Discrete Semiconductor Devices and Circuits

Class A BJT Amplifiers

62 questions By Tony R. Kuphaldt

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Question 10 of 62

The following schematic diagram shows a simple common-emitter transistor amplifier circuit:

Explain why the voltage gain (A_V) of such an amplifier is approximately $\frac{R_C}{R_E}$, using any or all of these general “rules” of transistor behavior:

$I_E = I_C+I_B$

$I_E \approx I_C$

$V_{BE} \approx 0.17\ volts$

$\beta = \frac{I_C}{I_B}$

Remember that (AC) voltage gain is defined as $\frac{\Delta V_{out}}{\Delta V_{in}}$. Hint: this question might be easier to answer if you first consider how to explain the unity-gain of a common-collector amplifier circuit (simply eliminate R_C, replacing it with a direct connection to −V, and consider V_E to be the output voltage).

Reveal answer

Since V_BE is relatively constant, ∆V_in ≈ ∆V_E. The next essential step in the explanation for the voltage gain formula is to couple this fact with I_E ≈ I_C. The rest I’ll leave for you to explain.

For your discussion response, be prepared to explain everything in mathematical terms. You will have to use Kirchhoff’s Voltage Law at least once to be able to do this completely.

Notes:
Although the given answer seems complete, what I’m looking for here is a good analytical understanding of why the voltage gain is what it is. Placing the requirement of using KVL on the students’ answers ensures that they will have to explore the concept further than the given answer does.
Question 11 of 62

Determine what would happen to the voltage gain of a common-emitter 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 . . .

Reveal answer

Resistance R_C increased; A_V increases

Resistance R_E increased; A_V decreases

Resistance R_bias1 increased; A_V does not change

Resistance R_bias2 increased; A_V does not change

Notes:
The purpose of this question is to get students to apply their knowledge of common-emitter amplifier voltage gain to hypothetical changes in resistance. These are important concepts, so be sure to discuss them adequately, challenging your students to explain why the voltage gain is affected as described, not just explained by blindly following a gain formula.

Many students experience difficulty understanding why voltage gain is directly proportional to collector resistance. What they visualize when they consider a greater collector resistance is less collector voltage, which they understandably equate to less output signal and thus less gain. While the quiescent (DC) output voltage does decrease with increasing R_C, what is not so obvious is that change in collector voltage (∆V_C) increases with increasing R_C.

Ask your students to explain why changes in bias resistor values do not (significantly) effect voltage gain. Does this mean the values are arbitrary? Discuss with them the purpose of bias resistors, if necessary, and what would happen if they were not there or if they were grossly mis-sized.
Question 12 of 62

A student attempts to calculate the voltage gain of the following common-emitter amplifier circuit, and arrives at an incalculable value (divide-by-zero 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 self-destruction of the transistor!).

Reveal answer

The emitter resistance of a transistor dynamically changes with emitter current and with semiconductor temperature, which is why it is often called the dynamic emitter resistance. A commonly approximation for its value is this:

$$r’_e \approx \frac{25mV}{I_E}$$

Follow-up question: explain why this dynamic emitter resistance is often ignored when calculating voltage gain in a common-emitter circuit such as this:

Notes:
This question may serve as a good starting point for a discussion on thermal runaway, discussing how r′_e decreases with temperature, increasing I_E, once again decreasing r′_e, an infinitum, ad destructum.

The follow-up question provides a good opportunity to discuss the engineering principle of swamping: when two quantities are unequal to the extent that one renders the other relatively insignificant. This concept is very important in analysis because it allows us to construct simpler models of realistic processes than we could if we had to take every factor into account. It is also important in design because it allows us to overshadow certain unwanted effects.