The slide contents are: 1.Introduction to analogue electronics 2.Operational Amplifier (op-amp) 3.Power Amplifiers

1. Course Name: Analogue and
Digital Electronics (ECTS:5)
Course code:
 By: Amanuel M.
 2014 E.C./2022 G.C.

2. 1. Introduction to analogue electronics
2. Operational Amplifier (op-amp)
3. Power Amplifiers
4. Multi Stage Amplifiers

3. To know analogue electronics systems and their difference
with digital electronics.
To recall OP-AMP and its applications in analogue electronics.
To use different amplifiers for different applications.
To compute the output power for amplifiers; operational,
power, multi-stage & differential amplifiers.

4. • The world of electronics is all about electrical circuits, electronic
components, and interconnected technologies.
• All these elements (electronics) can be primarily categorized as
digital, analog, or a combination of both.
• Analog electronics is a branch of electronics that deals with a
continuously variable signal. It’s widely used in radio and audio
equipment along with other applications where signals are derived
from analog sensors before being converted into digital signals for
subsequent storage and processing.

5. • Digital circuits are considered as a dominant part of today’s
technological world, some of the most fundamental components in a
digital system are actually analog in nature.
• Analog circuits can be defined as a complex combination of op-amps,
resistors, capacitors, and other basic electronic components.
• These circuits can be as simple as a combination of two resistors to
make a voltage divider or elegantly built with many components.
• Such circuits can attenuate, amplify, isolate, modify, distort the
signal, or even convert the original one (analog) into a digital signal.
https://www.engineersgarage.com/an-introduction-to-
analog-electronics/

6. • An operational amplifier(op-amp) is an active element with a very
high gain differential amplifier with high input impedance and low
output impedance.
• Used provide voltage amplitude changes (amplitude and polarity),
oscillators, filter circuits, and many types of instrumentation
circuits.
• An op-amp contains a number of differential amplifier stages to
achieve a very high voltage gain.

7. • Commercially available in integrated circuit (IC) packages in several
forms (mostly 8-pin in dual in-line package/DIP).
Fig. 1.1. A typical op-amp: (a) pin configuration, (b) circuit symbol

8.  Single-Ended Input
• Single-ended input operation results when the input signal is connected to
one input with the other input connected to ground.
• If an input applied to the non-inverting terminal will appear with the same
polarity at the output =>See Fig1.2.(a)
• An input applied to the inverting terminal will appear inverted at the
output. => See Fig1.2.(b)
Fig. 1.2. Single-ended operation: (a) Non-inverting op-amp & (b) Inverting op-amp

9.  Double-Ended (Differential) Input
• Apply signals (𝑽𝒅) at each input this being a double-ended operation.
• Two separate signals (𝒗𝟏 & 𝒗𝟐) are applied to the inputs, the difference
signal will be: Use Fig1.3(c):
• The output voltage section consists of a voltage-controlled voltage-source
(A𝒗𝒅) in series with the output resistance 𝑹𝒐.
𝒗𝒅 = 𝒗𝟏 − 𝒗𝟐
Use KVL: 𝒗𝟐 +𝒗𝒅 − 𝒗𝟏 = 𝟎
𝑽𝒅 = 𝑽𝟏 − 𝑽𝟐
Fig. 1.3. Double-ended (differential) operation (a & b) & (c) equivalent circuit for Op-amps.

10. • Noted that input resistance (𝑹𝒊) is the Thevenin equivalent resistance seen
at the input terminals, &
• The output resistance (𝑹𝒐) is the Thevenin equivalent resistance seen at the
output.
• The op-amp senses the difference between the two inputs, multiplies it by
the gain A, and causes the resulting voltage to appear at the output (𝒗𝒐);
• A is called the open-loop voltage gain because it is the gain of the op-amp
without any external feedback from output to input.
• A negative feedback is achieved when the output is fed back to the inverting
terminal of the op-amp called closed loop circuit. See Fig. 1.4. in next slide
𝒗𝑶 = 𝑨𝒗𝒅 = 𝑨(𝒗𝟏 − 𝒗𝟐)
Where;
A : open-loop voltage gain

11. • When there is a feedback path from output to input, the ratio of the
output voltage to the input voltage is called the closed-loop gain.
• A practical limitation of the op-amp is that the magnitude of its
output voltage cannot exceed |VCC|, i.e. the output voltage (𝑣𝑂) is
dependent on and is limited by the power supply voltage (𝑉𝐶𝐶).
Fig. 1.4. (a) Op amp output voltage 𝒗𝑶 as a function of the differential input voltage 𝒗𝒅
& Negative feedback Op-amp circuit

