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A certain op amp has a unity-gain bandwidth of \(f_{t}=15 \mathrm{MHz}\). If this op amp is used in a noninverting amplifier having a closed-loop dc gain of \(A_{0 C L}=10\), determine the closedloop break frequency \(f_{B C L}\). Repeat for a dc gain of 100 .

Short Answer

Expert verified
For \( A_{0 CL} = 10 \), \( f_{B CL} \) is 1.5 MHz; for \( A_{0 CL} = 100 \), \( f_{B CL} \) is 0.15 MHz.

Step by step solution

01

Define the Known Values

The problem provides the following values: a unity-gain bandwidth \( f_t = 15 \text{ MHz} \) and a desired closed-loop dc gain \( A_{0 CL} \). The solution will first calculate for \( A_{0 CL} = 10 \) and then for \( A_{0 CL} = 100 \).
02

Use the Gain-Bandwidth Product Formula

The gain-bandwidth product for an operational amplifier is given by the formula \( f_t = A_{0 CL} \times f_{B CL} \), where \( f_{B CL} \) is the break frequency for the closed-loop gain. This relationship holds true because the gain-bandwidth product remains constant regardless of the configuration.
03

Solve for Closed-loop Break Frequency when \( A_{0 CL} = 10 \)

Substitute \( A_{0 CL} = 10 \) into the gain-bandwidth product formula:\[ f_{B CL} = \frac{f_t}{A_{0 CL}} = \frac{15 \text{ MHz}}{10} = 1.5 \text{ MHz}\].So, for a closed-loop gain of 10, \( f_{B CL} = 1.5 \text{ MHz} \).
04

Solve for Closed-loop Break Frequency when \( A_{0 CL} = 100 \)

Substitute \( A_{0 CL} = 100 \) into the gain-bandwidth product formula:\[ f_{B CL} = \frac{f_t}{A_{0 CL}} = \frac{15 \text{ MHz}}{100} = 0.15 \text{ MHz}\].Thus, for a closed-loop gain of 100, \( f_{B CL} = 0.15 \text{ MHz} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unity-Gain Bandwidth
The unity-gain bandwidth is a vital specification of an operational amplifier (op-amp). It represents the frequency at which the gain of the op-amp drops to one, meaning it neither amplifies nor attenuates the input signal. Let's explore what this means in simpler terms.
Imagine you are listening to a song, and you have a volume knob that represents the gain. When the unity-gain bandwidth is reached, the knob is set such that the music is as loud as it was recorded; it neither gets softer nor louder.
In technical terms, the unity-gain bandwidth is denoted by the symbol \(f_t\). It tells us the maximum frequency the op-amp can handle without amplifying the signal. For example, if an op-amp has a unity-gain bandwidth of \(15 \text{ MHz}\), it means it can pass signals up to \(15 \text{ MHz}\) without any amplification.
Closed-Loop Gain
Closed-loop gain is a measure of how much an op-amp amplifies an input signal when negative feedback is applied. Negative feedback is a technique used to stabilize an amplifier and control its gain. In other words, closed-loop gain is what you get when the op-amp's output is fed back to its input.
In a closed-loop system, the gain of the amplifier is less than the open-loop gain, but it is far more predictable and stable. The term is represented by \(A_{0 CL}\) in formulas. For instance, a closed-loop gain of 10 means that the output signal is 10 times stronger than the input signal.
By using closed-loop configurations, op-amps can behave more reliably and linearly, ensuring consistent performance in various circuits.
Gain-Bandwidth Product
The gain-bandwidth product is an important concept when working with op-amps, as it helps predict how an op-amp will perform in different configurations. It is the product of the amplifier's gain and its bandwidth, essentially staying constant for a given op-amp.
Here's how it works: for any given op-amp, if you increase the gain, the bandwidth must decrease to maintain the same gain-bandwidth product. Conversely, if you decrease the gain, the bandwidth increases. This relationship is crucial when designing circuits so that you can balance between gain and bandwidth.
For example, if an op-amp has a gain-bandwidth product of \(15 \text{ MHz}\), and you configure it for a gain of 100, the bandwidth will be \(0.15 \text{ MHz}\) to maintain that constant product. This principle is vital in ensuring that the op-amp works efficiently across its operable frequency range.
Break Frequency
The break frequency, also known as cutoff frequency, is a point where the gain of an amplifier falls to \( \frac{1}{\sqrt{2}}\) of its maximum value, which is approximately 0.707 times the gain. This is the frequency beyond which the amplifier can no longer provide consistent gain.
In the context of a closed-loop system, the break frequency, \(f_{B CL}\), is a critical factor in determining how the op-amp can handle higher frequencies with a particular gain setting. If you set a gain, the break frequency can change, showing you the effective bandwidth your circuit can handle.
In practice, knowing the break frequency helps you ensure that your amplifier performs optimally within its limits. For instance, with \(A_{0 CL} = 10\), the break frequency would be 1.5 MHz for our 15 MHz unity-gain bandwidth example, indicating the maximum frequency span for reliable operation.

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Most popular questions from this chapter

We have an inverting amplifier using 5 -percent-tolerance resistors and an ideal op amp. The nominal amplifier gain is \(-10\). What are the minimum and maximum gains possible, assuming that the resistances are within the stated tolerance? What is the percentage tolerance of the gain?

Give the definitions of the differential input voltage and of the common-mode input voltage for a differential amplifier having input voltages \(v_{1}\) and \(v_{2}\).

A voltage follower is needed to amplify an input signal given by $$ \begin{aligned} v_{\text {in }}(t) &=0 & & t \leq 0 \\ &=t^{3} & & 0 \leq t \leq 2 \\ &=8 & & 2 \leq t \end{aligned} $$ in which \(t\) is in \(\mu \mathrm{s}\). Determine the minimum slew-rate specification required for the op amp if distortion must be avoided.

We have the following two alternatives for designing an amplifier having a de gain of 25 . The first alternative is to use a single noninverting stage, having a gain of 25 . The second alternative is to cascade two noninverting stages, each having a gain of \(5 . \mathrm{Op}\) amps having a gain-bandwidth product of \(10^{6}\) are to be used. Draw the circuit diagram, write an expression for the gain as a function of frequency, and find the \(3-\mathrm{dB}\) bandwidth for each alternative. Which alternative has the greater bandwidth?

A certain op amp has an open-loop dc gain of \(A_{0 O L}=200,000\) and an open- loop \(3-\mathrm{dB}\) bandwidth of \(f_{B O L}=5 \mathrm{~Hz}\). Sketch the Bode plot of the open-loop gain magnitude to scale. If this op amp is used in a noninverting amplifier having a closed-loop dc gain of 100 , sketch the Bode plot of the closedloop gain magnitude to scale. Repeat for a closed-loop dc gain of 10 .

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