ANALOGUE SYSTEM INTEGRATION

March 25th, 2017

Table of Contents Page:
1.1. Objectives 2
1.2 Introduction 3
1.3 Theory 3
1.3. DC calculation 3
1.4. AC calculations. 6
2.1. Second part 11
2.1.1. Simulation of the circuit DC analysis. 11
2.1.2. Simulate the circuit AC analysis. 13
2.1.3. Common gain mode 16
2.1.4. Practical DC analysis 17
3.1. Third part. 18
3.1.1. Improvement 18
3.1.2. Common mode gain 21
1.5. Conclusion 22
Objectives

Understanding and explanation of two AC & DC circuit analysis building blocks and their function. Finding the values
of DC collector currents, voltages, quiescent power dissipation and the total voltage gain. Then calculate the AC
characteristics of the circuit such as total voltage gain (dB), input-output impedance, CMRR and the maximum output voltage
swing.

Design the two AC & DC circuit analysis building blocks and simulate it using probes to measure the values of the DC
operating conditions and AC operating characteristics. Compare the results, and comment on any discrepancies.

Making an improvement for the original circuit to drive 8 oms load speaker.

Introduction
Nowadays, operational amplifier is the most useful single device in analogue electronics. The performance of the operational
amplifier is wide range of analog signal processing tasks and it is cheap. This report will discuss the simple integrated
circuit amplifier by three parts which is (hand calculation, software simulation and practical lab).

1.3. Theory

Figure 1 schematic of operational amplifier with four transistors
1.3. DC calculation
In this section the requirement is to find the three values in the DC analysis circuit which is;
Current Ic
Voltage Vc
Quiescent power dissipation

However there are some values to start the calculation and it is provided data in this section starting with;
Vcc = +10 V
VEE = -10 V
R1 = 4.7k?, R2 = 4.7k?, R3 = 1k?, R4 = 3.1k?, R5 = 1k?
ß = 120 (NPN) & 60 (PNP)
VBE = 0.7 V
Assume ß >> 1

1- Finding the value of IT
IT = (VEE – VBE) / R1
= (10 – 0.7) V / 4.7k?
= 9.3V/ 4.7k?
= 2 mA

2-Finding the value of IC
IC = IT/2 ? as IC ˜ IE
= 2 x 10-3/ 2
= 1 mA
ICQ1 ˜ ICQ2 ˜ 1 mA, because it is a differential amplifier so the current be equal

3-Finding the value of V2
V2 = ICQ2 x R2
= 1mA x 4.7k?
= 4.7V
4-Finding the value of VCQ2
Therefore both Q1 & Q2 are biased at 1 mA while collector voltages for both Q1 & Q2 are;
VCQ2 = VCC – V2
= 10V – 4.7V
= 5.3V

5-Finding the value of VEQ3
In this case the voltage at the emitter of PNP transistor Q3 equal;
VEQ3 = VCQ1 + VBE
= 5.3V + 0.7V
= 6V

6-Find the value of IEQ3
The emitter current of Q3 is equal to;
IEQ3 = VEE – VEQ3/R3
= 10V – 6V/1k?
= 4V/1k?
= 4mA since IEQ3 = ICQ

7-Finding the value of VCQ3
The voltage VCQ3 at the collector of Q3 or the base of Q4 is
VCQ3 = – VEE + (IEQ3 x R4)
= -10V + (4mA x 3.1k?)
= 2.4V

8-Finding the value of IEQ4
Therefore, the quiescent emitter current of Q4 is
IEQ4 = [VCQ3 – VBE – (-VEE)]/ R5
= [2.4V – 0.7 + 10]/ 1k?
= 11.7 mA

9-Finding the value of Vout
Vout = -VCC + (IEQ4 x R5)
= -10 + (11.7mA x 1k?)
= 1.7V

10-Finding the Quiescent power dissipation is Power in each resistor.
P_1= I^2 xR_1= ?(1.98m)?^2 x (4.7k) = 18.4mW
P_2= I^2 xR_2= ?(0.99m)?^2 x (4.7k) = 4.6mW
P_3= I^2 xR_3= ?(3.9m)?^2 x (1k) = 15.21mW
P_4= I^2 xR_4= ?(3.9m)?^2 x (3.1k) = 47.15mW
P_5= I^2 xR_5= ?(11.39m)?^2 x (1k) = 129.73mW

So the total power in each resistor is;
P_t=P_1+ P_2+ P_3+P_4 + P_5
P_t= (18.4m) + (4.6m) + (15.21) + (47.15m) + (129.73) = 215.1mw.

