Instrumentation Sensors Book
.pdf4.3 Types of Amplifiers |
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BW = GBP/Av = 1.5 MHz/180
BW = 8.3 kHz
4.3Types of Amplifiers
The op-amp can be configured for voltage or current signal amplification, conversion of voltage signals to current signals and vice versa, impedance matching, and comparison and summing applications in process control.
4.3.1Voltage Amplifiers
Inverting Amplifier
Consider the ideal op-amp shown in Figure 4.3, which is configured as an inverting voltage amplifier. Resistors R1 and R2 provide feedback; that is, some of the output signal is fed back to the input (see Section 15.2.1). The large amplification factor in op-amps tends to make some of them unstable, and causes dc drift of the operating point with temperature. Feedback stabilizes the amplifier, minimizes dc drift, and sets the gain to a known value.
When a voltage input signal is fed into the negative terminal of the op-amp, as in Figure 4.3, the output signal will be inverted. Because of the high input impedance of the op-amp, no current flows into the input, or I3 = 0. Then, since the sum of the currents at the negative input is zero:
I1 + I2 = 0 |
(4.3) |
The voltage at the junction of R1 and R2 is zero, the same as the positive input, and is termed a virtual ground from which:
Vout = |
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The negative sign is because the output signal is negative. In this configuration, the closed loop voltage gain of the stage is:
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Figure 4.3 Circuit diagrams of inverting amplifier.
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Unfortunately, ideal op-amps do not exist. Op-amps have finite open loop gain, an output impedance, and finite input impedance. Circuit analysis shows that in a typical amplifier circuit, the effect of these parameters on (4.5) is less than 0.1%, so that in most practical cases, the effects can be ignored.
Example 4.2
In Figure 4.2, if resistor R1 = 2,700Ω and resistor R2 = 470 kΩ, what is the gain, and what is the output voltage amplitude if the ac input voltage is 1.8 mV?
Gain = R2 = 470 = 174
R1 2.7
ac Output Voltage = −1.8 × 174 mV = −313.5 mV = −0.41V
Summing Amplifier
A summing amplifier is a common use of the inverting amplifier, which is used to sum two or more voltages (see Section 16.6.6). The summing circuit is shown in Figure 4.4. The transfer function is given by:
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Example 4.3
In Figure 4.3, R3 = R1 = 5.6 kΩ, and R3 = 220 kΩ. (a) If V1 = 27 mV and V2 = 0V, what is Vout? (b) If V2 is changed to 43 mV, what is the new Vout?
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(b)Vout |
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Figure 4.4 Summing amplifier.
4.3 Types of Amplifiers |
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Noninverting Amplifier
A noninverting amplifier configuration is shown in Figure 4.5. When the input signal is fed into the positive terminal, the circuit is noninverting [4]. Since the negative input is referenced to the positive input, the voltage at the negative terminal is Vin. Using Ohm’s Law, this gives;
Vin = Vout − Vin
R1 R2
From which the voltage gain is given by:
Gain = |
Eout |
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(4.7) |
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Ein R1
In this configuration, the amplifier gain is 1 plus the resistor ratio, so that the gain does not directly vary with the resistor ratio. However, this configuration does give a high input impedance (i.e., that of the op-amp), and a low output impedance.
Example 4.4
In Figures 4.3 and 4.5, R1 = 3.9 kΩ and R2 = 270 kΩ. If a dc voltage of 0.03V is applied to the inputs of each amplifier, what will be the output voltages?
From Figure 4.3:
Vout = −270 × 003.V = −2.077V
3.9
From Figure 4.5:
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An alternative noninverting amplifier using two inverters is shown in Figure 4.6. The second inverter has unity gain. In this case, the gain is the ratio of R2 to R1, as in the case of the inverting amplifier.
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Figure 4.5 Noninverting amplifier circuit using single op-amp.
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Figure 4.6 Noninverting amplifier.
