1. Passive Components Rayat Shikshan Sanstha’s
S.M. Joshi College Hadapsar-028
Department of Electronics Science
Passive Components
Presented by-
Dr. Kakade P.K

2. Op-Amp Introduction of Operation Amplifier (Op-Amp) Block Diagram
Analysis of ideal Op-Amp applications
Comparison of ideal and non-ideal Op-Amp
Non-ideal Op-Amp consideration
Operational Amplifier

3. BLOCK DIAGRAM Input Differential stage Gain stage
Output Buffering stage
Operational Amplifier

4. Stages Input Differential Stage
Intermediate Signal-Ended High-Gain Stage
Output Buffering Stage
Current Source / Short Circuit Protection

5. Input Differential Stage
The input stage consists of the transistors Q1 through Q7 with biasing performed by Q8, Q9, and Q10.
Transistors Q1 and Q2 are emitter followers which causes input resistance to be high and deliver the differential input signal to the common base amplifier formed by Q3 and Q4.
Transistors Q5, Q6, and Q7, and resistors R1, R2, and R3 form the load circuit of the input stage. This portion of the circuit provides a high resistance load.
Transistors Q3 and Q4 also serve as protection for Q1 and Q2. The emitter-base junction of Q1 and Q2 breaks down at around 7V but the pnp transistors have breakdown voltages around 50V. So, having them in series with Q1 and Q2 protects Q1 and Q2 from an accidental connection between the input terminals.

6. Intermediate Single-Ended High-Gain Stage
The second stage is composed of Q16, Q17,
Q13B, and the resistors R8 and R9.
Transistor Q16 acts as an emitter follower
giving the second stage a high input resis-
tance.
Transistor Q17 is a common-emitter amplifier
with a 100-Ώ resistor in the emitter. The load
of this amplifier is composed of the output
resistance of Q13B. This use of a transistor
as a load resistance is called active load.
The output of this amplifier (the collector of
Q17) has a feedback loop through Cc. This
capacitor causes the op-amp to have a pole
at about 4Hz.

7. Output Buffering Stage
The Output Stage consists of the
complimentary pair Q14 and Q20,
and a class AB output stage
composed of Q18 and Q19. Q15
and Q21 give short circuit protection
(described later) and Q13A supplies
current to the output stage.
The purpose of the Output Stage is
to provide the amplifier with a low
output resistance. Another requirement
of the Output Stage is the ability to
dissipate large load currents without
dissipating large quantities of power.
This is done through the class AB Out-
put Stage.

8. Operational Amplifier (Op-Amp)
Very high differential gain
High input impedance
Low output impedance
Provide voltage changes (amplitude and polarity)
Used in oscillator, filter and instrumentation
Accumulate a very high gain by multiple stages
Operational Amplifier

9. IC Product
DIP-741
Dual op-amp 1458 device
Operational Amplifier

10. Single-Ended Input + terminal : Source – terminal : Ground
0o phase change
+ terminal : Ground
– terminal : Source
180o phase change
Operational Amplifier

11. Double-Ended Input Differential input 0o phase shift change
between Vo and Vd
Qu: What Vo should be if,
Ans: (A or B) ?
(A)
(B)
Operational Amplifier

12. Distortion
The output voltage never excess the DC voltage supply of the Op-Amp
Operational Amplifier

13. Common-Mode Operation
Same voltage source is applied
at both terminals
Ideally, two input are equally
amplified
Output voltage is ideally zero
due to differential voltage is
zero
Practically, a small output
signal can still be measured
Note for differential circuits:
Opposite inputs : highly amplified
Common inputs : slightly amplified
 Common-Mode Rejection
Operational Amplifier

14. Common-Mode Rejection Ratio (CMRR)
Differential voltage input :
Common voltage input :
Common-mode rejection ratio:
Output voltage :
Note:
When Gd >> Gc or CMRR 
Vo = GdVd
Gd : Differential gain
Gc : Common mode gain
Operational Amplifier

15. CMRR Example What is the CMRR? Solution : (2) (1)
NB: This method is Not work! Why?
Operational Amplifier

16. Op-Amp Properties Infinite Open Loop gain (2) Infinite Input impedance
The gain without feedback
Equal to differential gain
Zero common-mode gain
Pratically, Gd = 20,000 to 200,000
(2) Infinite Input impedance
Input current ii ~0A
T- in high-grade op-amp
m-A input current in low-grade op-amp
(3) Zero Output Impedance
act as perfect internal voltage source
No internal resistance
Output impedance in series with load
Reducing output voltage to the load
Practically, Rout ~ 
Operational Amplifier

