Login or Register for FREE!
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 op-amp?
  • Correct Answer
    Very low
  • Very high
  • 100 ohms
  • 1000 ohms

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.

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

Last edited by r1j1s1. Register to edit

Tags: arrl chapter 6 arrl module 6b

What is 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
  • Correct Answer
    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.

REMEMBER: Ringing is Undesired

Last edited by w9mpx. Register to edit

Tags: arrl chapter 6 arrl module 6h

What is the typical input impedance of an op-amp?
  • 100 ohms
  • 1000 ohms
  • Very low
  • Correct Answer
    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 megohms. Op-amps with JFET inputs may have an input impedance in the range of a few terohms (very high!).


MEMORY HINTS:

  • Think - (I)nput pointing to Very h(i)gh
  • Think - (O)utput pointing to Very l(o)w

Rhyme - "I-I-Op-Amp Very High" where "input impedance" = I-I.

Last edited by ironcal67. Register to edit

Tags: arrl chapter 6 arrl module 6b

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
  • Correct Answer
    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.

Hint: The correct answer has the words "Differential & zero ".

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.

Last edited by kd7bbc. Register to edit

Tags: arrl chapter 6 arrl module 6b

How can unwanted ringing and audio instability be prevented in an op-amp RC audio filter circuit?
  • Correct Answer
    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.

Hint: To prevent, you "restrict"

Last edited by josephdfarkas. Register to edit

Tags: arrl chapter 6 arrl module 6h

What is the gain-bandwidth of an operational amplifier?
  • The maximum frequency for a filter circuit using that type of amplifier
  • Correct Answer
    The frequency at which the open-loop gain of the amplifier equals one
  • The gain of the amplifier at a filter's cutoff frequency
  • The frequency at which the amplifier's offset voltage is zero

Op-amps have a trade off between frequency and gain. This is expressed on datasheets as a gain-bandwidth product, the frequency where the op-amp has unity gain. Circuits where the op amp is operated near this frequency won't work as well as ones where the frequency range is lower.

— wubbles

Hint: (op)-amp, (op)en loop; the only answer with "op" in it

— serif

Silly Hint: 'an operational amplifier' an = one, 'one' in the answer

Hint: Only answer with hyphenated term, and question has one too.

Last edited by kk4wbm. Register to edit

Tags: none

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?
  • 0.21
  • 94
  • Correct Answer
    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.

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 6 arrl module 6b

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
  • Correct Answer
    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.

Last edited by orochimaru. Register to edit

Tags: arrl chapter 6 arrl module 6b

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?
  • 0.23 volts
  • 2.3 volts
  • -0.23 volts
  • 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

Tags: arrl chapter 6 arrl module 6b

What absolute voltage gain can be expected from the circuit in Figure E7-3 when R1 is 1800 ohms and RF is 68 kilohms?
  • 1
  • 0.03
  • Correct Answer
    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:

\[\left| G\right| =37.7777...\approx38\]

Test trick: To remember the formula, note that in the diagram, the \(R_\text{F}\) is 'over' the \(R_\text{1}\).

Last edited by myzhang1029. Register to edit

Tags: arrl chapter 6 arrl module 6b

What absolute voltage gain can be expected from the circuit in Figure E7-3 when R1 is 3300 ohms and RF is 47 kilohms?
  • 28
  • Correct Answer
    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 E7-3 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.

Last edited by yr7 - extra. Register to edit

Tags: arrl chapter 6 arrl module 6b

What is an operational amplifier?
  • Correct Answer
    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)

HINT: I would search HIGH and LOW for a good OPERATIONAL AMPLIFIER.

Last edited by yr7 - extra. Register to edit

Tags: arrl chapter 6 arrl module 6b

Go to E7F Go to E7H