SchematicProblem:Solution outline:Part 1: Designing the feedback amplifier configurationPart 2: Designing the feedback networkImplementationPart 1: Designing the feedback amplifier configurationPart 2: Designing the feedback networkFirst iterationSecond iterationThird iterationECE 304: Iterative Design of Feedback Network Schematic OUTFFFBIN+R7{RB}-++-E1GAIN = -920V/V+R4{Ro}+R6{RC}00+R2{Ri}AMP PARAMSRi = 11kRo = 3k0+R5{RA} FIGURE 1 Given voltage amplifier and feedback circuit Problem: Design a feedback amplifier using the parts in Figure 1. The feedback amplifier is to satisfy these impedance conditions: 1. Output resistance with feedback RO(FB) ≤ 20 Ω 2. Input resistance with feedback RI(FB) ≤ 680 Ω Solution outline: PART 1: DESIGNING THE FEEDBACK AMPLIFIER CONFIGURATION 1. We find the type of feedback connection needed to meet the goals 2. We find the type of amplifier the feedback amplifier will be 3. We find the type of dependent source that will be used in an ideal feedback circuit; in other words, the dimensions of the feedback factor βFB 4. We set up the circuit for the ideal feedback amplifier 5. We find the performance factor and determine which of the two impedance conditions is most limiting upon βFB 6. We find the ideal βFB-value PART 2: DESIGNING THE FEEDBACK NETWORK 1. We select a two-port for the T-section 2. We find the βFB for this two-port 3. We eliminate unnecessary resistors 4. We select resistor values that meet the βFB 5. If there is only one resistor value, it is determined by βFB. If there are several resistor values compatible with the value of βFB, roughly select these values to get the biggest performance factor. 6. We find the loaded gain with the two-port in place and βFB turned off 7. We find the new performance factor 8. We find a new formula for the limiting impedance (determined in PART 1, ITEM 5 as either RI(FB) or RO(FB)) that uses the new performance factor. The value of RO for the loaded amplifier is changed to include the loading effect of the feedback circuit. 9. We solve for the new βFB 10. We go back to STEP 4, using new βFB Unpublished work © 11/6/2004 J R Brews Page 1 11/7/200411. When convergence is reached, we stop Implementation PART 1: DESIGNING THE FEEDBACK AMPLIFIER CONFIGURATION 1. We find the type of feedback connection needed to meet the goals The requested RO below 20 Ω is less than the open-loop amplifier RO of 3 kΩ, so this resistance must be reduced. Therefore, the output FB connection is shunt. The requested RI below 680 Ω is less than the open-loop amplifier RI of 11 kΩ, so this resistance must be reduced. Therefore, the input FB connection is shunt. 2. We find the type of amplifier the feedback amplifier will be A shunt output implies the amplifier is to deliver voltage. A shunt input implies the amplifier is driven by current. Therefore the feedback amplifier has a gain of dimensions OUTPUT/INPUT = V/A = Ω. It is a transresistance amplifier. Notice that the original open-loop amplifier is a voltage amplifier, and the feedback amplifier is a different type of amplifier. 3. We find the type of dependent source that will be used in an ideal feedback circuit; in other words, the dimensions of the feedback factor βFB The performance factor contains the product βFB × Gain that must be unit-less. Therefore, the dimensions of βFB are reciprocal to the Gain of the feedback amplifier, or βFB = 1/Ω = A/V = VCCS. 4. We set up the circuit for the ideal feedback amplifier +R2{Ri}0bFB+R4{Ro}0+-G3GAIN = ??-++-E1GAIN = -920V/V0AMP PARAMSRi = 11kRo = 3k0I31A IN OUTFIGURE 2 Ideal feedback amplifier; the GAIN of the dependent feedback source is βFB, but PSPICE does not allow variable names for the gain, so to fill this in we need a numerical value for βFB 5. We find the performance factor and determine which of the two impedance conditions is most limiting upon βFB The loaded gain is found by turning off βFB. Because the input is current, we use a current driver. We find the input voltage is VI = ISRI. We find the output voltage is VO = –920VI = –920RI VS, so the loaded gain is ALD = –920RI. The performance factor is then 1 + βFB(-920RI). For negative feedback, βFB must be negative to make the performance factor positive. The impedance targets become EQ. 1 RO(FB) = )R920(1k3IFB−β+Ω ≤ 20 Ω EQ. 2 RI(FB) = )R920(1k11IFB−β+Ω ≤ 680 Ω. We recast these inequalities as limitations upon βFB: Unpublished work © 11/6/2004 J R Brews Page 2 11/7/2004EQ. 3 (OUTPUT RESTRICTION) S72.14R920120k3IFBµ=−≥β−ΩΩ EQ. 4 (INPUT RESTRICTION) S50.1R9201680k11IFBµ=−≥β−ΩΩ. Because βFB is negative, in these inequalities (–βFB) is a positive number. If EQ. 3 is satisfied, EQ. 4 automatically is satisfied, so the condition on the output resistance is the more restrictive. 6. We find the ideal βFB-value We adopt –βFB = 14.72 µS as the initial feedback value. We take the lower limit to get the best gain from the feedback amplifier, which will be 1/βFB if the performance factor is large. OUTIN20.00V0+R2{Ri}294.5uAAMP PARAMSRi = 11kRo = 3k03.239VbFB0I30A0A+R4{Ro}I41A1.000A+-G3GAIN = -14.7233uA/V0-++-E1GAIN = -920V/V FIGURE 3 Ideal feedback amplifier output resistance is 20 Ω when βFB = -14.72 µA Figure 3 shows a test current inserted at the output of the ideal feedback amplifier. The input current sources is turned off. The voltage at the output for a test current of 1A has a value the same as the output resistance of the amplifier, which is found to be RO(FB) = 20 Ω. PART 2: DESIGNING THE FEEDBACK NETWORK 1. We select a two-port for the T-section We want a two-port for the T-section that has a VCCS on the left side. Because current is the variable delivered by feedback on the left, voltage must be the independent variable on the left. Because voltage is the control variable for this VCCS, voltage must be the independent variable on the right. Therefore, the dependent source on the right also is a VCCS. Following the usual procedure we find the values below for the two-port parameters. EQ. 5 +−=βACBACAFBR//RRR//RR1 R CBA11R//RR +=+−=γBCABCBFBR//RRR//RR1 R . CAB22R//RR += Notice that the T-section is symmetrical if it is reflected horizontally and RA exchanged with RB. This symmetry shows up in R11 and R22 being the same if RA ↔ RB. Likewise for βFB and γFB. Unpublished work