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The negative feedback in an amplifier leads to which one of the following?
The negative feedback in amplifiers causes:
Av = V_{o}/V_{in}
Here, V_{o} is the output voltage of an amplifier and V_{in} is the input voltage of an amplifier.
A_{v} = V_{o} / V_{in }= 1/(1+A_{o}β)
Here, β = feedback factor,
A_{o} = openloop gain of the amplifier.
(gain × bandwidth = 0.35)
Applications of negative feedback to a certain amplifier reduced its gain from 200 to 100. If the gain with the same feedback is to be raised to 150, in the case of another such appliance, the gain of the amplifier without feedback must have been
Concept:
A negative feedback network is shown below.
Where A is the gain of the amplifier without feedback and feedback gain = β
Calculation:
Case 1:
Gain without feedback, A = 200
Gain with feedback A_{CL} = 100
Case 2:
Gain with feedback, A_{CL} = 150
Feedback,
An amplifier has a Open Loop voltage gain of –500. This gain is reduced to –100 when negative feedback is applied. The reverse transmission factor,β of this system is:
Concept:
The gain of a feedback system is given by:
A = Open Loop gain
A_{f} = Closed Loop Gain
β = Feedback/Transmission factor
Calculation:
Given A_{f} = 100 and A = 500
1 + Aβ = 5
Aβ = 4
β = 4 / 500 = 0.008
The feedback topology in the amplifier circuit (the base bias circuit is not shown for simplicity) in the figure is
In a current series feedback, current is sampled from the output and voltage is feedback to the source.
In the given amplifier circuit, the feedback signal becomes zero by opening the output feedback.
Hence, it is a current series feedback.
It is desired to reduce total harmonic of amplifier from 8% to 1% by use of 10% negative feedback. If the gain of the amplifier with original distortion and with reduced distortion is A_{1} and A_{2} then, A_{1} + A_{2} = −−−−−−
Given that β = 0.1and we know that
So we can write
∴ A_{1} + A_{2} =78.75
In a series shunt feedback network, feedback is connected in series with signal source but in shunt with the load. Error voltage from feedback network is in series with the input. Voltage fed back from output is proportional to output voltage, hence parallel or shunt connected. The current in and voltage out connection refers to a shuntshunt connection.
Consider a voltage series feedback network, where amplifier gain = 100, feedback factor = 5. For the basic amplifier, input voltage = 4V, input current=2mA. Find the input resistance of the network.
R_{I} = V_{I}/I_{I} = 4/2m = 2kΩ
R_{IF} = R_{I}(1+A.β) = 2k(1+500)
= 1002kΩ.
In seriesseries feedback, the output is current sampled, that is it is in series with the load. Also, input is a voltage mixer, which is in series with signal source. So feedback factor
Β = V_{F}/I_{L} in Ohms.
Consider given circuit.
What is the feedback configuration?
The resistance R4 is the feedback network resistance. There is no bypass capacitor being used. The resistance is not directly connected to either the input node or output node. Hence it’s a current series feedback.
In a feedback network, input voltage is 14V, feedback voltage is 6V and source voltage is 20V. β is in ohms. What is its configuration?
Given that input is 14V, feedback is 6V and source is 20 V, we can see
V_{I} = V_{S} – V_{F}, which is voltage mixing. Also, β is in ohms that is voltage/current.
Since output of feedback is voltage and input is current, the output has current sampling.
Thus, configuration is a seriesseries feedback/current – series feedback.
Find the relative change in gain with negative feedback given that return ratio is 24, and feedback factor is 3, when the change in open loop gain is 2.
_{F} = A/(1+Aβ)
Aβ = 24
A = 8
Relative change in gain = dA_{F}/A_{F} = dA/A(1+Aβ)
dA_{F}/A_{F} = 2/8*25 = 0.01.
In feedback systems, the feedback signal is in proportion with the output signal.
X_{F} ∝ X_{O}
X_{F} = βX_{O}, where β is the feedback factor or reverse transmission factor.
Which factor determines the gain of the voltage series feedback amplifier?
In setting the gain of the voltage series feedback amplifier, the ratio of two resistors is important and not the absolute value of these resistors. For example: If a gain of 11 is desired, we choose R_{1 }= 1kΩ and R_{1} = 10kΩ or R_{1} = 100Ω and R_{F} = 1kΩ.
A voltage shunt feedback amplifier forms a negative feedback because, any increase in the output signal results in a feedback signal into the inverting input causing a decrease in the output signal.
Which among the following is not a special case of voltage shut feedback amplifier?
A voltage follower is a special case of noninverting amplifier ( or voltage series feedback amplifier) and it has a gain of unity.
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