Instrumentation Sensors Book

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:

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:

The negative sign is because the output signal is negative. In this configuration, the closed loop voltage gain of the stage is:

R2

+

Vout

Figure 4.3 Circuit diagrams of inverting amplifier.

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:

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?

+

Vout

Figure 4.4 Summing amplifier.

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:

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:

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.

Figure 4.5 Noninverting amplifier circuit using single op-amp.

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:

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:

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

−

+

Vin Vout = Vin

Figure 4.7 Voltage follower.

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:

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.

The CMRR is normally expressed as the Common-Mode Rejection (CMR) in decibels.

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?

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

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:

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?

Figure 4.9 Instrument differencing amplifier.

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:

where the resistors are related by the equation:

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?

Figure 4.12 Current amplifier.