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Subelement E7
PRACTICAL CIRCUITS
Section E7G
Active filters and op-amp circuits: active audio filters; characteristics; basic circuit design; operational amplifiers
What is the typical output impedance of an integrated circuit op-amp?
  • Very low
  • Very high
  • 100 ohms
  • 1000 ohms

A theoretical op-amp 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.

With very low output impedances, all the voltage drop occurs in the load (picture an equivalent circuit with the output impedance in series with the load). This means that almost all of the voltage amplification is delivered to the load, which is a nice feature to have in an amplifier.

MEMORY HINT:

Think - (I)nput pointing to Very h(i)gh

Think - (O)utput pointing to Very l(o)w

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What is the effect of ringing in a filter?
  • An echo caused by a long time delay
  • A reduction in high frequency response
  • Partial cancellation of the signal over a range of frequencies
  • Undesired oscillations added to the desired signal

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. (hwwyy7nbf8wryhnyoahzhokg6gm=)

REMEMBER: Ringing is Undesired kg5kou

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What is the typical input impedance of an integrated circuit op-amp?
  • 100 ohms
  • 1000 ohms
  • Very low
  • Very high

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 HINT:

Think - (I)nput pointing to Very h(i)gh

Think - (O)utput pointing to Very l(o)w

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What is meant by the term op-amp input offset voltage?
  • The output voltage of the op-amp minus its input voltage
  • The difference between the output voltage of the op-amp and the input voltage required in the immediately following stage
  • The differential input voltage needed to bring the open loop output voltage to zero
  • The potential between the amplifier input terminals of the op-amp in an open loop condition

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.

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Hint: The correct answer is the only one with the word "Differential" in it.

-KE0IPR

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How can unwanted ringing and audio instability be prevented in a multi-section op-amp RC audio filter circuit?
  • Restrict both gain and Q
  • Restrict gain but increase Q
  • Restrict Q but increase gain
  • Increase both gain and Q

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.

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Which of the following is the most appropriate use of an op-amp active filter?
  • As a high-pass filter used to block RFI at the input to receivers
  • As a low-pass filter used between a transmitter and a transmission line
  • For smoothing power supply output
  • As an audio filter in a receiver

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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What magnitude of voltage gain can be expected from the circuit in Figure E7-4 when R1 is 10 ohms and RF is 470 ohms?
  • 0.21
  • 94
  • 47
  • 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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How does the gain of an ideal operational amplifier vary with frequency?
  • It increases linearly with increasing frequency
  • It decreases linearly with increasing frequency
  • It decreases logarithmically with increasing frequency
  • It does not vary with frequency

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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What will be the output voltage of the circuit shown in Figure E7-4 if R1 is 1000 ohms, RF is 10,000 ohms, and 0.23 volts DC is applied to the input?
  • 0.23 volts
  • 2.3 volts
  • -0.23 volts
  • -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.

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What absolute voltage gain can be expected from the circuit in Figure E7-4 when R1 is 1800 ohms and RF is 68 kilohms?
  • 1
  • 0.03
  • 38
  • 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:

\[\mid G\mid=37.7777...\approx38\]

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What absolute voltage gain can be expected from the circuit in Figure E7-4 when R1 is 3300 ohms and RF is 47 kilohms?
  • 28
  • 14
  • 7
  • 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 E7-4 when R1 is 3300 ohms and \(R_\text{F}\) is 47 kilohms is:

\[G = \frac{47,000\:Ω}{3,300\:Ω} =14.2424...\approx14\]

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What is an integrated circuit operational amplifier?
  • A high-gain, direct-coupled differential amplifier with very high input impedance and very low output impedance
  • A digital audio amplifier whose characteristics are determined by components external to the amplifier
  • An amplifier used to increase the average output of frequency modulated amateur signals to the legal limit
  • A RF amplifier used in the UHF and microwave regions

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)

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