Circuits with Feedback - Assignment Example

Running Header: CIRCUITS Circuits with Feedback . .. Feedback Questions Problem 1: Show how the equation for the gain of a feedback amplifier is obtained Solution: We know that when a portion of output signal (voltage or current) reaches back to the input side of an amplifier and if it influence it then we say that a a feedback has occurred. Also if the output signal is out of phase of in put signal then a negative feedback occurs and if output signal reinforces the input signal i.e. if both are in phase then a positive feedback results (Saeed, 1998, pp.615). Derivation of Feedback Equation for Non-inverting Amplifier Let we consider a non-inverting amplifier along with its block diagram as shown in fig. 1 (Bogart, 1997, p. 665). Let the block labeled 'A, represents the amplifier and its open loop gain and block ', is the feedback path with ', feed back ratio and let ', be the summing junction with 'vin, be input voltage and 've, output voltage also called error voltage (Bogart, 1997, pp. 664-665). Since the output voltage've, of junction is the input of amplifier hence if 'vo, be output voltage of amplifier and 'vf , then, A = , A = Where ve = vin - vf Therefore, vo = A(vi - vf) . ... (1) Also = Or vf = vo . (2) Therefore substituting equation 2 in equation 1 we have, vo = A(vin - vo) vo = Avin -A vo vo(1 + A )= Avin Thus = .. (3) This is required gain of non-inverting amplifier. Fig. 1(a) Non-inverting amplifier Fig. 1(b) Block diagram of Non-inverting amplifier Now if we apply we apply the above derived result to non-inverting amplifier of Fig. 1(a) then as, vi'= vo As 'vi', is the voltage fed back from the output, and vf = vo, therefore it implies that, = for non-inverting amplifier. Now if we substitute this value in equation (3) then it turns to be, = = . (4) Equation (4) is the closed-loop voltage gain of a no-inverting amplifier with feedback in terms of open-loop gain A. Derivation of Feedback Equation for Inverting Amplifier Let we consider an inverting amplifier as shown in fig. 2 (Bogart, 1997, p. 670). Then if 'A, is the gain then, as vi+ = 0, there fore vo = A(vin+ - vin) vo = - A vin Fig. 2(a) Also from Ohm's Law the current is simply the difference in the voltage across R1 divided R1, i1 = (vin+ - vin)/R1 (5) Similarly, if = ( vi - vo)/Rf (6) By Kerchief's current Law at the inverting input, i1 = if + i where i is the current entering the amplifier at it's inverting point, yet for an idea amplifier it reaches to zero hence, i1 = if (7) Substituting the values from equations (5) and (6) we have, (vin+ - vin)/R1= (vi - vo)/Rf (8) Or vin+/R1 - vin/R1= vi/Rf - vo/Rf Solving for , = (9) This is the required voltage gain expression for inverting amplifier with feedback in terms of open loop gain. Problem 2: Each of the amplifiers shown below incorporate series feedback. The gains, input resistances and output resistances are quoted without feedback. For each amplifier determine: - a) The feedback fraction. b) The gain with feedback. c) The input impedance with feedback. d) The output impedance with feedback. Solution: (i) a) If we consider the given cct. then for feedback fraction ', for given cct. is defined as, = (2ia) Substituting the values in equation (2ia) from given cct. we have, = = 0.175 2ia Fig. (2i) (2ib) If 'Avf, be the gain with feedback the as 'Avf, in terms of feedback fraction ', and open given loop gain 'Av, is given as, Avf == Substituting the values of Av= -250 (given), and = 0.174 (from result 2a) Avf = Avf =-5.574 ..2ib (2ic) The input impedance 'Zi, with feedback for an inverting amplifier is given as Zi = R9+ Substituting the values in above equation, Zi = 1K+ 1K Zi = 1K .2ic (2id) The output impedance 'Zo, with feedback for an inverting amplifier is given as, Zo(stage) = Substituting the values in above equation, Zo(stage) = Zo(stage) = 105 .2id (2ii) a) Let we consider fig. (2ii) then for feedback friction ', for given cct. is given as, = or after substituting the values in above equation we have, = = 0.047 2iia (2ii) b). Let we consider fig. (2ii) then for 'Avf, in terms of feedback fraction ', and open given loop gain 'Av, is given as, Avf == Substituting the values of Av= 180 (given), and = 0.047 (from result 2aii) Avf = Avf = 19.2 ..2iib (2ii) c) The input impedance 'Zi, with feedback for an inverting