17. Frequency-Gain Relation
Ideally, signals are amplified from DC to the highest AC frequency
Practically, bandwidth is limited
741 family op-amp have an limit bandwidth of few KHz.
20log(0.707)=3dB
Unity Gain frequency f1: the gain at unity
Cutoff frequency fc: the gain drop by 3dB from dc gain Gd
GB Product : f1 = Gd fc
Operational Amplifier

18. GB Product
Example: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV
Sol:
Since f1 = 10 MHz
By using GB production equation
f1 = Gd fc
fc = f1 / Gd = 10 MHz / 20 V/mV
= 10  106 / 20  103
= 500 Hz
10MHz
? Hz
Operational Amplifier

19. Ideal Vs Practical Op-Amp
Open Loop gain A 
105
Bandwidth BW
10-100Hz
Input Impedance Zin
>1M
Output Impedance Zout
0  
Output Voltage Vout
Depends only on Vd = (V+V)
Differential mode signal
Depends slightly on average input Vc = (V++V)/2 Common-Mode signal
CMRR
10-100dB
Operational Amplifier

20. Ideal Op-Amp Applications
Analysis Method :
Two ideal Op-Amp Properties:
The voltage between V+ and V is zero V+ = V
The current into both V+ and V termainals is zero
For ideal Op-Amp circuit:
Write the kirchhoff node equation at the noninverting terminal V+
Write the kirchhoff node eqaution at the inverting terminal V
Set V+ = V and solve for the desired closed- loop gain
Operational Amplifier

21. Noninverting Amplifier
Kirchhoff node equation at V+ yields,
Kirchhoff node equation at V yields,
Setting V+ = V– yields or
Operational Amplifier

22. Noninverting amplifier Noninverting input with voltage divider
Less than unity gain
Voltage follower
Operational Amplifier

23. Inverting Amplifier Kirchhoff node equation at V+ yields,
Setting V+ = V– yields
Notice: The closed-loop gain Vo/Vin is dependent upon the ratio of two resistors, and is independent of the open-loop gain. This is caused by the use of feedback output voltage to subtract from the input voltage.
Operational Amplifier

24. Multiple Inputs Kirchhoff node equation at V+ yields,
Setting V+ = V– yields
Operational Amplifier

25. Inverting Integrator
Now replace resistors Ra and Rf by complex components Za and Zf, respectively, therefore
Supposing
The feedback component is a capacitor C, i.e.,
The input component is a resistor R, Za = R
Therefore, the closed-loop gain (Vo/Vin) become:
where
What happens if Za = 1/jC whereas, Zf = R?
Inverting differentiator
Operational Amplifier

26. Op-Amp Integrator Example: Determine the rate of change
of the output voltage.
Draw the output waveform.
Solution:
(a) Rate of change of the output voltage
(b) In 100 s, the voltage decrease
Operational Amplifier

27. Op-Amp Differentiator
Operational Amplifier

28. Non-ideal case (Inverting Amplifier)
Equivalent Circuit
3 categories are considering
Close-Loop Voltage Gain
Input impedance
Output impedance
Operational Amplifier

29. Close-Loop Gain Applied KCL at V– terminal,
By using the open loop gain,  
The Close-Loop Gain, Av
Operational Amplifier

30. Close-Loop Gain
When the open loop gain is very large, the above equation become,
The close-loop gain now reduce to the same form
as an ideal case
Operational Amplifier

31. Input Impedance Input Impedance can be regarded as,
where R is the equivalent impedance
of the red box circuit, that is
However, with the below circuit,
Operational Amplifier

32. Input Impedance Finally, we find the input impedance as,  Since,
, Rin become,
Again with
The op-amp can provide an impedance isolated from
input to output
Operational Amplifier

33. Output Impedance
Only source-free output impedance would be considered,
i.e. Vi is assumed to be 0
Firstly, with figure (a),
By using KCL, io = i1+ i2
By substitute the equation from Fig. (a),
R and A comparably large,
Operational Amplifier