12. • Thus op-amps can operate in three modes, depending on the differential
input voltage 𝒗𝒅:
• If we attempt to increase vd beyond the linear range, the op-amp becomes
saturated and yields:
• I.e.
1.2.2. Common-Mode Operation
• Occurs when the same input signals are applied to both inputs, common-
mode operation results
1. Positive Saturation; 𝒗𝒐 = 𝑽𝑪𝑪
2. Linear region; −𝑽𝑪𝑪 ≤ 𝒗𝒐 = A𝒗𝒅 ≤ 𝑽𝑪𝑪
3. Negative Saturation; 𝒗𝒐 = −𝑽𝑪𝑪
𝒗𝑶 = 𝑽𝑪𝑪 𝑜𝑟 𝒗𝑶 = −𝑽𝑪𝑪
−𝑽𝑪𝑪 ≤ 𝒗𝑶 ≤ 𝑽𝑪𝑪
Reading assignment: Double-Ended
Output

13. • Common-Mode Rejection (CMRR): for differential connection, amplification
of the opposite input signals is much greater than that of the common input
signals, the circuit provides a common mode rejection as described by a
numerical value called the common-mode rejection ratio (CMRR).
• Differential Inputs (𝑽𝒅): separate inputs (𝑽𝒊𝟏 & 𝑽𝒊𝟐) are applied to the op-
amp, the resulting difference signal is the difference between the two
inputs. See Fig. 1.3 (a) & (b)
𝑽𝒅 = 𝑽𝒊𝟏 − 𝑽𝒊𝟐
Fig. 1.5. Common mode operation

14. • Common Inputs (𝑽𝒄): both input signals are the same (in-phase), a common
signal element due to the two inputs can be defined as the average of the sum
of the two signals. See Fig 1.5.
• Output Voltage (𝑽𝒐): any signals applied to an op-amp in general have both in-
phase and out-of phase components, the resulting output can be expressed as;
• Opposite polarity (out-phased) as input; 𝑽𝒊𝟏 = −𝑽𝒊𝟐 = 𝑽𝒔
The resulting difference voltage:
For common voltage:
∴ The output voltage:
𝑽𝒄 =
(𝑽𝒊𝟏 + 𝑽𝒊𝟐)
𝟐
𝑽𝒐 = 𝑨𝒅𝑽𝒅 + 𝑨𝒄𝑽𝒄
Where;
A : gain of the op-amp
𝑽𝒅 = 𝑽𝒊𝟏 − 𝑽𝒊𝟐 = 𝑽𝒔 − −𝑽𝒔 = 𝟐𝑽𝒔
𝑽𝒄 =
(𝑽𝒊𝟏 + 𝑽𝒊𝟐)
𝟐
=
[𝑽𝒔 − (−𝑽𝒔)]
𝟐
= 𝟎
𝑽𝒐 = 𝑨𝒅𝑽𝒅 + 𝑨𝒄𝑽𝒄 = 𝑨𝒅(𝟐𝑽𝒔) + 𝑨𝒄 × 𝟎 = 𝟐𝐀𝐝𝐕𝐬

15. • Same polarity (in-phased) as input; 𝑽𝒊𝟏 = 𝑽𝒊𝟐 = 𝑽𝒔
The resulting difference voltage:
For common voltage:
∴ The output voltage:
Common-Mode Rejection Ratio (CMRR)
• Can be expresses mathematically as:
• Or can be expresses using logarithmic mathematical as;
𝑽𝒅 = 𝑽𝒊𝟏 − 𝑽𝒊𝟐 = 𝑽𝒔 − 𝑽𝒔 = 𝟎
𝑽𝒄 =
(𝑽𝒊𝟏 + 𝑽𝒊𝟐)
𝟐
=
[𝑽𝒔 + 𝑽𝒔]
𝟐
= 𝑽𝒔
𝑽𝒐 = 𝑨𝒅𝑽𝒅 + 𝑨𝒄𝑽𝒄 = 𝑨𝒅 × 𝟎 + 𝑨𝒄𝑽𝒔 = 𝐀𝐜𝐕𝐬
𝐂𝐌𝐑𝐑 =
𝑨𝒅
𝑨𝒄
𝐂𝐌𝐑𝐑 𝐥𝐨𝐠 = 𝟐𝟎 𝐥𝐨𝐠𝟏𝟎
𝑨𝒅
𝑨𝒄

16.  Example-1: Calculate the CMRR for the circuit measurements shown
in Fig.1.6
Fig. 1.6. For Ex-1; (a) differential-mode; (b)common-
mode

