A theoretical op-amp (operational amplifier) is the ideal voltage amplifier, with infinite input impedance and zero output impedance (in real devices, change "infinite" and "zero" to "very high" and "very low").
Very high input impedances create light loads, as the current is small through them. This means that the op-amp does not load its driver circuit.
MEMORY HINT:
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Ringing simply means that part of your circuit is at resonance and oscillating (like a bell). In some cases this is good - without oscillators there would be no CW! In other cases however, the ringing interferes with the function of the circuit. In this case, the filter is adding its own component to the output signal. This can sometimes be heard as a tone or whistle over the desired audio.
REMEMBER: Ringing is Undesired
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An ideal op-amp has infinite input impedance, but a real op-amp just has a very high impedance, typically at least 100 Megaohms. Op-amps with JFET inputs may have an input impedance in the range of thousands of Gigaohms (very high!).
MEMORY HINTS:
Rhyme - "I-I-Op-Amp Very High" where "input impedance" = I-I.
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An op-amp is DC powered and when simply connected and powered-up with no other components the output will contain a DC bias. However, applying an input voltage will result in it being amplified and combined with the DC bias leaving a new output voltage.
Because the op-amp will amplify a positive or negative voltage then it is possible to choose a specific input voltage that results in the combined amplified and bias voltage at the output equaling zero volts. This input voltage that leads to a zero volt output with no other circuitry is known as the input-offset voltage.
kd7bbc
Hint: The correct answer has the words "Differential & zero ".
-KE0IPR
Hint: Think about a scale. If it is "off-set" you try to get the scale to show zero when nothing is on it. (This is not just with a scale, lots of things are "off-set" from zero). In this case the answer has zero, and the question is talking about input offset voltage.
-KN4DJK
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To prevent unwanted ringing and audio instability in a multi-section op-amp RC audio filter circuit, restrict both gain and Q.
In order to keep an op-amp from going into oscillation, both the gain and the Q are restricted and set by the feedback circuit.
Hint: To prevent, you "restrict"
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In the 4 answers, only 3 are filters (contain the word filter). We know that a perfect (theoretical) op-amp would have a flat response so this is neither a low-pass or a high-pass.
We are left with the correct answer: As an audio filter in a receiver.
-chevdor
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By the voltage gain formula where \(A\) represents the operational amplifier (op-amp) voltage gain,
\begin{align} A_{\text{voltage}}&=\frac{V_\text{out}}{V_\text{in}}=-\frac{R_F}{R_1}=-\frac{470\:\Omega}{10\:\Omega}\\ &=-47 \end{align}
The magnitude is all that is required, so \(\mid A_{\text{voltage}}\mid=47\) is the answer.
Applying the ideal operational amplifier constraints we can determine circuit behavior.
Infinite gain implies zero difference between the inputs at steady state. Therefore the node at the negative input will be ground or 0 V. (a)
Infinite input impedance means that all current that flows through \(R1\) must flow out through \(R_F\). (b)
Putting (a) and (b) together for an input voltage V, will induce a current in R1 which is \(I = \frac{(V_{\text{in}}-0)}{R1}\), which must equal the current in \(R_F\) which is \(I = \frac{(0-V_{\text{out}})}{R_F}\).
From this, \(\frac{V_{\text{in}}}{R1} = -\frac{V_{\text{out}}}{R_F}\). Rearranging by multiply everything by \(R_F\) and dividing everything by \(V_{\text{in}}\) to get \(\frac{R_F}{R1} = -\frac{V_{\text{out}}}{V_{\text{in}}}\). Therefore in this configuration the gain is \(-\frac{R_F}{R1}\) which is \(-\frac{470\:\Omega}{10\:\Omega}\) or \(-47\). Again, because the only the magnitude is requested, the negative sign can be dropped.
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Ideal operational amplifiers can be found at your local physics store, right next to frictionless surfaces and massless springs...
They don't really exist!
If they did, they would have infinite input impedance (draw no current), zero output impedance, no offset, stable at all gains, and have flat frequency response from DC to Cosmic Rays...
When analyzing the behavior of an op-amp circuit, it is often useful to start with looking at it with an ideal op-amp in place to understand its theoretical "best" behavior. Included with that is the fact that an ideal op-amp does not have frequency-dependent gain.
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This schematic is of a simple one-stage inverting op-amp amplifier.
The gain, G, is defined as:
\[G = \frac{-R_\text{F}}{R_\text{in}}\]
Where:
\(R_\text{F}\) is the "feedback resistor" and
\(R_\text{in}\) is the "input resistor" (R1 in our
diagram).
\[G=\frac{-R_\text{F}}{{R1}}=\frac{-10000\:\Omega}{1000\:\Omega}=-10\]
The output voltage is the gain multiplied by the input voltage. Therefore, \[0.23\:\text{V} \times -10 = -2.3\:\text{V} \]
Explanation for the negative sign in the gain equation: The gain here is 10 times the input voltage, but the input is applied to the negative input of the amplifier and R_{F} from the output to the negative input to generate negative feedback to limit gain, making the voltage gain -10.
Another way to think about it is that with negative feedback, the opamp will try to make the two inputs equal, so the negative input will be at \(0V\). If there is a voltage divider of \(R1\) and \(R_F\) with \(.23V\) on one side and the middle is \(0V\), the far end of the divider needs to be \(-2.3V\); \(R1\) must drop \(.23V\), so Rf must drop \(10\times\) the voltage. Since \(IR1=IR_F\), \(\frac{Vin-0V}{R1} = \frac{0-V_\text{out}}{R_F}\); \(V_\text{out} = \frac{-R_F}{R1} \times V_\text{in} = -2.3V\)
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This schematic is of a simple one-stage inverting op-amp amplifier.
The gain, G, is defined as:
\[G = \frac{-R_\text{F}}{R_\text{in}}\]
Where:
\(R_\text{F}\) is the "feedback resistor" and
\(R_\text{in}\) is the "input resistor" (R1 in our
diagram).
The gain is negative because the inverting input is used. So, for this question:
\begin{align}
R_\text{in} = R1 &= 1800\:\Omega \\
R_\text{F} = 68\:\text{k}\Omega &= 68000\:\Omega
\end{align}
Therefore the gain is: \(G = \frac{-68000\:Ω}{1800\:Ω} = -37.7\)
The question asks for absolute voltage gain, so the sign of the answer is eliminated:
$\(\mid G\mid=37.7777...\approx38\)
Test trick: To remember the formula, note that in the diagram, the \(R_\text{F}\) is 'over' the \(R_\text{1}\).
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The gain, G, is defined as:
\[G = \frac{R_\text{F}}{R_\text{in}}\]
Where:
\(R_\text{F}\) is the "feedback resistor" and
\(R_\text{in}\) is the "input resistor" (R1 in our
diagram).
The absolute voltage gain that can be expected from the circuit in Figure E7-3 when R1 is 3300 ohms and \(R_\text{F}\) is 47 kilohms is:
\[G = \frac{47,000\:Ω}{3,300\:Ω} =14.2424...\approx14\]
Tip: Note that in the diagram, \(R_\text{F}\) is over R1, which gives you a hint of how to work the formula.
Hint: There are 4's in both the question and the answer. The answer has the only duplicated number from the question.
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An operational amplifier (op-amp) is a DC-coupled high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output. In this configuration, an op-amp produces an output potential (relative to circuit ground) that is typically hundreds of thousands of times larger than the potential difference between its input terminals. Its input impedance is very high and its output impedance is very low. (From Wikipedia)
HINT: I would search HIGH and LOW for a good OPERATIONAL AMPLIFIER.
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