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Chapter 8 - Feedback 1 - Desensitize The Gain 2 - Reduce Nonlinear Distortions 3 - Reduce The Effect of Noise 4 – Control The Input And Output Impedances 5 – Extend The Bandwidth Of The Amplifier The General Feedback Structure Basic structure of a feedback amplifier. To make it general, the figure shows signal flow as opposed to voltages or currents (i.e., signals can be either current or voltage). Source xs xi A xf b xo Load The open-loop amplifier has gain A xo = A*xi Output is fed back through a feedback network which produces a sample (xf) of the output (xo) xf = bxo Where b is called the feedback factor The input to the amplifier is xi = xs – xf (the subtraction makes feedback negative) Implicit to the above analysis is that neither the feedback block nor the load affect the amplifier’s gain (A). This not generally true and so we will later see how to deal with it. The overall gain (closed-loop gain) can be solved to be: xo A Af xs 1 Ab Ab is called the loop gain 1+Ab is called the “amount of feedback” The General Feedback Structure (Intro to Simulink) A xo A xi xi xs xf xf b xo b Af xs xi xf xo A xs 1 Ab xo b A b 1 A b feedabck factor loop gain amount of feedabck xf Ab xs 1 Ab Finding Loop Gain Generally, we can find the loop gain with the following steps: – – – – Break the feedback loop anywhere (at the output in the ex. below) Zero out the input signal xs Apply a test signal to the input of the feedback circuit Solve for the resulting signal xo at the output If xo is a voltage signal, xtst is a voltage and measure the open-circuit voltage If xo is a current signal, xtst is a current and measure the short-circuit current – The negative sign comes from the fact that we are apply negative feedback x f b xtst xs=0 xi xi 0 x f A xo Axi Ax f bAxtst xf loop gain b xtst xo xo bA xtst Negative Feedback Properties Negative feedback takes a sample of the output signal and applies it to the input to get several desirable properties. In amplifiers, negative feedback can be applied to get the following properties –Desensitized gain – gain less sensitive to circuit component variations –Reduce nonlinear distortion – output proportional to input (constant gain independent of signal level) –Reduce effect of noise –Control input and output impedances – by applying appropriate feedback topologies –Extend bandwidth of amplifier These properties can be achieved by trading off gain Gain Desensitivity Feedback can be used to desensitize the closed-loop gain to variations in the basic amplifier. Let’s see how. Assume beta is constant. Taking differentials of the closed-loop gain equation gives… Af A 1 Ab dA f Divide by Af dA f Af dA 1 Ab 2 dA 1 Ab 1 dA 1 Ab A 1 Ab 2 A This result shows the effects of variations in A on Af is mitigated by the feedback amount. 1+Abeta is also called the desensitivity amount We will see through examples that feedback also affects the input and resistance of the amplifier (increases Ri and decreases Ro by 1+Abeta factor) Bandwidth Extension We’ve mentioned several times in the past that we can trade gain for bandwidth. Finally, we see how to do so with feedback… Consider an amplifier with a high-frequency response characterized by a single pole and the expression: Apply negative feedback beta and the resulting closed-loop gain is: As A f s AM 1 s H As AM 1 AM b 1 bAs 1 s H 1 AM b •Notice that the midband gain reduces by (1+AMbeta) while the 3-dB roll-off frequency increases by (1+AMbeta) Basic Feedback Topologies series-shunt Depending on the input signal (voltage or current) to be amplified and form of the output (voltage or current), amplifiers can be classified into four categories. Depending on the amplifier category, one of four types of feedback structures should be used (series-shunt, seriesseries, shunt-shunt, or shunt-series) Voltage amplifier – voltage-controlled voltage source Requires high input impedance, low output impedance Use series-shunt feedback (voltage-voltage feedback) Current amplifier – current-controlled current source Use shunt-series feedback (current-current feedback) Transconductance amplifier – voltage-controlled current source Use series-series feedback (current-voltage feedback) Transimpedance amplifier – current-controlled voltage source Use shunt-shunt feedback (voltage-current feedback) shunt-series series-series shunt-shunt Examples of the Four Types of Amplifiers RD RD vOUT vIN vOUT vIN Vb Vb iIN Voltage Amp • iOUT iOUT iIN Transimpedance Amp Transconductance Amp Current Amp Shown above are simple examples of the four types of amplifiers. Often, these amplifiers alone do not have good performance (high output impedance, low gain, etc.) and are augmented by additional amplifier stages (see below) or different configurations (e.g., cascoding). iOUT RD RD vOUT vIN Vb RD RD Vb vOUT v IN iIN lower Zout iOUT iIN lower Zout higher gain higher gain Series-Shunt Feedback Amplifier (Voltage-Voltage Feedback) Samples the output voltage and returns a feedback voltage signal Ideal feedback network has infinite input impedance and zero output resistance Find the closed-loop gain and input resistance The output resistance can be found by applying a test voltage to the output So, increases input resistance and reduces output resistance makes amplifier closer to ideal VCVS V f bVo Vi Vs V f Vo AVs bVo Af Vo A Vs 1 bA Rif Vs V V V bAVi s Ri s Ri i Ri 1 Ab I i Vi Ri Vi Vi Rof Ro 1 bA The Series-Shunt Feedback Amplifier The Ideal Situation The series-shunt feedback amplifier: (a) ideal structure; (b) equivalent circuit. Z of ( s ) Af Rif Vo A Vs 1 Ab Vs Vs Ii Vi Ri Vs Vi Ri Ri Rif Ri 1 A b Zif ( s ) Zi( s ) 1 A ( s ) b ( s ) Vi b A Vi Vi Z o( s ) 1 A(s ) b (s ) Series-Series Feedback Amplifier (Current-Voltage Feedback) For a transconductance amplifier (voltage input, current output), we must apply the appropriate feedback circuit Sense the output current and feedback a voltage signal. So, the feedback current is a transimpedance block that converts the current signal into a voltage. To solve for the loop gain: Break the feedback, short out the break in the current sense and applying a test current To solve for Rif and Rof Apply a test voltage Vtst across O and O’ I A Af o Vs 1 Ab Rif I A o (also called Gm ) Vi I Loop Gain Ab out Gm R f I tst Vs Vi V f Ri I i bI o Ri I i AbVi Ri 1 Ab Ii Ii Ii Ii Rof Vtst I tst AVi Ro I tst AbI tst Ro 1 Ab Ro I tst I tst I tst Iout Gm ZL Itst Vf RF Shunt-Shunt Feedback Amplifier (Voltage-Current Feedback • When voltage-current FB is applied to a transimpedance amplifier, output voltage is sensed and current is subtracted from the input – The gain stage has some resistance – The feedback stage is a transconductor – Input and output resistances (Rif and Rof) follow the same form as before based on values for A and beta A Vo Ii Vo bVo A V A Af o I s 1 Ab I s Ii I f Rif Ri 1 Ab Ro Rof 1 Ab Shunt-Series Feedback Amplifier (Current-Current Feedback) • A current-current FB circuit is used for current amplifiers – For the b circuit – input resistance should be low and output resistance be high • A circuit example is shown – RS and RF constitute the FB circuit • RS should be small and RF large – The same steps can be taken to solve for A, Abeta, Af, Rif, and Rof • Remember that both A and b circuits are current controlled current sources Iout RD Iin RF RS The General Feedback Structure MATLAB / SIMULINK Anyone interested to put together an Introduction to MATLAB / Simulink PowerPoint Presentation plus some examples? The General Feedback Structure Exercise 8.1 4 A f 10 A 10 a) b b) Af c) Amount_Feedback 