11-Finding the total power in each transistor.
P_Q1= I_C xV_CE= (9.3 )x (0.99m) = 9.2mW
P_Q2= I_C xV_CE= (4.7 )x (0.99m) = 4.6mW
P_Q3= I_C xV_CE= (4.1 )x (3.9m) = 15.6mW
P_Q4= I_C xV_CE= (861 )x 11.39m) = 98.1mW

So the total power in each resistor
P_t=P_1+ P_2+ P_3+P_4
P_t= (9.2m) + (4.6m) + (15.6) + (98.1m) = 127.5mw.

1.4. AC calculations.
In this section the requirement is to find the four values in the AC analysis circuit which is;
Total voltage Gain in (dB)
Input impedance
Output impedance
(dB Common Mode Rejection Ratio) CMRR

In this section of an AC circuit: due to AC small signal gain assuming VT =26mV, consequently the AC emitter resistance of
the transistor Q1 – Q2 is
1- Finding the value of re
re = VT / IE
re = 26mV/IE mA
= 26mV/1mA
= 26?
The voltage gain of the first stage can now be calculated after finding
Rin (Q3) ||R2
2- Finding the value of Rin (Q3)
Rin (Q3) = (1 + ß). (re3 + R3)re3 = VT/ IEQ3
= 26mV/4mA = 6.5?
= (1 + 60). (6.5? + 1k?) As PNP ß = 60
= (61) (1k?)
= 61.4k?
3- Finding the value of RC1
RC1 = Rin (Q3) ||R2 = (Rin (Q3) x R2)/ (Rin (Q3) + R2)
= (61.4k x 4.7k)/ (61.4k + 4.7k)
= 4.4k?
4- Finding the value of AV1
AV1 = RC1/ 2. re
= 4.4k/ 2. (26 ?)
= 4.4k/ 52?
= 85
Consequently in dB,AdB1 = 20.log10 (85)
= 38.6dB

5- Finding the value of Rin (Q4)
In the second stage the calculation of the first dynamic emitter resistance of Q4.
Rin (Q4) = (1 + ß). (re4 + R5)
re4 = VT/ IEQ3 = 26mV/11.7mA = 2.2?
Rin (Q4 = (1 + 120). (2.2? + 1k?) As NPN ß = 120
= (121) (1k?)
= 121.3k?

6- Finding the value of RC2
RC2 = Rin (Q4) ||R4 = (Rin (Q4) x R4)/ (Rin (Q4) + R4)
= (121.3k x 3.1k)/ (121.3k + 3.1k)
= 3k?

7- Finding the value of AV2
AV2 = -RC2/ re3 + R3
= -3k/ 6.5? + 1k?
= -4.4k/ 1k?
= -3
Consequently in dB, AdB2 = 20.log10 (3)
= 9.5dB

8- Finding the value of AV3
In the output stage the calculation of the first dynamic emitter resistance of Q4.
AV3 = Vout4/ V in4
= R5/ re4 + R5
= 1k/ (2.2? + 1k?)
= 0.997
Consequently in dB, AdB3 = 20.log10 (1) = 0dB
9- Finding the value of Av (total)
The calculation of the total voltage gain equal to multiply the gain of the stage
Av (total) = AV1 x AV2 x AV3
= 85 x -3 x 1
= -255

10- Finding the value of AdB (total)
The calculation of the total voltage gain equal by adding the gain of the stage in dB
AdB (total) = AdB1 + AdB2 + AdB3
= 38.6dB + 9.5dB + 0dB
= 48.1dB

output voltage swing – AV (swing)
AV (swing) = -RC/ 2. re
= -4.7k/ 2. (26?)
= -90.38
In dB
AdB (swing) = 20.log10 (90.38)
= 39.16dB

Common Mode Rejection Ratio – CMRR
CMRR = RT/ re
= 4.7k? / 26?
= 180.8 In dB
CMRRdB = 20.log10 (180.7)
= 45.14dB

Common Mode Gain – CMG
CMG = -RC/ 2. RT
= – 4.7k? / 2. (4.7k?)
= -0.5 In dB
CMGdB = 20.log10 (0.5)
= -6.02dB

2.1. Second part
2.1.1. Simulation of the circuit DC analysis.
This figure (2) below describe the simulation of the circuit using multisim program.

