Operational amplifiers, or op-amps for short, are integrated circuits that include a high-gain, direct-coupled differential amplifier with very high input and very low output impedance. They can be used to build amplifiers, filter circuits, and many other types of circuits that do analog signal processing.
QUESTION: What is an operational amplifier? (E7G12)
ANSWER: A high-gain, direct-coupled differential amplifier with very high input impedance and very low output impedance
QUESTION: What is the typical input impedance of an op-amp? (E7G03)
ANSWER: Very high
QUESTION: What is the typical output impedance of an op-amp? (E7G01)
ANSWER: Very low
While the gain of an ideal operational amplifier does not vary with frequency, op amps in the real world do have a finite bandwidth. Some modern op amps can be used at high frequencies, but many of the older ones can’t be used at frequencies above a couple of MHz. To find out if you can use an op amp at the frequency of your signals, check out the gain-bandwidth specification. The gain-bandwidth specification is the frequency at which the open-loop gain of the amplifier equals one.
QUESTION: How does the gain of an ideal operational amplifier vary with frequency? (E7G08)
ANSWER: It does not vary with frequency
QUESTION: What is the gain-bandwidth of an operational amplifier? (E7G06)
ANSWER: The frequency at which the open-loop gain of the amplifier equals one
Ideally, with no input signal, there should be no voltage difference between the two input terminals, and the output voltage should also be zero. Since no electronic component is ideal, there will be a voltage between these two terminals. We call this the input offset voltage. Put another way, the op-amp input-offset voltage is the differential input voltage needed to bring the open-loop output voltage to zero.
QUESTION: What is meant by the “op-amp input offset voltage”? (E7G04)
ANSWER: The differential input voltage needed to bring the open loop output voltage to zero
Because they are active components—that is to say that they amplify—filters made with op amps are called active filters. One use for an op-amp active filter is as an audio filter in a receiver. The values of capacitors and resistors external to the op-amp primarily determine the gain and frequency characteristics of an op-amp RC active filter.
Ringing is one undesirable characteristic of an op-amp filter. One effect of ringing in a filter is that it adds undesired oscillations to the desired signal. One way to prevent unwanted ringing and audio instability in a multi-section op-amp RC audio filter circuit is to restrict both gain and Q.
QUESTION: What is ringing in a filter? (E7G02)
ANSWER: Undesired oscillations added to the desired signal
QUESTION: How can unwanted ringing and audio instability be prevented in an op-amp RC audio filter circuit? (E7G05)
ANSWER: Restrict both gain and Q
Calculating the gain and output voltage of an op amp circuit is relatively straightforward. The gain is simply RF/Rin. In the op amp circuit shown in Figure E7-3, Rin = R1. The output voltage of a circuit is then the input voltage times the gain.
QUESTION: What magnitude of voltage gain can be expected from the circuit in Figure E7-3 when R1 is 10 ohms and RF is 470 ohms? (E7G07)
ANSWER: 47
If R1 is 10 ohms and RF is 470 ohms, the gain is 470/10, or 47.
QUESTION: What absolute voltage gain can be expected from the circuit in Figure E7-3 when R1 is 1800 ohms and RF is 68 kilohms? (E7G10)
ANSWER: 38
If R1 is 1800 ohms and RF is 68 kilohms, the gain is 68,000/1,800, or about 38.
QUESTION: What absolute voltage gain can be expected from the circuit in Figure E7-3 when R1 is 3300 ohms and RF is 47 kilohms? (E7G11)
ANSWER: 14
If R1 is 3300 ohms and RF is 47 kilohms, the gain is 47,000/3,300, or about 14.
QUESTION: What will be the output voltage of the circuit shown in Figure E7-3 if R1 is 1000 ohms, RF is 10,000 ohms, and 0.23 volts DC is applied to the input? (E7G09)
ANSWER: –2.3 volts
If R1 is 1000 ohms, RF is 10,000 ohms, the gain of the circuit will be 10,000/1,000 or 10, and the output voltage will be equal to the input voltage times the gain. 0.23 V × 10 = 2.3 V, but since the input voltage is being applied to the negative input, the output voltage will be negative.
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