20 log 1 A b Amount_Feedback 60 R1 R1 R2 A d) 1 Ab Vs 1 Vo A f Vs Vf b Vo b 1 given Vo 10 Vf 0.999 Vi Vs Vf 4 Vi 10 10 A Af 1 Ab e) b Find b R1 R1 R2 0.1 4 A 0.8 10 b 0.1 R2 R1 9 10 9.998 10 100 0.02 A f A 1 Ab A f 9.998 The General Feedback Structure Exercise 8.1 Some Properties of Negative Feedback Gain Desensitivity Af A 1 Ab deriving dAf dA (1 A b) dividing by 2 dAf 1 Af (1 A b) Af A 1 Ab dA A The percentage change in Af (due to variations in some circuit parameter) is smaller than the pecentage cahnge in A by the amount of feedback. For this reason the amount of feedback 1 Ab is also known as the desensitivity factor. Some Properties of Negative Feedback Bandwidth Extension High frequency response with a single pole A ( s) AM 1 s H AM denotes the midband gain and H the upper 3-dB frequency Af ( s ) A ( s) 1 b A ( s) AM 1 AMb Af ( s ) 1 Hf s H 1 A M b H 1 A M b Lf L 1 AM b Some Properties of Negative Feedback Noise Reduction, Reduction of Nonlinear Distortion Read and discuss in class The Shunt-Series Feedback Amplifier Example 8.1 [ RL ( R1 R2) ] A Vo' Vi' RL R1 R2 [ RL ( R1 R2) ] RL R1 R2 b Af Rof Vf R1 Vo'' R1 R2 Vo A Vf 1 Ab Ro 1 Ab ro Ro ro RL ( R1 R2) RL R1 R2 RL ( R1 R2) RL R1 R2 Rid ro Rid Rs Rif R1 R2 R1 R2 Ri 1 A b Ri Rs Rid Rin Rif Rs R1 R2 R1 R2 The Shunt-Series Feedback Amplifier Example 8.1 3 6 R1 10 4 R2 10 5 Rs 10 Rid 10 3 ro 10 4 10 [ RL ( R1 R2) ] A RL R1 R2 [RL(R1 R2)] RL R1 R2 b R1 R1 R2 Rid R1R2 ro Rid Rs R1 R2 4 b 9.99 10 R1R2 Ri Rs Rid R1 R2 5 Ri 1.11 10 Rif Ri1 Ab Rif 7.766 10 Rin Rif Rs Rin 7.666 10 Af A 1 Ab 3 A 6.002 10 5 5 Af 857.92 3 RL 210 The Shunt-Series Feedback Amplifier Example 8.1 Ro RL ( R1 R2) ro RL R1 R2 RL ( R1 R2) ro RL R1 R2 Rout 10 given Rout RL Rof Rout RL Find Rout 99.989 Ro 666.223 Rof Ro 1 A b Rof 95.228 The Series-Shunt Feedback Amplifier Exercise 8.5 The Series-Shunt Feedback Amplifier Example 8.2 The Series-Shunt Feedback Amplifier Exercise 8.6 The Shunt-Shunt Feedback Amplifier Exercise 8.3 The Shunt-Shunt Feedback Amplifier Example 8.4 The Shunt-Shunt Feedback Amplifier Example 8.4 The Shunt-Shunt Feedback Amplifier Exercise 8.7 Determining The Loop Gain Exercise 8.8-9 The Stability Problem The Stability Problem and Margins Polar Plot Af(s ) Nyquist Closed-Loop Transfer Function Root-Locus Bode A(s ) 1 A(s ) b (s ) The Nyquist Plot intersects the negative real axis at 180. If this intersection occurs to the left of the point (-1, 0), we know that the magnitude of the loop gain at this frequency is greater than the unity and the s ystem will be unstable. If the intersection occurs to the right of the point (-1,0) the sys tem will be stable. It follows that if the Nyquist encircles the point (-1,0) the amplifier will be unstable. The Stability Problem Nyquist Closed-Loop Transfer Function Root-Locus Magnitude and Phase Bode Gain Margin Phase Margin - If at the frequency of unity loop-gain magnitude, the phase lag is in excess of 180 degrees, the amplifier is unstable. The Stability Problem The Nyquist Plot w 100 99.9 100 G( w) j 1 s ( w) j w f ( w) 1 50 4.6 3 2 s ( w) 9 s ( w) 30 s ( w) 40 5 Im ( G( w) ) 0 0 5 2 1 0 1 2 Re( G( w) ) 3 4 5 6 Stability Study Using Bode Plots Bode plot for loop gain Abeta – Gain margin is an indication of excess gain before instability – Phase margin is an indication of excess phase before -180° phase shift at unity gain Stability Study Using Bode Plots w .1 .11 2 G( w) K 2 K j w ( j w 1) ( j w 2) j 1 Bode1( w) 20 log G( w) 20 20 Open Loop Bode DiagramBode1( w) Bode1( w) 0 0 20 20 0 0.5 1 w T( w) G( w) 1.5 0.5 1 1.5 2 w 2 T( w) 0 G( w) Bode2( w) 20 log T( w) 1 G( w) 1 G( w) 10 0 Bode2( w) 10 Closed-Loop Bode Diagram 20 0.5 1 1.5 w 2 The Stability Problem Exercise 8.10 Effect of Feedback on the Amplifier Poles Effect of Feedback on the Amplifier Poles Exercise 8.11