Figure 2 DC analysis circuit

According to the figure (2) above the values of collector current are;
I_c1= 1.05 mA
I_c2=919 µA
I_c3=3.59 mA
I_c4=10.4 mA

The values of voltage are;
V_1=0 v
V_2=5.68 v
V_3=1.13 v
V_4=399 mv
The measurement of the reading in probe 1
Voltage : 0 mV
Voltage peak to peak : 0 V
Voltage (rms) : 0V
Voltage (dc) : 1.7 µV
Current : -3.45 µA
Current (p-p) : 0A
Current (rms) : 3.45 µA
Current (dc) : -3.45 µA
Frequency : 0

The measurement of the reading in probe 2
Voltage : 5.68 V
Voltage peak to peak : 52.3 pV
Voltage (rms) : 5.68 V
Voltage (dc) : 5.63 V
Current : 919 uA
Current (p-p) : 0 A
Current (rms) : 919 µA
Current (dc) : 919 µA
Frequency : 23.4 KHz

The measurement of the reading in probe 3
Voltage : 1.13 V
Voltage peak to peak : 150 PV
Voltage (rms) : 1.13 V
Voltage (dc) : 1.13 V
Current = 3.59 mA
Current (p-p) : 0 A
Current (rms) : 3.59 mA
Current (dc) : 3.59 mA
Frequency : 23.4 KHz

The measurement of the reading in probe 4
Voltage : 399 mV
Voltage peak to peak : 149 pV
Voltage (rms) : 399 mV
Voltage (dc) : 399 mV
Current = 10.4 mA
Current (p-p) : 0 A
Current (rms) : 10.4 mA
Current (dc) : 10.4 mA
Frequency : 23.4 KHz
2.1.2. Simulate the circuit AC analysis.
Figure 3 AC analysis circuit
The measurement of the reading in probe 1
Voltage : -14.1 mV
Voltage peak to peak : 28.2 mV
Voltage (rms) : 10 mV
Voltage (dc) : 1.7 µV
Current : 3.02 µA
Current (p-p) : 850 µA
Current (rms) : 3.46 µA
Current (dc) : 3.45 µA
Frequency : 1 KHz

The measurement of the reading in probe 2
Voltage : 6.17 V
Voltage peak to peak : 1.46 V
Voltage (rms) : 5.7 V
Voltage (dc) : 5.68 V
Current : 816 uA
Current (p-p) : 310 µA
Current (rms) : 926 µA
Current (dc) : 919 µA
Frequency : 1 KHz

The measurement of the reading in probe 3
Voltage : -330 mV
Voltage peak to peak : 4.4 V
Voltage (rms) : 1.93 V
Voltage (dc) : 1.14 V
Current = 3.12 mA
Current (p-p) : 1.42 mA
Current (rms) : 3.36 mA
Current (dc) : 3.59 mA
Frequency : 1 KHz

The measurement of the reading in probe 4
Voltage : -1.06V
Voltage peak to peak : 4.37 V
Voltage (rms) : 1.6 V
Voltage (dc) : 405 mV
Current = 8.94 mA
Current (p-p) : 4.37 mA
Current (rms) : 10.5 mA
Current (dc) : 10.4 mA
Frequency : 1 KHz

The calculation of total voltage Gain (dB):

V_out/V_in = 4.37/(28.2 ×?10?^(-3) )=155

So the gain will be 20?LOG?_10 155=43.8dB

The calculation of total input impedance:
For R_in1 is
R_in1=(V(p-p))/(I(p-p))= (28.2 ×?10?^(-3))/(850 ×?10?^(-9) )=33058 ?

For R_in2 is
R_in2=(V(p-p))/(I(p-p))= (1.46 )/(310×?10?^(-6) )=4710 ?

For R_in3 is
R_in3=(V(p-p))/(I(p-p))= (4.4 )/(1.42 ×?10?^(-3) )=3099 ?

