$35.00

## Description

**1 **** ****Objectives**

The purpose of the lab is to study some of the advanced opamp configurations commonly found in practical appli- cations. The circuits studied will include the summing amplifier, the differential amplifier and the instrumentation amplifier.

**2 **** ****Introduction**

**2.1 **** ****Summing**** ****Amplifier**

An inverting amplifier can be modified to accommodate multiple input signals as shown in Fig. 1. Since the circuit is linear, the output voltage can easily be found by applying the superposition principle: the output voltage is a weighted sum of the two input signals. The weighting factor is determined by applying one of the input signals while the other is grounded and analyzing the resulting circuit. Since the circuit is linear, the analysis is repeated for the other input, and the final result is the addition of both signals. The advantage of this approach is that we can easily recognize the effect of each signal on the circuit’s performance, and the overall output can be obtained in most of the cases by inspection. For the circuit in Fig. 1, the output voltage can be found as

V_{o} = −

_{R}_{3}

R_{1}

_{V } _{+} ^{R}^{3}

^{i} ^{1 } ^{R}_{2}

^{V}i 2

(1)

^{R}_{3}

Figure 1: Summing amplifier circuit

The summing amplifier can be extended to have any number of input signals. Consider a two-bit digital signal applied to the inputs of the circuit in Fig. 1, resulting in an analog voltage at the output that is determined by the binary input. A more general configuration based on this circuit can be used to build digital-to-analog converters (DAC).

**2.2 **** ****Di****f****ferential**** ****Amplifier**

The differential amplifier is designed to amplify the difference of the two inputs. The simplest configuration is shown in Fig. 2. If the resistor values are chosen such that R_{2} /R_{1} = R_{4} /R_{3} , then the output of the amplifier is given by:

_{V } _{=} ^{R}^{2}

^{o } ^{R}_{1}

(V_{i} _{2} − V_{i} _{1} ) (2)

R_{2}

+5V R_{1}

^{2 } ^{7 } _{1}

^{V+ } _{N1 } _{6}

_{N2 } ^{V}_{o}

_{V−}

_{3 } _{4 } _{5}

^{V}i1

^{V}i2

R_{3 } _{R}_{4}

_{−5V}

Figure2:Differentialamplifiercircuit

c Department of Electrical and Computer Engineering, Texas A&M University

This expression shows that the circuit amplifies the difference between the two input signals V_{i} _{2} − V_{i} _{1 } and rejects the common mode input signals (V_{o } = 0 if V_{i} _{1 } = V_{i} _{2} ). Therefore, the differential amplifier can be used in a very noisy environment to reject common noise that appears at both inputs. When the same signal is applied to both inputs, the voltage gain is defined as common-mode gain (A_{CM} ), which is zero for an ideal differential amplifier. The common-mode rejection ratio is defined as,

CMRR =

__A___{DM}__ ____(Di____f____fe____r____ential-mode____ ____gain)__

A_{CM} (Common-mode gain)

(3)

Substituting A_{CM} = 0 to the above expression, CMRR for an ideal differential amplifier becomes infinite. In practice, resistors have a tolerance of typically 5%, and the common-mode gain will not be zero, resulting in finite CMRR .

**2.3 **** ****Instrumentation**** ****Amplifier**

The instrumentation amplifier is a differential amplifier that has high input impedance and the capability of gain adjustment through the variation of a single resistor. A typical instrumentation amplifier is shown in Fig. 3.

+5V

_{3 } _{7}

^{1 } _{R } _{R}

^{V+ } N1 6

_{V−}

^{2 } _{4 } ^{5 } ^{R}

−5V

+5V

_{2 } 7 _{1}

^{R}gain

^{V+ } _{N1 } _{6}

_{N2 } ^{V}_{o}

V−

^{3 } _{4 } ^{5}

+5V

−5V

^{2 } ^{7 } _{1 } _{R}

^{V+ } _{N1 } _{6}

N2

V−

3 _{4 } ^{5 } ^{R } ^{R}

^{V}_{i2}

−5V

Figure 3: Typical instrumentation amplifier circuit

The voltage drop across R_{gain} is equal equal to the voltage difference of the two input signals. Therefore, the current through R_{gain } caused by the voltage drop must flow through the two R resistors above and below R_{gain} . The output voltage can be calculated as

V_{o} =

_{2R}

_{1} _{+}

^{R}_{gain}

(V_{i} _{2} − V_{i} _{1} ) (4)

Though this configuration looks cumbersome to build a differential amplifier, the circuit has several properties that make it very attractive. It presents high input impedance at both terminals because the inputs connect into non-inverting terminals. Also a single resistor R_{gain } can be used to adjust the voltage gain.

