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HA5020 Operational Amplifier Feedback Resistor Selection
Application Note January 1995 AN9305
Optimum AC performance of current feedback amplifiers in general and of the HA-5020 in particular depends upon careful selection of the feedback resistor, RF. The benefit of higher usable bandwidth (compared with conventional voltage feedback amplifiers) and the ability to control the frequency response (by choosing the value of RF) carries an expense in that the design process becomes more complicated. This is particularly true if an intuitive knowledge of how the device will behave in the end application is lacking. The purpose of this App Note is to provide a conceptual foundation on which this intuitive knowledge can be built. The choice of the optimum resistor value depends upon design goals for the application subject to conditions of closed loop gain, source impedance, and load. As a point of reference, typical curves are provided in the data sheet that show how the frequency response is affected by closed loop gain, feedback resistor value, and load resistance. Source impedance, if it is large, becomes a factor only in conjunction with capacitance at the inputs. The data sheet curves are all generated with a 50 source impedance. To illustrate how one might approach the problem of selecting a feedback resistor based on closed loop gain, consider the simple model of Figure 1. Between the inputs is a unity gain voltage buffer with non-zero output impedance indicated by RI. The transimpedance gain, RZ, is a function of frequency having a high DC value that forces I to zero. The model's behavior is influenced by external elements consisting of a feedback network (RF and RG), source and load impedances (RS and RL), and stray capacitance at the amplifier's inputs (CS).
RS VS CS +IN +1 I RF CS VOUT
gain peaking particularly at low gains (intuitively, CS is in parallel with RG causing the gain as determined by the feedback network to increase with frequency). Gain peaking due to capacitance at the inverting input is most easily dealt with by placing a resistor in series with the positive input. If we assume that the stray capacitance at the positive input equals the stray at the negative input, we can choose RS equal to the parallel combination of RF and RG. This introduces a pole at the positive input which cancels the zero at the negative input, thereby eliminating the gain peak. Note that any remaining gain peaking is a result of excessive phase shift around the loop. Excess phase shift around the loop can be reduced by increasing RF. Bandwidth degradation due to non-zero inverting input resistance is also easy to deal with as long as the product of the closed loop gain and the inverting input resistance does not exceed the optimum value for RF in unity gain. By solving the transfer function for constant bandwidth, we arrive at the following equations for RF and RG: RF = RFO - ACL * RI RI = RF / (ACL - 1) Where, RFO is the optimum value for unity gain (1000), RI is the inverting input impedance (75), and ACL is the desired closed loop gain. A comparison between actual measured results in Figures 2 and 3 provides graphic reinforcement for the utility of these equations. Figure 2 illustrates the failure to consider stray input capacitance and inverting input resistance, while Figure 3 incorporates the lessons learned from analyzing our simple model. In Figure 2, a family of closed loop gain curves was obtained on a representative unit using RS = 50 and constant RF (RF = RFO = 1000). The measured stray capacitance at either input was 2pF. The results in Figure 3 were obtained from the same unit, except that (within the constraints of available standard resistor values) RF and RG were chosen according to the equations above and RS was chosen to be equal to the parallel combination of RF and RG. One limitation of the above model is that it does not include the effects of the load. In general if RL is 400 or above, the response is independent of the load. If RL is less than 400, the response becomes more damped and the bandwidth degrades. Here again the bandwidth degradation can be compensated for by lowering the value of RF. (EQ. 1) (EQ. 2)
+ -
RZ(S) I
RL
-IN
RI
RG
FIGURE 1. SIMPLE CURRENT FEEDBACK AMPLIFIER MODEL
Derivation of the transfer function will confirm that the nonzero inverting input impedance, RI, causes the circuit's bandwidth to degrade as the closed loop gain increases, while stray capacitance at the negative input gives rise to
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Application Note 9305
3 2 NORMALIZED GAIN (dB) 1 0 -1 -2 -3 -4 -5 -6 -7 0 12 24 36 48 60 72 84 FREQUENCY (MHz) 96 108 120 AV = +10 AV = +6 AV = +1 AV = +2 3 2 NORMALIZED GAIN (dB) 1 0 -1 -2 -3 -4 -5 -6 -7 0 12 24 36 48 60 72 FREQUENCY (MHz) 84 96 108 120 AV = +10 AV = +2 AV = +1 AV = +6
FIGURE 2. FREQUENCY RESPONSE vs CLOSED LOOP GAIN USING FIXED RF = 1k, RS = 50, RL = 402 TABLE 1. RESISTOR VALUES FOR FIGURE 2 BW (MHz) 97 84 22 16 PEAKING (dB) 1 <0 <0 <0
FIGURE 3. FREQUENCY RESPONSE vs CLOSED LOOP GAIN RF = 1000-AV(75), RS = RF || RG, RL = 402 TABLE 2. RESISTOR VALUES FOR FIGURE 3 BW (MHz) 87 87 74 62 PEAKING (dB) 2 2 0.5 0.5
AV +1 +2 +6 +10
RF () 1K 1K 1K 1K
RG () 1K 200 110
RS () 50 50 50 50
AV +1 +2 +6 +10
RF () 909 825 562 237
RG () 825 110 26.1
RS () 909 422 90.9 23.7
NOTE: RF = 1000-AV(75), RS = RF || RG, RL = 402
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