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Question

The voltage gain of a certain amplifier without feedback is 10. With feedback its voltage gain reduces to 5. Determine its feedback factor.

The correct answer is

β = 0.1

To determine the feedback factor ($\beta$) of an amplifier, we use the relationship between its voltage gain without feedback (open-loop gain) and its voltage gain with feedback (closed-loop gain). This problem deals with an amplifier where the voltage gain is reduced due to negative feedback, which is a common characteristic of feedback systems.

Amplifier Gain Concepts

In amplifier circuits, two important gain parameters are considered:

  • Open-loop gain ($A$): This is the voltage gain of the amplifier circuit without any feedback applied. In this problem, the amplifier's voltage gain without feedback is given as $A = 10$.
  • Closed-loop gain ($A_f$): This is the voltage gain of the amplifier when a feedback path is introduced. For a negative feedback system, the closed-loop gain is always less than the open-loop gain. Here, the voltage gain with feedback is given as $A_f = 5$.

The feedback factor ($\beta$) represents the fraction of the output signal that is fed back to the input. For negative feedback, the general formula relating these quantities is:

$A_f = \frac{A}{1 + A\beta}$

Where:

  • $A_f$ = Closed-loop voltage gain
  • $A$ = Open-loop voltage gain
  • $\beta$ = Feedback factor

Feedback Factor Calculation

We are provided with the following values related to the amplifier's voltage gain and feedback:

  • Open-loop voltage gain, $A = 10$
  • Closed-loop voltage gain, $A_f = 5$

We need to find the feedback factor, $\beta$.

Let's substitute the given values into the formula for closed-loop gain:

$5 = \frac{10}{1 + 10\beta}$

Now, we will solve this equation for $\beta$ step-by-step:

Step 1: Multiply both sides by the denominator $(1 + 10\beta)$:

$5 \times (1 + 10\beta) = 10$

Step 2: Distribute the 5 on the left side of the equation:

$5 + 50\beta = 10$

Step 3: Subtract 5 from both sides of the equation to isolate the term with $\beta$:

$50\beta = 10 - 5$

$50\beta = 5$

Step 4: Divide both sides by 50 to find the value of $\beta$:

$\beta = \frac{5}{50}$

$\beta = \frac{1}{10}$

$\beta = 0.1$

Feedback Factor Result

The calculated feedback factor for the amplifier is $0.1$. This value indicates the fraction of the output signal that is fed back to the input to achieve the reduced voltage gain of 5 from an initial gain of 10.

The final answer is $\beta = 0.1$.

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Important Questions from Feedback Amplifier

  1. Arrange the following amplifiers in the order of maximum achievable -3dB frequency. Assume the transconductance ($g_m$) of the driver transistor (M1) same in all the circuits. 

    Choose the correct answer from the options given below:

  2. Arrange the following amplifiers in the order of maximum achievable voltage gain. Assume the geometry and overdrive voltage is same for the transistor M1 in all the circuits and $R_D < <r_{o1}$.

    choose the correct answer from the options given below:

  3. Match the following lists :

    List – I List – II 
    a. Voltage series feedbacki. Trans resistance amplifier
    b. Current series feedbackii. Current shunt feedback
    c. Current amplifieriii. Trans-conductance
    d. Voltage shunt feedbackiv. Voltage amplifier

    Codes :

  4. Match the following :

    List – I List – II 
    a. voltage shunt negative feedbacki. increase of CMRR
    b. constant current source differential amplifierii. O/P voltage attenuated by a factor 1/29
    c. Phase shift oscillatoriii. FSK decoder
    d. PLLiv. decrease of O/P impedance

    Codes :

  5. Negative feedback in amplifier results in

    1. reduced voltage gain
    2. reduced bandwidth
    3. increased S/N ratio
    4. reduced distortion

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