Voltage Follower
The voltage follower circuit, as shown in Figure 4.7, is an impedance matching device with unity gain (see Section 5.4.1). The configuration has a very high input impedance (>100 MΩ with MOS devices), and low output impedance (<50Ω). The output drivers of the op-amp limit the maximum output current.
Differencing Amplifier
There are many applications for differencing amplifiers in instrumentation, such as amplifying the output from a Wheatstone bridge. Ideally, the output voltage is given by:
Vout = A(V1 − V2) |
(4.8) |
where A is the gain of the amplifier.
The basic differencing amplifier is shown in Figure 4.8. The resistors should be matched as shown. Then, for an ideal op-amp, the transfer function is given by:
Vout |
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(V2 − V1 ) |
(4.9) |
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This equation indicates that the output voltage depends only on the resistor ratio and the input voltage difference ∆V(V2 − V1), and is independent of the input operating point. That is, V2 can vary from ±10V, and if ∆V does not change, then
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Vin Vout = Vin
Figure 4.7 Voltage follower.
4.3 Types of Amplifiers |
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Vout
R2
Figure 4.8 Basic differencing amplifier.
the output will not change. In practice, real differential amplifiers only can approach the ideal case.
The Common-Mode (CM) input voltage is defined as the average of the applied input voltages:
CM = (V1 + V2)/2 |
(4.10) |
The CM gain (ACM) is the ratio of change in output voltage for a change in CM input voltage (ideally zero) [5].
The gain (A) of the op-amp is the ratio of the change in output voltage for a differential input voltage; see (4.8).
The Common-Mode Rejection Ratio (CMRR) is the ratio of the differential gain to the CM gain.
CMRR = A/ACM |
(4.11) |
The CMRR is normally expressed as the Common-Mode Rejection (CMR) in decibels.
CMR = 20 log10 (CMRR) |
(4.12) |
The higher the CMR value, the better the circuit performance. Typically, values of CMR range from 60 to 100 dB.
Example 4.5
In the dc amplifier shown in Figure 4.8, an input of 130 mV is applied to terminal A and −85 mV is applied to terminal B. What is the output voltage, assuming the amplifier was zeroed with 0V at the inputs?
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× 25.5V = −5.5V |
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The disadvantage of this circuit is that the input impedance is not very high, and is different for the two inputs. To overcome these drawbacks, voltage followers are
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normally used to buffer the inputs, resulting in the circuit shown in Figure 4.9, which is used in instrument differencing amplifiers (see Section 15.3.2).
Example 4.6
In the amplifier shown in Figure 4.9, R1 = 4.7 k?, and R2 = 560 k?. If V1 and V2 are both taken from 0V to 3.7V, and the output changes from 0V to 0.19V, then what is the CMR of the circuit?
ACM = 0.19/3.7 = 0.051
A = 560/4.7 = 119.15
CMR = 20 log 119.15/0.051 = 20 log 2340 = 67.37 dB
4.3.2Converters
The circuits shown above were for voltage amplifiers. Op-amps also can be used as current amplifiers, voltage to current converters, current to voltage converters, and special purpose amplifiers.
A current to voltage converter is shown in Figure 4.10. When used as a converter, the relation between input and output is called the Transfer function (or Ratio). These devices do not have gain as such, because of the different input and output units. In Figure 4.10, the transfer ratio is given by:
µ = |
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Example 4.7
In Figure 4.10, the input current is 83A and the output voltage is 3.7V. What is the transfer ratio, and the value of R1?
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Figure 4.9 Instrument differencing amplifier.
4.3 Types of Amplifiers |
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R1
IIN
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VOUT
Figure 4.10 Current to voltage converter.
