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barcode reader vb.net codeproject Voltage Supply Limits in an OpAmp Integrator in Software
EXAMPLE 1210 Voltage Supply Limits in an OpAmp Integrator Denso QR Bar Code Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Quick Response Code Encoder In None Using Barcode generation for Software Control to generate, create QR image in Software applications. Problem
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t vS (t )dt =
1 02 t 05 + 03 cos(10t ) dt
= 25t + 15 sin(10t) However, the supply voltage is limited to 15 V, and the integrator output voltage will therefore saturate at the lower supply voltage value of 15 V as the term 25t increases with time Figure 1244 depicts the output voltage waveform 16 14 12 10 8 Ideal integrator output 6 Integrator output 4 affected by DC offset 2 0 2 0 1 2 3 4 5 6 7 8 9 10 Time (s) Figure 1244 Effect of DC offset on integrator
Comments: Note that the DC offset in the waveform causes the integrator output voltage
to increase linearly with time The presence of even a very small DC offset will always cause integrator saturation One solution to this problem is to include a large feedback resistor in parallel with the capacitor; this solution is explored in the homework problems Focus on ComputerAided Solutions: An Electronics WorkbenchTM simulation of a
practical integrator (including the feedback resistance mentioned in the comments above) can be found in the accompanying CDROM Try removing the feedback resistor to verify that saturation at the supply voltages is a real problem Frequency Response Limits Another property of all ampli ers that may pose severe limitations to the opamp is their nite bandwidth We have so far assumed, in our ideal opamp model, that the openloop gain is a very large constant In reality, AV (OL) is a function of Integrator voltage (V) 12
Operational Ampli ers
frequency and is characterized by a lowpass response For a typical opamp, AV (OL) (j ) = A0 1 + j / 0 (1282) The cutoff frequency of the opamp openloop gain, 0 , represents approximately the point where the ampli er response starts to drop off as a function of frequency, and is analogous to the cutoff frequencies of the RC and RL circuits of 6 Figure 1245 depicts AV (OL) (j ) in both linear and dB plots for the fairly typical values A0 = 106 and 0 = 10 It should be apparent from this gure that the assumption of a very large openloop gain becomes less and less accurate for increasing frequency Recall the initial derivation of the closedloop gain for the inverting ampli er: In obtaining the nal result, Vout /VS = RF /RS , it was assumed that AV (OL) This assumption is clearly inadequate at the higher frequencies 105 10 8 Gain
Openloop gain of practical opamp 120 100
Gain, dB
Openloop gain of practical opamp (dB plot) 6 4 2 0 10 1 100 101 102 103 104 Radian frequency (logarithmic scale) 105 80 60 40 10 1 100 102 103 104 101 Radian frequency (logarithmic scale) 105 Figure 1245 Openloop gain of practical opamp
AV dB 20 log10 A0 20 log10 A1
20 log10 A2 0 1 2 Log scale
The nite bandwidth of the practical opamp results in a xed gainbandwidth product for any given ampli er The effect of a constant gainbandwidth product is that as the closedloop gain of the ampli er is increased, its 3dB bandwidth is proportionally reduced, until, in the limit, if the ampli er were used in the openloop mode, its gain would be equal to A0 and its 3dB bandwidth would be equal to 0 The constant gainbandwidth product is therefore equal to the product of the openloop gain and the openloop bandwidth of the ampli er: A0 0 = K When the ampli er is connected in a closedloop con guration (eg, as an inverting ampli er), its gain is typically much less than the openloop gain and the 3dB bandwidth of the ampli er is proportionally increased To explain this further, Figure 1246 depicts the case in which two different linear ampli ers (achieved through any two different negative feedback con gurations) have been designed for the same opamp The rst has closedloop gain A1 , and the second has closedloop gain A2 The bold line in the gure indicates the openloop frequency response, with gain A0 and cutoff frequency 0 As the gain decreases from the openloop gain, A0 , to A1 , we see that the cutoff frequency increases from 0 to 1 If we further reduce the gain to A2 , we can expect the bandwidth to increase to 2 Thus, the product of gain and bandwidth in any given opamp is constant That is, A0 0 = A1 1 = A2 2 = K (1283)

