PRACTICAL CIRCUITS
What is the frequency response of the circuit in E7-3 if a capacitor is added across the feedback resistor?
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High-pass filter
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Correct AnswerLow-pass filter
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Band-pass filter
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Notch filter
The impedance of a capacitor is infinite at DC and decreases as the frequency increases. Putting a capacitor across Rf would thus act more and more like a short circuit as the frequency of this circuit increases. Therefore, this is a Low-pass filter.
Hint: Coach would want Feedback if it was a Low Pass
Last edited by jerryfaughn. Register to edit
What voltage gain can be expected from the circuit in Figure E7-3 when R1 is 10 ohms and RF is 470 ohms?
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0.21
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4700
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Correct Answer47
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24
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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What will be the output voltage of the circuit shown in Figure E7-3 if R1 is 1,000 ohms, RF is 10,000 ohms, and 0.23 volts DC is applied to the input?
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0.23 volts
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2.3 volts
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-0.23 volts
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Correct Answer-2.3 volts
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 RF 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 \(IR1=IR_F\), \(\frac{Vin-0V}{R1} = \frac{0-V_\text{out}}{R_F}\); \(V_\text{out} = \frac{-R_F}{R1} \times V_\text{in} = -2.3V\)
Last edited by kd7bbc. Register to edit
What absolute voltage gain can be expected from the circuit in Figure E7-3 when R1 is 1,800 ohms and RF is 68 kilohms?
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1
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0.03
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Correct Answer38
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76
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:
\[\left| G\right| =37.7777...\approx38\]
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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What absolute voltage gain can be expected from the circuit in Figure E7-3 when R1 is 3,300 ohms and RF is 47 kilohms?
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28
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Correct Answer14
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7
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0.07
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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