17.  Solution:
Step-1: find 𝑨𝒅 using the differential-mode circuit given in Fig. 1.6. (a) &
above equations;
𝑨𝒅 =
𝑽𝒐
𝑽𝒅
=
𝟖 𝑽
𝟏 𝒎𝑽
= 𝟖𝟎𝟎𝟎
Step-2: find 𝑨𝒄 using the common-mode circuit given in Fig. 1.6. (b) & above
equations;
𝑨𝒅 =
𝑽𝒐
𝑽𝒅
=
𝟏𝟐 𝒎𝑽
𝟏 𝒎𝑽
= 𝟏𝟐
Step-3: using
Or in logarithmic form;
𝑨𝒅 =
𝑽𝒐
𝑽𝒅
NB: notice that (differential);
𝑽𝒅 = 𝟐𝑽𝒔 &
𝑽𝒐 = 𝟐𝐀𝐝𝐕𝐬 = (𝟐𝐕𝐬)𝐀𝐝
Thus; 𝑽𝒐 = 𝑽𝒅𝑨𝒅
𝑨𝒄 =
𝑽𝒐
𝑽𝒄
NB: notice that (common mode);
𝑽𝒄 = 𝑽𝒔 &
𝑽𝒐 = 𝐀𝒄𝐕𝐬
Thus; 𝑽𝒐 = 𝐀𝒄𝐕𝒄
𝐂𝐌𝐑𝐑 =
𝑨𝒅
𝑨𝒄
=
𝟖𝟎𝟎𝟎
𝟏𝟐
= 𝟔𝟔𝟔. 𝟕
𝐂𝐌𝐑𝐑 𝐥𝐨𝐠 = 𝟐𝟎 𝐥𝐨𝐠𝟏𝟎
𝑨𝒅
𝑨𝒄
= 𝟐𝟎 𝐥𝐨𝐠𝟏𝟎 𝟔𝟔𝟔. 𝟕 = 𝟓𝟔. 𝟒𝟖 𝒅𝑩

18. • Noted that; it should be clear that the desired operation will have
very large 𝐀𝐝 with very small 𝐀𝐜. i.e.
• Thus we can express the output voltage (𝐕𝐨) in terms of the value of
CMRR as follows:
 Example-2: Determine the output voltage of an op-amp for input
voltages of 𝑉𝑖1 = 150 𝜇𝑉, 𝑉𝑖2 = 140 𝜇𝑉. The amplifier has a differential
gain of 𝐴𝑑 = 4000 and the value of CMRR is: (a) 100 (b) 105
𝐂𝐌𝐑𝐑 =
𝑨𝒅
𝑨𝒄
⇒ 𝒍𝒂𝒓𝒈𝒆 𝒗𝒂𝒍𝒖𝒆
𝑽𝒐 = 𝑨𝒅 𝑽𝒅(𝟏 +
𝟏
𝑪𝑴𝑹𝑹
𝑽𝒄
𝑽𝒅
)

19.  Solution:
Step-1: find 𝑽𝒅 & 𝑽𝒄;
𝑽𝒅 = 𝑽𝒊𝟏 − 𝑽𝒊𝟐 = 𝟏𝟓𝟎 − 𝟏𝟒𝟎 𝛍𝑽 = 𝟏𝟎 𝝁𝑽 &
𝑽𝒄 =
𝑽𝒊𝟏+ 𝑽𝒊𝟐
𝟐
=
𝟏𝟓𝟎+𝟏𝟒𝟎 𝛍𝑽
𝟐
= 𝟏𝟒𝟓 𝝁𝑽
Step-2: find 𝑽𝒐 for each cases;
(a) 𝐕𝐨 = 𝐀𝐝 𝐕𝐝(𝟏 +
𝟏
𝐂𝐌𝐑𝐑
𝐕𝐜
𝐕𝐝
) = 𝟒𝟎𝟎𝟎 (𝟏𝟎 𝝁𝑽) 𝟏 +
𝟏
𝟏𝟎𝟎
𝟏𝟒𝟓 𝛍𝐕
𝟏𝟎 𝛍𝐕
= 𝟒𝟓. 𝟖 𝐦𝐕
(b) 𝐕𝐨 = 𝐀𝐝 𝐕𝐝(𝟏 +
𝟏
𝐂𝐌𝐑𝐑
𝐕𝐜
𝐕𝐝
) = 𝟒𝟎𝟎𝟎 𝟏𝟎 𝛍𝐕 𝟏 +
𝟏
𝟏𝟎𝟓
𝟏𝟒𝟓 𝛍𝐕
𝟏𝟎 𝛍𝐕
= 𝟒𝟎. 𝟎𝟎𝟔 𝐦𝐕