amplifier is given as Zi = R17+ Substituting the values in above equation, Zi = 470+ Zi = 515 .2iic Fig. (2ii) (2ii) d) The output impedance 'Zo, with feedback for an inverting amplifier is given as Zo(stage) = Substituting the values in above equation, Zo(stage) = Zo(stage) = 866 .2iid Problem 3 a) The amplifier shown below has a gain of 150. Negative feedback is to be Introduced by splitting the emitter resistor and connecting the bypass capacitor across the top half. Calculate the value of resistors required to reduce the gain to 20. (NB the total resistance between emitter and ground must not be changed). b) It is hoped that the amplifier will be able to be made using several different transistor types without having to change the design. If the transistor type were changed for one that gave a gain of 180 without feedback, what would be the effect on the amplifier with feedback Solution 3a) Let Rf be the required resistance the as we know feed back voltage gain in terms of open loop gain and feed back friction is given as Avf == . 3ai Substituting the values, 20 = Solving for , we have, = 0.043. Since for given cct. is defined as, = or after substituting values, 0.043 = Solving for Rf, Rf = 957 be required resistance reduce the gain to 20. 3 b) Let we consider an inverting amplifier as shown in fig. 2. Then if 'A, is the gain then, as vi+ = 0, there fore vo = A(vin+ - vin) vo = - A vin . (3b1) Fig. 3(b) Also from Ohm's Law the current is simply the difference in the voltage across R1 divided R1, i1 = (vin+ - vin)/R1 (3b2) Similarly, if = ( vi - vo)/Rf (3b3) By Kerchief's current Law at the inverting input, i1 = if + i where i is the current entering the amplifier at its inverting point, yet for an idea amplifier it reaches to zero hence, i1 = if (3b4) Substituting the values from equations (5) and (6) we have, (vin+ - vin)/R1= (vi - vo)/Rf Or vin+/R1 - vin/R1= vi/Rf - vo/Rf (3b5) Now from equation (3b1), vin= (3b6) Since for an ideal amplifier 'A, should be very large and approaches to infinity, Therefore, vin= = 0 = or = (3b7) Equation (3b7) reveals that magnitude of depends on the ratio of external resistors values and not on the amplifier (provided amplifier gain and impedance remain quite large). Hence variations in amplifier characteristics do not affect and feed back (Bogart, 1997, pp. 660). Problem 4 Produce a voltage-series feedback network for each of the following amplifiers to give the gain Avf required. Solution: The figure (4) (Saeed,1998, p.620) is the required generalized diagram of required feedback network. Here the signal feedback to input side of amplifier is series and in the form of voltage signal and given as, vfb = vo with same current through the network. The signal Sampled from output side is in the form of output voltage (taken as shunt) and proportionate signal vfb = vo is applied to the in put side as negative feed back signal. Here we see that output of the basic amplifier and feedback network are in parallel having with the same voltage vo that is why the resultant arrangement is called as series- shunt type feedback or voltage amplifier with voltage series feed back. Fig. (4) a). Since Av = - 300 and Avf = - 25 are given values, To have required gain the input resistance and output resistance must be adjusted again. Since for voltage series feedback network input feed back resistance (Saeed, 1998, pp. 619-620), Rif = ria (1+ Avf) Fig. (4a) Where Rif = Input feedback resistance ria = input amplifier resistance Substituting the Rif = ria (1+ 25) ( Avf = Avf ) Hence for required gain, Rif = 26ria Also out put resistance for desired results be, Rof = roa / (1+ Avf) Where Rof = output feedback resistance roa = output amplifier resistance Rof =0.038roa b). Similarly for Av = 25000 and Avf = 150 Rif = 151ria Rof = 6.62roa Fig. (4b) c.) Input feedback resistance in terms of open loop gain and input resistance are given as Rif = ria (1+ Av) Where = =0.97 Therefore, Rif = ria (1+0.97120000)=116401ria And Rof = roa/ (1+ Av) Rof = roa/ (1+0.97120000)0 Fig. 4(c) References Bogart, T. Jr. (1997). Electronic Devices and Circuits. Upper Saddle River New Jersey: Prentice Hall. Saeed, M. (1998). Electronic Devices and Circuits. Lahore: A-1 Publishers Sedra, S.A. & Smith, K. C. (1998) Microelectronic Circuits. 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