1

XMUT303 Analogue Electronics

Mid-Term Test Revision Questions (Answer)

A. Operational Amplifier

1. An engineer in an electronic manufacturing company implements an inverting amplifier using an op amp, as shown in Figure 1, but finds that the circuit picks up interference, presumably due to a ground loop.

Figure 1: Inverting amplifier circuit

a. Derive the transfer function equation of the circuit given above. [2 marks]

( )

b. Calculate the output voltage of the circuit, given signal at the input of the amplifier is

2 Volts and the values of the resistors are =12 k and = 2 k . [4 marks]

( ) (

)

c. With the aid of diagrams, suggest two ways that the engineer could do to reduce the noise level on the output. [2 marks]

The solution could be by: using differential input amplifier, using twisted pair cable, or adding ferrite-bead choke.

2

2. Consider the difference amplifier shown in Figure 2.

Figure 2: Difference amplifier circuit.

a. Derive equation that best explain the output voltage, , expressed in terms of the input voltages, and of the circuit given above. [2 marks]

( )

( )

b. What is the common-mode gain (ACM), differential-mode gain (ADM) and common-mode rejection ratio (CMRR). [3 marks]

ACM is gain of difference amplifier as average of its two input signals.

ADM is gain of difference amplifier as a difference of its two input signals.

CMRR is the ability of a differential amplifier to reject a common mode signal.

c. Calculate the common-mode gain, ACM (dB), the differential-mode gain, ADM (dB), and the common-mode rejection ratio, CMRR (dB) for the circuit given above. [10 marks]

Common mode gain (ACM):

( )

Differential mode gain (ADM):

[

( )

( )

]

Common Mode Rejection Ratio (CMRR):

|

| |(

)

|

3

3. Consider the circuit diagram shown in Figure 3 as given below.

Figure 3: Nulling of a differential amplifier

The op amp used is an LM324, which has parasitic parameters as follows:

nominal input offset voltage, ≤ 3 mV,

input bias current, ≤ 100 nA,

input offset current, ≤ 30 nA.

a. Describe nulling circuit. [2 marks]

Nulling is a circuit arrangement for eliminating imperfections that existed in the circuit such as biasing and offsetting requirements of the components and/or the circuits.

b. Describe three types of nulling circuit in practice. [3 marks]

Using external components and circuit, through built-in feature of the electronic devices (e.g. adjusting the nulling pin of the op amp), or designing at circuit level that already taking into account biasing and offsetting requirements.

c. The resistors , , , and are included in the circuit to reduce the effects of these parasitic parameters. Give appropriate values for , , , and . Show your working and explain your reasoning. [15 marks]

The value of resistor is found from:

4

The value of resistor is found from:

Note that:

But, the value of is found from:

( )

( )

Say that we use 5 mV to be safe.

Rearranging and entering values into the equation, the value of resistor is:

(

)

(

)

Then the value of resistor is found from (note that is a variable resistor):

4. You are given the following two-op amp instrumentation circuit in the Figure 4 below.

Figure 4: Two-op amp instrumentation amplifier circuit

5

a. Why instrumentation amplifiers are commonly manufactured as monolithic integrated circuits? [2 marks]

Instrumentation amplifiers are typically packaged as monolithic integrated circuits, to ensure that they have high common mode rejection ratio.

b. If you have to design an op amp based amplifier in cascading arrangement, what circuit feature usually you have at the output stage? Give two reasons for your answer . [2 marks]

The output stage of a multistage amplifier usually employs a push-pull amplifier. It is to ensure sufficient coverage of the signal bandwidth and adequate driving of the load.

c. Derive an expression for the gain,

, for the circuit shown in the circuit above.

[5 marks]

d. Calculate the minimum gain of the circuit shown above if R1 = 1 k and R2 = 10 k. [2 marks]

Providing Rg is considerably bigger than R2, the minimum gain of the circuit is 11 times.