=37.V = =
µ 44.5V mA 44.5kV A 83µA
R1 = 37.V = 44.5kΩ 83µA
A voltage to current converter is shown in Figure 4.11. The op-amp converts a voltage into a current. In this case, the transfer ratio is given by:
Iout |
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−R2 |
mhos |
(4.14) |
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In this case, the units are in mhos (1/ohms), and the resistors are related by the equation:
R1 (R3 +R5) = R2R4 |
(4.15) |
However, in industrial instrumentation, a voltage to current converter is sometimes referred to as a current amplifier.
Example 4.8
In Figure 4.11, R1 = R4 = 6.5 kΩ, and R2 = 97 kΩ. What is the value of R3 and R5 if the op-amp is needed to convert an input of 2.2V to an output of 15 mA?
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65. × 103 |
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Figure 4.11 Voltage to current converter. |
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4.3.3Current Amplifiers
Devices that amplify currents are referred to as current amplifiers. Figure 4.12 shows a basic current amplifier. The gain is given by:
Iout |
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(4.16) |
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where the resistors are related by the equation:
R2(R4 + R6) = R3R5 |
(4.17) |
Example 4.9
In Figure 4.12, I1 =1.23A, R1 = 12 kΩ, R4 = 3.3 kΩ, R5 = 4.6 kΩ, and R6 = 120 kΩ. What is the gain and output current?
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Figure 4.12 Current amplifier.
4.3 Types of Amplifiers |
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From (4.17) R3/R2 = (R4 + R6)/R5 = (3.3 + 120)/4.6 = 26.8
From (4.16) Iout = Iin R3 × R1/R2 × R4 = 1.23 × 26.8 × 12/3.3A
Iout = 119.9A = 0.12 mA
Gain = 119.9/1.23 = 97.45
4.3.4Integrating and Differentiating Amplifiers
Two important functions that are used in process control (see Section 16.6.5) are integration and differentiation [6].
An integrating amplifier is shown in Figure 4.13(a). In this configuration, the feedback resistor is replaced with an integrating capacitor. Using the ideal case, the currents at the input to the amplifier can be summed as follows:
Vin/R1 + C dVout/dt = 0 |
(4.18) |
Solving the equation gives:
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(4.19) |
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This shows that the output voltage is the integral of the input voltage, with a scale factor of −1/RC. If Vin is a step function, then the circuit can be used as a linear ramp generator. Assume Vin steps E volts and stays at that value, the equation reduces to:
Vout = −(E/RC)t |
(4.20) |
which is a linear ramp with a negative slope of E/RC. This circuit in practice is used with outer circuits for limiting and resetting.
A differentiating circuit is shown in Figure 4.13(b). In this configuration, the input resistor is replaced with a capacitor. The currents can be summed in the ideal case as follows:
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Figure 4.13 Circuits showing (a) integrating amplifier and (b) differentiating amplifier.
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Solving for the output voltage, the response is; |
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Showing that Vout is the derivative of Vin, the circuit tends to be unstable, but in practice is easily stabilized when used with other circuits.
4.3.5Nonlinear Amplifiers
Many sensors have a logarithmic or nonlinear transfer characteristic. Such devices require signal linearization. This can be implemented using amplifiers with nonlinear characteristics. These are achieved by the use of nonlinear elements, such as diodes or transistors in the feedback loop [7].
Two examples of logarithmic amplifiers are shown in Figure 4.14. Figure 4.14(a) shows a logarithmic amplifier using a diode in the feedback loop, and Figure 4.14(b) shows a logarithmic amplifier using a bipolar junction transistor in the feedback loop. The summation of the currents in an ideal case gives:
Vin/R1 = ID |
(4.23) |
where ID is the current flowing through the diode, or the collector/emitter current of the transistor.
The relation between the diode voltage (VD) and the diode current is given by:
VD = A logC ID/IR |
(4.24) |
where A is the proportionality constant depending on c the base of the logarithm, and IR is the diode leakage current.
Because the voltage across the diode is also the output voltage (Vout), (4.23) and (4.24) can be combined, giving:
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Figure 4.14 Circuits of logarithmic amplifiers using (a) a diode feedback, and (b) an NPN transistor.