**3 **** ****Calculations**

**1.**** **For the summing amplifier in Fig. 1, find R_{1} and R_{2} to have V_{o} = −(V_{i} _{1} + 2V_{i} _{2} ), if R_{3} = 15k Ω.

**2.**** **For the differential amplifier in Fig. 2, find R_{1} to have V_{o} = V_{i} _{2} − V_{i} _{1} , if R_{2} = R_{3} = R_{4} = 10k Ω.

**3.**** **For the instrumentation amplifier in Fig. 3, find R to have V_{o} = 3(V_{i} _{2} − V_{i} _{1} ), if R_{gain } = 1k Ω.

**4.**** **For each circuit, find V_{o} if V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.3V .

**4 **** ****Simulations**

**For**** ****all**** ****simulations,**** ****provide **** ****screenshots**** ****showing**** ****the**** ****schematics **** ****and**** ****the**** ****plots**** ****with**** ****the**** ****simulated**** ****values **** ****prop- ****erly**** ****labeled.**

**1.**** **Draw the schematics for the circuits in Figs. 1, 2, and 3 with the calculated component values using the UA741 opamp model.

**2.**** **Apply the inputs V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.3V , and obtain the **time-domain**** ****waveforms **for the input and output voltages using **transient**** ****simulation**. Confirm that the circuits operate as designed.

**5 **** ****Measurements**

**For**** ****all**** ****measurements,**** ****provide **** ****screenshots**** ****showing**** ****the**** ****plots**** ****with**** ****the**** ****measured**** ****values **** ****properly**** ****labeled.**

**5.1 **** ****Summing**** ****Amplifier**

**1.**** **Build the circuit in Fig. 1 with the simulated component values.

**2.**** **Apply the inputs V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.3V , and obtain the **time-domain**** ****waveforms **for the input and the output voltages using the **scope**** **to confirm that the circuit is a summing amplifier.

**3.**** **Raise the DC input voltage V_{i} _{2} until clipping at the output is observed.

**5.2 **** ****Di****f****ferential**** ****Amplifier**

**1.**** **Build the circuit in Fig. 2 with the simulated component values.

**2.**** **Apply the inputs V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.3V , and obtain the **time-domain**** ****waveforms **for the input and the output voltages using the **scope**** **to confirm that the circuit is a differential amplifier.

**3.**** **Apply V_{i} _{1} = 0.2 sin(2π1000t ) and connect V_{i} _{2} to ground. Measure A_{DM } = V_{o} /V_{i} .

**4.**** **Apply V_{i} _{1} = V_{i} _{2} = 0.2 sin(2π1000t ). Measure A_{CM} = V_{o} /V_{i} .

**5.**** **Calculate the common-mode rejection ratio (CMRR).

**5.3 **** ****Instrumentation**** ****Amplifier**

**1.**** **Build the circuit in Fig. 3 with the simulated component values.

**2.**** **Apply the inputs V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.3V , and obtain the **time-domain**** ****waveforms **for the input and the output voltages using the **scope**** **to confirm that the circuit is a instrumentation amplifier.

**6 **** ****Report**

**1.**** **Include calculations, schematics, simulation plots, and measurement plots.

**2.**** **Prepare a table showing calculated, simulated and measured results.

**3.**** **Compare the results and comment on the differences.

**7 **** ****Demonstration**

**1.**** **Build the circuits in Figs. 2 and 3 on your breadboard and bring it to your lab session.

**2.**** **Your name and UIN must be written on the side of your breadboard.

**3.**** **Submit your report to your TA at the beginning of your lab session.

**4.**** **For the differential amplifier in Fig. 2:

• Show the time-domain waveforms with V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.3V .

• Measure A_{dm} with V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.

• Measure A_{cm} with V_{i} _{1} = V_{i} _{2} = 0.2 sin(2π1000t ).

• Calculate CMRR

**5.**** **For the instrumentation amplifier in Fig. 3:

• Show the time-domain waveforms with V_{i} _{1} = 0.2 sin(2π1000t ) and V_{i} _{2} = 0.3V .