B. Op Amp Analysis

5. Given in the Figure 5 below a circuit diagram of op amp loaded with a resistive load and an imperfect voltage source.

Figure 5: circuit diagram of loading of an op amp

a. Describe the difference between op amp with FET input stages with those with BJT transistors. [2 marks]

Compared to op amps with FET input stages, op amps with BJT input stages have

6

higher bias current but lower offset voltage.

b. The amplifier above has = 100k , = 100 V/V, and = 1 and is driven by a

source with = 25k and drives a load = 3 . Calculate the overall gain as well as the amount of input and output loading. [10 marks]

By using the equation given in (a), we obtain the overall gain of the amplifier:

As a result, the overall gain for the first case is

(

) (

)

This is less than 100 V/V because of loading.

6. Given in Figure 6 below is a frequency response of an open loop voltage amplifier.

Figure 6: Frequency response plots of the amplifier

a. Estimate the phase margin and gain margin of the amplifier above. [10 marks]

The phase margin in the frequency response of an op amp is given as:

( )

7

We have the phase margin of the op amp circuit above as

The gain margin in the frequency response of an op amp is given as:

We have the gain margin of the op amp circuit above as -20 dB

b. With your estimation on the phase and gain margin, is the amplifier stable or it is not

stable? [2 marks]

Based on the estimation of the phase margin and gain margin the amplifier is not

stable, as at 180 its gain and phase are already in the negative zone.

c. If you have to redesign the above amplifier what do you do to make the amplifier to be stable? [2 marks]

Form a closed loop or feedback circuit with a considerable gain or, add a controller to compensate for the extra gain introduced to the amplifier.

7. For a given op amp circuit diagram shown in Figure 7 below.

Figure 7: Circuit for output error of an op amp due to bias current

a. What is bias current and what is offset current in an op amp. [2 marks]

Bias current is the average of the currents in the input pins of the op amp and offset current is the difference between the currents at the input pins of the op amp.

b. Let = 22 k and = 2.2 M, and let the op amp ratings be = 80 nA and = 20 nA. Calculate the output error of the amplifier given above due to input bias current. [10 marks]

8

The DC noise gain of the op amp is 1 + R2/R1 = 101 V/V

Moreover, ( ) 22 k, with = 0, we have:

( )

( )

mV

As a result the output error is calculated to be 175 mV

C. Op Amp based Circuits I

8. Given in the Figure 8 below is a summing amplifier.

Figure 8: Summing amplifier

a. Describe the summing amplifier. [2 marks]

It is a type operational amplifier circuit which can be used to sum signals. The sum of the input signal is amplified by a certain factor and made available at the output. Any number of input signal can be summed using an op amp. The circuit shown above is a three input summing amplifier in the inverting mode.

b. Describe its operation. [2 marks]

9

In the circuit, the input signals Va, Vb, and Vc are applied to the inverting input of the op amp through input resistors Ra, Rb, and Rc respectively.

Any number of input signals can be applied to the inverting input in the above manner.

Non-inverting input of the op amp is grounded using resistor Rm. In this circuit Rf is the feedback resistor whereas RL is the load resistor.

c. Derive its transfer function. [8 marks]

By applying Kirchhoff’s current law at node V2 we get:

Since the input resistance of an ideal op amp is close to infinity and has infinite gain. We can neglect Ib & V2, therefore:

(1)

Equation (1) can be rewritten as:

(

) (

) (

)

Neglecting Vo, we get:

Rearrange the equation:

[(

) (

) (

)]

Also,

[(

) (

) (

) ] (2)

If resistor Ra, Rb, Rc has same value i.e. , then equation (2) can be written as:

(

)( ) (3)

If the values of Rf and R are made equal, then the equation becomes:

( )

10

d. Describe two applications of summing amplifier. [4 marks]

Averaging Circuit :

An averaging circuit can be made from the summing amplifier circuit by making the all input resistor equal in value i.e.

and the gain must be selected such that if there are m inputs, then

Scaling amplifier :

In a scaling amplifier each input will be multiplied by a different factor and then summed together.

Scaling amplifier is also called a weighted amplifier.

