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Elementary Amplifier Theory
10 questions By Tony R. Kuphaldt
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Question 7 of 10
A Class-A transistor amplifier uses a single transistor to generate an output signal to a load. The amplifier shown here happens to be of the “common collector” topology, one of three configurations common to single-transistor circuits:
An analogue for this electronic circuit is this water-pressure control, consisting of a variable valve passing water through an orifice (a restriction), then on to a drain:
The “input” to this amplifier is the positioning of the valve control handle. The “output” of this amplifier is water pressure measured at the end of the horizontal “output” pipe.
Explain how either of these “circuits” meets the criteria of being an amplifier. In other words, explain how power is boosted from input to output in both these systems. Also, describe how efficient each of these amplifiers is, “efficiency” being a measure of how much current (or water) goes to the load device, as compared to how much just goes straight through the controlling element and back to ground (the drain).
Reveal answerIn both systems, a small amount of energy (current through the “base” terminal of the transistor, mechanical motion of the valve handle) exerts control over a larger amount of energy (current to the load, water to the load). The systems shown here are rather wasteful, especially at high output voltage (pressure).
Notes:Wasteful they may be, but “Class-A” transistor circuits find very common use in modern electronics. Explain to your students that its inefficiency restricts its practical use to low-power applications.
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Question 8 of 10
A class-B transistor amplifier (sometimes called a push-pull amplifier) uses a pair of transistors to generate an output signal to a load. The circuit shown here has been simplified for the sake of illustrating the basic concept:
An analogue for this electronic circuit is this water-pressure control, consisting of two variable valves. One valve connects the output pipe to a supply of pressurized water, and the other connects the output pipe to a source of vacuum (suction):
The “input” to this amplifier is the positioning of the valve control handle. The “output” of this amplifier is water pressure measured at the end of the horizontal “output” pipe. Valve action is synchronized such that only one valve is open at any given time, just as no more than one transistor will be “on” at any given time in the class-B electronic circuit.
Explain how either of these “circuits” meets the criteria of being an amplifier. In other words, explain how power is boosted from input to output in both these systems. Also, describe how efficient each of these amplifiers is, “efficiency” being a measure of how much current (or water) goes to the load device, as compared to how much just goes straight from one supply “rail” to the other (from pressure to vacuum).
Reveal answerIn both systems, a small amount of energy (current through the “base” terminal of the transistor, mechanical motion of the valve handle) exerts control over a larger amount of energy (current to the load, water to the load). Both systems are very energy efficient, with little flow wasted by flowing from supply to vacuum (from V to -V) and bypassing the load.
Notes:Push-pull amplifiers are a bit more difficult to understand than simple class-A (single-ended), so be sure to take whatever time is necessary to discuss this concept with your students. Ask them to trace current through the load resistor for different input voltage conditions. Your students need not know any details of transistor operation, except that a positive input voltage turns on the upper transistor, and a negative input voltage turns on the lower transistor.
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Question 9 of 10
An amplifier has a voltage gain of 5 and a current gain of 75, both figures being ratios. Calculate the following gains:
- Power gain (as a ratio)
- Power gain (dB)
- Voltage gain (dB)
- Current gain (dB)
Reveal answer- Power gain (as a ratio) = 375
- Power gain (dB) = 25.74 dB
- Voltage gain (dB) = 13.98 dB
- Current gain (dB) = 37.50 dB
Notes:Some of your students will probably get the power calculations correct, but be off by a factor of two on the voltage and current gain (dB) calculations. Remind them that a different equation is used to calculate voltage and current gain in dB than is used in power calculations.
Question #10: “What is the overall voltage gain of two cascaded amplifiers (the output of the first amplifier going into the input of the second), each with an individual voltage gain of 3 dB?” Thought the answer to question #10, as 6 dB voltage expressed as a ratio, would be 3.98:1.
Correction: I thought a 6 dB voltage gain would be expressed in ratio as 1.99:1.
I’m also confused by the answer for the second half of question 10