The calculation of total output impedance:
For R_outis
R_out=(V(p-p))/(I(p-p))= (4.37 )/(4.37 ×?10?^(-3) )=1k ?
2.1.3. Common gain mode

Figure 4 AC analysis with two sources
The Com Gain of the total Gain (dB):

V_out/V_in = (35.3 ×?10?^(-3))/(28.2 )=1.252

So the gain will be 20?LOG?_10 1.252=1.952 dB

The Common Mode Rejection Ratio
CMRR = gain/(com gain) = 155/(1.252 )=123.8
So the gain will be 20?LOG?_10 123.8=41.85 dB

2.1.4. Practical DC analysis.

The procedure of practical DC analysis part configure the operational amplifier of the four transistors were demonstrated on
the lab using components which is;
R1 = 4.7 K ?
R2 = 4.7 K ?
R3 = 1 k ?
R4 = 3.1K ?
R5 = 1 k ?
Transistors NPN, BC548 and PNP BC559

The measurement of the voltage from the strip board circuit;
In the first probe the voltage equal 5.30 V
In the second probe the voltage equal 5.30 V
In the third probe the voltage equal 2.5V
In the fourth probe the voltage equal 1.22V

2.1.5 Practical AC analysis.
The procedure of practical AC analysis part is connect the circuit with the AC power supply to calculate the gain;

V_out (p-p)= 6 V
Vin(p-p) = 28.3 mV
The gain = (V_out (p-p))/(Vin(p-p))= 6/(28.3 ×?10?^(-3) )=212

To sum up, the result of the hand writing and multisim result it was close together
3.1. Third part.
3.1.1. Improvement
In this part there is a change in the circuit analysis to show the improvement through using current source and cascade
deferential amplifier following by the single ended with common emitter output as in figure 5.

Figure 5 improvement circuit

The measurement of the reading in probe 1
Voltage : -65.7 µV
Voltage peak to peak : 282 µV
Voltage (rms) : 100 µV
Voltage (dc) : 12.7 nV
Current : 1.92 µA
Current (p-p) : 9.26 nA
Current (rms) : 1.93 µA
Current (dc) : 1.93 µA
Frequency : 1 KHz

The measurement of the reading in probe 2
Voltage : 8.4 V
Voltage peak to peak : 6.86 mV
Voltage (rms) : 8.4 V
Voltage (dc) : 8.4 V
Current : 533 µA
Current (p-p) : 2.29 µA
Current (rms) : 533 µA
Current (dc) : 533 µA
Frequency : 1 KHz

The measurement of the reading in probe 3
Voltage : -3.85 V
Voltage peak to peak : 2.09 V
Voltage (rms) : 4.4 V
Voltage (dc) : -4.34 V
Current = 392 µA
Current (p-p) : 133 µA
Current (rms) : 364 µA
Current (dc) : 361 µA
Frequency : 1 KHz

The measurement of the reading in probe 4
Voltage : -4.5 V
Voltage peak to peak : 2.08 V
Voltage (rms) : 5.04 V
Voltage (dc) : -4.98 V
Current = 1.83 mA
Current (p-p) : 694 µA
Current (rms) : 1.69 mA
Current (dc) : 1.67 mA
Frequency : 1 KHz

The calculation of total voltage Gain (dB):

V_out/V_in = 2.08/(282 ×?10?^(-6) )=7375.8

So the gain will be 20?LOG?_10 7375.8 =77.3 dB

The calculation of total input impedance:
For R_in1 is
R_in1=(V(p-p))/(I(p-p))= (282 ×?10?^(-6))/(9.26×?10?^(-9) )=30453 ?

For R_in2 is
R_in2=(V(p-p))/(I(p-p))= (6.86 ×?10?^(-3))/(2.29×?10?^(-6) )=2995.6 ?

For R_in3 is
R_in3=(V(p-p))/(I(p-p))= (2.09 )/(133×?10?^(-6) )=15714.2 ?

The calculation of total output impedance:
For R_outis
R_out=(V(p-p))/(I(p-p))= 2.08/(694 ×?10?^(-6) )=2997 ?

3.1.2. Common mode gain

Figure 6common mode gain
The Com Gain of the total Gain (dB):

V_out/V_in = (423 ×?10?^(-9))/(281 ×?10?^(-6) )=1.505 m

So the gain will be 20?LOG?_10 1.252=-56.448 dB

The Common Mode Rejection Ratio
CMRR = gain/(com gain) = 7375.8/(1.505×?10?^(-3) )=4900863
So the gain will be 20?LOG?_10 4900863=133.9 dB

Conclusion
The simple integrated circuit amplifier has been done using three parts which is (hand calculation, software simulation and
practical lab). Then the result was approximately equal.