessary presence of R_{1} and R_{2}.
A clever way to recover some of that naturally large input impedance is to add a bit of regenerative (positive) feedback to the circuit in the form of a capacitor and another resistor. This technique is given an equally clever name: bootstrapping.

Explain how bootstrapping works, and why that particular name is given to the technique.
A common set of equations for calculating input and output impedances of commoncollector amplifier circuits is as follows:


If precision is not required, we may greatly simplify these equations by assuming the transistor to be ideal; i.e. having an infinite current gain (β = ∞). Rewrite these equations accordingly, and explain how you simplified each one.
Explain each of the mathematical approximations for this typical commonemitter amplifier circuit (with a bypass capacitor):




What does each term in each expression represent, and why do they relate to one another as shown?
Explain each of the mathematical approximations for this typical commonemitter amplifier circuit (with the dynamic emitter resistance ßwamped” by R_{E}):




What does each term in each expression represent, and why do they relate to one another as shown?
A common set of equations for calculating input and output impedances of bypassed commonemitter amplifier circuits is as follows:


If precision is not required, we may greatly simplify the first equation by assuming the transistor to be ideal; i.e. having an infinite current gain (β = ∞). Rewrite the first equation accordingly, and explain how you simplified it.
A parasitic property of semiconductor PN junctions is capacitance across the depletion regions. This is often referred to as the Miller Effect. In transistor circuits, the Miller effect contributes to a decrease in voltage gain as signal frequency increases.
Explain why junction capacitances make the voltage gain of an amplifier decrease with increasing frequency.
The BJT amplifier configuration most affected by the Miller effect at high frequencies is the commonemitter. Commoncollector and commonbase amplifier configurations do not suffer the same great losses of voltage gain at high frequency as the commonemitter circuit does. After examining the following amplifier circuits (with the Miller effect capacitance shown external to the transistors), explain why:


The “Miller capacitance” of a transistor in a commonemitter configuration is often expressed as the product of the transistor’s basetocollector junction capacitance (C_{BC}) and β+1:

Why is this? What purpose does it serve to include the transistor’s gain into the calculation, rather than just expressing the junction capacitance as it is?
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