Nunit 2 feedback
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The document discusses the principles and benefits of feedback amplifiers, including their characteristics such as gain de-sensitivity, bandwidth extension, noise reduction, and reduction in nonlinear distortion. It explores various types of feedback configurations and their effects on amplifier performance, emphasizing that negative feedback can significantly improve the linearity and performance of amplifiers. Additionally, it provides a mathematical analysis of gain changes due to feedback and describes various feedback amplifier types and their configurations.
Nunit 2 feedback
- 1. UNIT-2 Mohammad Asif Iqbal Assistant Professor, Deptt of ECE, JETGI, Barabanki
- 2. General Feedback Structure Source (ME) ∑ A (YOU) Load (Result) B (Feed back) xs + xi xo xf xo =Axi xf=Bxo xi=xs -xf
- 3. Continue… The gain of the feedback amplifier can be obtained by combining all three equations Loop gain Close loop gain open loop gain •From the above equation we can observe that the gain with feedback will be smaller than the open loop gain by the quantity 1+ AB, which is called the amount of feedback. •If , as in the case in many circuits the loop gain AB is large, AB 1, then from above equation it follows that This is something very interesting: the gain is almost entirely determined by the feedback network. Since the feedback network usually consist passive components, which can be chosen to be as accurate as one wishes
- 4. Some properties of negative feedback •Gain de-sensitivity •Bandwidth extension •Noise reduction •Reduction in nonlinear distortion
- 5. 1#Gain de-sensitivity We have Assume that β is constant. Taking differentials of both sides of above Eq. results in Dividing Eq. 2 by Eq. 1 yields Eq -1 Eq -2 which says that the percentage change in 𝐴 𝑓 (due to variations in some circuit parameter) is smaller than the percentage change in A by a factor equal to the amount of feedback. For this reason, the amount of feedback, 1 + Aβ, is also known as the desensitivity factor.
- 6. Upper 3-dB frequency 2#Bandwidth extension Consider an amplifier whose high-frequency response is characterized by a single pole. Its gain at mid and high frequencies can be expressed as Mid-band gain Upper 3-dB frequency What are these? Now suppose if we apply a negative feedback, with a frequency independent factor B, around this amplifier, we will get a closed-loop gain Put the value of A(s) from above equation
- 7. Continue.. After a little manipulation, we will get. Recall the gain equation without feedback Comparing both, you may observe Midband gain has changed from And upper 3-dB frequency has changed
- 8. More about bandwidth extension
- 9. 3#Noise reduction Negative feedback can be employed to reduce the noise or interference in an amplifier or, more precisely, to increase the ratio of signal to noise. Lets consider following amplifier with gain 𝐴1, an input signal 𝑉𝑠, and noise, or interference, 𝑉𝑛. So the signal to noise ratio for this amplifier is S/N = 𝑉𝑠/𝑉𝑛
- 10. Continued… Now consider this We have assume that it is possible to build another amplifier stage with gain A2 that does not suffer from the noise problem. This feedback is connected across the overall cascade of such an amount to keep the overall gainconstant. Now lets find the output voltage of this circuit byusingsuperposition: WhatwillbenewS/N?
- 11. But this is not the case when nonlinearity comesinthepicture 4#Reduction in nonlinear distortion Because of negative feedback, system becomes more liner, it can handle larger signals at output without distortion occurring. For understanding these thing, consider the output of following system A (YOU) xi xo xo =Axi xo =xoffset + Axi + K2xi 2 + K3xi 3…. A function which is independentof xi Nonlinearities Any amplifier which is working in active reason, then the term Axi is dominating and other factors are forced to zero by choosing an appropriate operating point. But this is possible for certain signal levels only, as the signal levels increases these other factors will play there part
- 12. Now lets connect the feedbcak ∑ A (Nonlinear) B (Linear) xs + xo =Axi xf=Bxo xi=xs -xf - xo =xoffset + Axi + K2xi 2 + K3xi 3….
- 13. NOW, LETS TALK ABOUT FOUR BASIC TYPES OF FEEDBACK AMPLIFIER… •The series-shunt or voltage series or series voltage feedback amplifier •The series-series or current series or series current feedback amplifier •The shunt-shunt or voltage shunt or shunt voltage feedback amplifier •The shunt-series or current shunt or shunt current feedback amplifier Shunt:-voltage Series:-current These represents the type of connection between output- feedback These represents the type of connection between input- feedback In input we will decide the parameter by consideringthefollowingfact: We will take that parameter which get divided in the givenconnection type.Meaninseries:-voltage & in shunt-current In these cases the given connection type is for input, and the parameter type tells you abouttheoutputconnection
- 14. Continue.. xo =xoffset + Axi + K2xi 2 + K3xi 3…. xi=xs -xf xf=Bxo xi=xs –Bxo = Є xo =xoffset+A(xs –Bxo)+K2Є2 +K3Є3 xo+ABxo=xoffset+Axs +K2Є2 +K3Є3 xo=xoffset /(1+AB)+Axs/(1+AB)+(K2Є2 +K3Є3)/(1+AB) Upper 3-dB frequency Replacing xi by these values Replacing xi by these values As these values are large enough, we can ignore these fractions xo= Axs/(1+AB) Finally we get an expression free from nonlinearities
- 15. 1#The series-shunt feedback amplifier Vs + Vo - + Vo - + Vf =BVo - + Vi - Ri + AVi - RoIi An amplifier, with gain A, Input resistance Ri output resistance Ro - Vf + This is a mixer, Feedback network, with gain B.
- 16. 1#The series-shunt feedback amplifier •The output signal voltage is sampled and is applied as a voltage into the mixer •Both sampled and feedback signals are voltage •It is also known as voltage amplifier whose gain is •An ideal voltage amplifier has infinite input resistance, and zero output resistance. •The best practical examples are emitter follower, source follower, voltage follower.
- 17. To find input and output resistances, Rif with feedback so, Rif = Vs / Ii =(Vi +Vf)/ Ii but, Vf = BVo = BAVi Then, Rif = (Vi + BAVi ) / Ii Rif = (1+AB) Vi/ Ii Rif = DRi Ri = Vf / Ii We know, the input resistance of the amplifier without feedback is Ri And from the figure it is,
- 18. To find Rof The standard procedure to find the output resistance of amplifier is to reduce the Vs to zero, and apply a test voltage Vt at the output, as shown in fig. Rof = Vt / I From figure we can write I = And since Vs =0 we can conclude that Vi = −Vf = −BVo = −BVt Thus Leading to + AVi - Ro Vt I 0