barcode reader vb.net codeproject Operational Ampli ers in Software

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12
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Operational Ampli ers
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Why Feedback (continued) Figure 125, the output voltage, vout , causes a current, iF , to ow through the feedback resistance so that KCL is satis ed at the inverting node Assume, for a moment, that the differential voltage v = v+ v is identically zero Then, the output voltage will continue to be such that KCL is satis ed at the inverting node, that is, such that the current iF is equal to the current iS Suppose, now, that a small imbalance in voltage, v, is present at the input to the op-amp Then the output voltage will be increased by an amount AV (OL) v Thus, an incremental current approximately equal to AV (OL) v/RF will ow from output to input via the feedback resistor The effect of this incremental current is to reduce the voltage difference v to zero, so as to restore the original balance in the circuit One way of viewing negative feedback, then, is to consider it a self-balancing mechanism, which allows the ampli er to preserve zero potential difference between its input terminals A practical example that illustrates a common application of negative feedback is the thermostat This simple temperature control system operates by comparing the desired ambient temperature and the temperature measured by a thermometer and turns a heat source on and off to maintain the difference between actual and desired temperature as close to zero as possible An analogy may be made with the inverting ampli er if we consider that, in this case, negative feedback is used to keep the inverting-terminal voltage as close as possible to the noninverting-terminal voltage The latter voltage is analogous to the desired ambient temperature in your home, while the former plays a role akin to that of the actual ambient temperature The open-loop gain of the ampli er forces the two voltages to be close to each other, much the way the furnace raises the heat in the house to match the desired ambient temperature It is also possible to con gure operational ampli ers in a positive feedback con guration if the output connection is tied to the noninverting input We do not discuss this con guration in the present chapter, but present an example of it, the voltage comparator, in 15
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EXAMPLE 121 Inverting Ampli er Circuit
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Determine the voltage gain and output voltage for the inverting ampli er circuit of Figure 125 What will the uncertainty in the gain be if 5 and 10 percent tolerance resistors are used, respectively
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Solution
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Known Quantities: Feedback and source resistances, source voltage Find: AV = vout /vin ; maximum percent change in AV for 5 and 10 percent tolerance resistors
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vS (t) = A cos( t); A = 0015 V; = 50 rad/s
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Schematics, Diagrams, Circuits, and Given Data: RS = 1 k ; RF = 10 k ;
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Assumptions: The ampli er behaves ideally; that is, the input current into the op-amp is zero and negative feedback forces v + = v
Part II
Electronics
Analysis: Using equation 1219, we calculate the output voltage:
vout (t) = AV vS (t) =
RF vS (t) = 10 0015 cos( t) = 015 cos( t) RS
The input and output waveforms are sketched in Figure 126
015 01 005 Volts 0
vS(t) vout(t)
005 01 015 0 01 02 03 04 05 06 Time (s) 07 08 09 10
The nominal gain of the ampli er is AV nom = 10 If 5 percent tolerance resistors are employed, the worst-case error would occur at the extremes: AV min = RF min 9,500 = 905 = RS max 1,050 AV min
AV max =
RF max 10,500 = 1105 = RS min 950
The percentage error is therefore computed as: 100 100 AV AV
= 100
10 905 = 95% 10
10 1105 nom AV max = 100 = 105% AV nom 10
Thus, the ampli er gain could vary by as much as 10 percent (approximately) when 5 percent resistors are used If 10 percent resistors were used, we would calculate a percent error of approximately 20 percent, as shown below AV min = 100 100 AV AV RF min 9,000 = = 818 RS max 1,100 AV max = RF max 11,000 = = 122 RS min 900
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