In this case different values are chosen for Ra, Rb and Rc. The governing equation is:

(( ) ( ) ( ) )

9. Given in the Figure 9 below is a difference amplifier

Figure 9: Difference amplifier

a. Describe the function of the difference amplifier. [2 marks]

Differential amplifiers amplify the difference between two voltages making this type

11

of operational amplifier circuit a subtractor unlike a summing amplifier which adds or sums together the input voltages.

b. Derive its transfer function equation. [8 marks]

By connecting each input in turn to 0 V ground we can use superposition to solve for the output voltage Vout. Then the transfer function for a differential amplifier circuit is given as:

At summing point, and

(

)

If , then

( ) ( )

If , then

( ) (

) (

)

But,

( ) ( )

Finally

( ) (

) (

)

When resistors, R1 = R2 and R3 = R4 the above transfer function for the differential amplifier can be simplified to the following expression:

( )

c. Describe two of its application in practice. [4 marks]

Wheatstone Bridge Differential Amplifier

12

The standard Differential Amplifier circuit now becomes a differential voltage comparator by “comparing” one input voltage to the other. By connecting one input to a fixed voltage reference set up on one leg of the resistive bridge network and the other to either a “Thermistor” or a “Light Dependant Resistor”, the amplifier circuit can be used to detect either low or high levels of temperature or light. In this case the output voltage becomes a linear function of the changes in the active leg of the resistive bridge.

Light Activated Differential Amplifier

The circuit above acts as a light-activated switch which turns the output relay either “ON” or “OFF” as the light level detected by the LDR resistor exceeds or falls below some pre-set value. A fixed voltage reference is applied to the non-inverting input terminal of the op-amp via the R1–R2 voltage divider network.

10. For an instrumentation amplifier,

a. Describe the function of the instrumentation amplifier. [2 marks]

Instrumentation Amplifiers (in-amps) are very high gain differential amplifiers which have a high input impedance and a single ended output. Instrumentation amplifiers are mainly used to amplify very small differential signals from strain gauges, thermocouples or current sensing devices in motor control systems.

Unlike standard operational amplifiers in which their closed-loop gain is determined by an external resistive feedback connected between their output terminal and one input terminal, either positive or negative.

Instrumentation amplifiers have an internal feedback resistor that is effectively isolated from its input terminals as the input signal is applied across two differential inputs, V1 and V2.

13

The instrumentation amplifier also has a very good common mode rejection ratio, CMRR (zero output when V1 = V2) well in excess of 100dB at DC.

b. Describe its operation. [4 marks]

The two non-inverting amplifiers form a differential input stage acting as buffer amplifiers with a gain of 1 + 2R2/R1 for differential input signals and unity gain for common mode input signals. Since amplifiers A1 and A2 are closed loop negative feedback amplifiers, we can expect the voltage at Va to be equal to the input voltage V1. Likewise, the voltage at Vb to be equal to the value at V2. As the op-amps take no current at their input terminals (virtual earth), the same current must flow through the three resistors network of R2, R1 and R2 connected across the op-amp outputs. This means then that the voltage on the upper end of R1 will be equal to V1 and the voltage at the lower end of R1 to be equal to V2. This produces a voltage drop across resistor R1 which is equal to the voltage difference between inputs V1 and V2, the differential input voltage. Because the voltage at the summing junction of each amplifier, Va and Vb is equal to the voltage applied to its positive inputs.

14

However, if a common-mode voltage is applied to the amplifiers inputs, the voltages on each side of R1 will be equal, and no current will flow through this resistor. Since no current flows through R1 (nor, therefore, through both R2 resistors, amplifiers A1 and A2 will operate as unity-gain followers (buffers). Since the input voltage at the outputs of amplifiers A1 and A2 appears differentially across the three resistors network, the differential gain of the circuit can be varied by just changing the value of R1. The voltage output from the differential op-amp A3 acting as a subtractor, is simply the difference between its two inputs ( V2–V1 ) and which is amplified by the gain of A3 which may be one, unity, (assuming that R3=R4). Then we have a general expression for overall voltage gain of the instrumentation amplifier circuit as:

c. State its transfer function equation. [ 2 marks]

( ) ( )( )

D. Op Amp based Circuits II

11. Consider the comparator circuit shown in Figure 10 below, whose output saturates at ±13 V.

Figure 10: Voltage comparator circuit

a. Calculate values for R1 and R2 to give a hysteresis of 0.5 V. [5 marks]

15

Hysteresis is found from:

The hysteresis of the circuit above is given as

(

) ( )

( ( ))

As a result . Pick to be equal to 1 k, as a result is 51 k.

b. Plot the voltage transfer curve of the comparator assuming hysteresis of 0.5 V. [5 marks]

c. Plot the expected output waveform, Vo against time when Vi is a Vpk-pk, 100 Hz ac-coupled triangle wave. Include the input waveform in your plot, as well as coordinates to clearly identify when transitions occur in the output waveform. [5 marks]

The expected output waveform against time when the input voltage is ac-coupled triangle wave is illustrated below:

16

Shown above is a hysteresis waveform plot with the low threshold ( ) and high threshold ( ) voltages and low output ( ) and high output ( ) voltages as indicated on the diagram.

12. Given in Figure 11 below is a Wien-Bridge oscillator circuit.

Figure 11: Wien-Bridge oscillator circuit

a. Describe the effect of reducing resistance of the Wien-Bridge oscillator on its frequency. [2 marks]

In a Wien-Bridge oscillator, if the resistances in the positive feedback circuit are decreased, the frequency increases.

b. Calculate the resonant frequency and gain of the circuit given above. [10 marks]

The transfer function of the oscillator is given as:

17

( )

( )

The condition for zero phase shift is given as:

Equating R1 to R2 to be equal to 10 k and C1 to C2 to be equal to 10 nF, this results in the resonance frequency of the oscillator,

√

√

Finally, the transfer function becomes:

At resonance frequency, the gain of the oscillator is:

c. Describe two applications of the oscillator given above. [2 marks]

Oscillator is used in function generator producing reference signals and providing timing feature for digital transmission application.

13. Given in the Figure 12 below is an RC filter circuit. The Bode plot of the circuit shown in Figure 13, with the output taken across capacitor C2 and selected values from the Bode plot from the circuit measurements in the lab are listed in Table 1.

Figure 12: RC filter circuit.

18

Figure 13: Bode plot. Solid curve is gain (left axis); dashed curve is phase (right axis)

Frequency (Hz) Gain (dB) Phase (degree)

9.5 -4.5 45

15 -3 32

109 -1.5 0

910 -3 -38

1.1 k -3.6 -45

1.4 k -4.5 -53

Table 1: Gain and phase values at specific frequencies for the circuit shown in Figure 12.

a. Calculate the cut off frequency of the filter circuit above. [10 marks]

The band-pass filter as given above can be treated as two stages filter that consists of high-pass and low-pass filters.

High-pass filter stage:

Low-pass filter stage:

As a result the lower frequency cut off of the band-pass filter is 11.37 Hz

b. From the table and graph, find the high-pass cut-off frequency for this filter is. [2 marks]

19

The cut off frequency of the filter is 9.5 Hz from the table and graph.

c. Explain the difference between the calculated and the measured values. [2 marks]

The differences are due to tolerances of the components used in the circuit and also due to errors of testing instruments used in measuring the variables.

14. For a given active filter shown in the Figure 14 below.

Figure 14: Sallen-key based low-pass filter circuit

a. For an active filter used for attenuating signal application, which of the standardized filter design e.g. Butterworth, Bessel, Chebyshev, and Elliptic will be best to realize a stop band filter. Give a reason for your answer. [2 marks]

For a given filter order the best design for signal attenuation application is the Chebyshev Type II. The reason is the ripples that exist in the stop band that could be useful for maximizing signal attenuation.

b. Compare passive and active filters. Give an advantage for each and state applications were each type of filter are particularly well suited. [5 marks]

Advantage:

Passive filter: more precise design outcome and cost effective.

Active filter: design flexibilities and application beyond conventional usages.

Application:

Passive filter: application that requires more precise result

Active filter: application that requires advanced signal treatment e.g. amplification, conversion, rectification, etc.

c. Calculate the cut-off frequency and Q-factor for the low-pass filter shown in figure given above, where = 1600 Ω, = 400 Ω, = 40 nF, and = 10 nF. [10 marks]

20

Cut-off frequency of the low pass filter:

√

√

Q factor of the low pass filter:

√

√

21

SELECTIVE FORMULAE IN ANALOGUE ELECTRONICS

Op Amp

[

( )

( )] and

( )

|

|

Inverting Amplifier

( )

Non-Inverting Amplifier

(

)

Summing Amplifier

(

)

Comparator

(

) ( )

Oscillator (Wien-Bridge)

( )

( )

Low/High Pass Filter

Sallen-Key Filter

√ and

√