20. 1.1.3. Op-Amp Basics
• An op-amp is a very high gain amplifier having very high input impedance
(typically a few mega-ohms) and low output impedance (less than 100 ).
• The basic circuit is made using a difference amplifier having two inputs
(plus and minus) and at least one output.
• The plus(+) input produces an output that is in phase with the signal
applied, while an input to the minus(-) input results in an opposite polarity
output.
Fig. 1.7. Basic op-amp

21. • The AC equivalent circuit of the op-amp is;
• The input signal applied between input terminals(+ & -) sees an input
impedance, 𝐑𝐢, typically very high.
• The output voltage is shown to be the amplifier gain times the input signal
taken through an output impedance, 𝐑𝒐, which is typically very low.
Fig. 1.8. Ac equivalent of op-amp circuit: (a) practical; (b) ideal.

22. • The basic op-amp connection will be, an input signal, 𝑽𝟏, is applied through
resistor 𝐑𝟏 to the minus input. The output (𝑽𝒐) is then connected back to
the same minus input through resistor 𝐑𝒇. The plus input is connected to
ground.
Fig. 1.9. (a) Basic op-amp connection & (b) its op-amp ac
equivalent circuit.

23. • But for ideal op-amp;
• The above circuit can be more simplified and solved for the voltage 𝐕𝟏 in
terms of the components due to each of the sources by using superposition
principle.
Fig. 1.10. (a) Ideal op-amp equivalent circuit & (b) its
simplified ac-circuit.

24. • For source 𝐕𝟏 only(−𝑨𝒗𝑽𝒊 = 𝟎) , shown in Fig.1.11 (a);
• Apply voltage division (VD) to get 𝐕𝐢𝟏 in terms of 𝐕𝟏;
• For source −𝐀𝐯𝐕𝐢, only (𝑽𝟏 = 𝟎), shown in Fig.1.11 (b);
• Apply VD;
Fig. 1.11. Ideal op-amp simplified
ac-circuit.
𝑽𝒊𝟏 =
𝑹𝒇
𝑹𝟏 + 𝑹𝒇
𝑽𝟏
𝑽𝒊𝟐 =
𝑹𝟏
𝑹𝟏 + 𝑹𝒇
(−𝑨𝒗𝑽𝒊)
Where;
𝑨𝒗 : voltage gain

25. • The total voltage 𝑽𝒊 is then;
• Solve for 𝑽𝒊;
• If 𝑨𝒗 ≫ 𝟏 & 𝑨𝒗𝑹𝟏 ≫ 𝑹𝒇, then the above equation can be expressed as;
• Solve for
𝐕𝐨
𝐕𝐢
, see Fig. 1.11 (b):
𝑽𝒊 = 𝑽𝒊𝟏 + 𝑽𝒊𝟐 =
𝑹𝒇
𝑹𝟏 + 𝑹𝒇
𝑽𝟏 +
𝑹𝟏
𝑹𝟏 + 𝑹𝒇
(−𝑨𝒗𝑽𝒊)
𝑽𝒊 =
𝑹𝒇
𝑹𝒇 + (𝟏 + 𝑨𝒗)𝑹𝟏
𝑽𝟏 =
𝑹𝒇
𝑹𝒇 + 𝑹𝟏 + 𝑨𝒗𝑹𝟏
𝑽𝟏
𝑽𝒊 =
𝑹𝒇𝑽𝟏
𝑨𝒗𝑹𝟏
𝑽𝒐
𝑽𝒊
=
−𝑨𝒗𝑽𝒊
𝑽𝒊
=
−𝑨𝒗
𝑽𝒊
𝑽𝒊 =
−𝑨𝒗
𝑽𝒊
𝑹𝒇𝑽𝟏
𝑨𝒗𝑹𝟏
=
−𝑹𝒇
𝑹𝟏
𝑽𝟏
𝑽𝒊
But;
𝑽𝒐 = −𝑨𝒗𝑽𝒊
substitute
𝑽𝟏
𝑹𝒇
𝑨𝒗𝑹𝟏
𝑽𝒊

26. • Thus; after eliminating 𝑽𝒊;
• The above equations shows that the ratio of overall output to input voltage
is dependent only on the values of resistors 𝑹𝟏 & 𝑹𝒇 gives => 𝑨𝒗 is very
large.
 Unity-gain: for 𝑹𝟏 = 𝑹𝒇, the gain will be;
 Constant Magnitude Gain: If 𝑹𝒇 is some multiple (constant 𝒂) of 𝑹𝟏, i.e. 𝑹𝒇
= 𝒂𝑹𝟏, the overall amplifier gain is a constant.
𝑽𝒐
𝑽𝟏
=
−𝑹𝒇
𝑹𝟏
𝒗𝒐𝒍𝒕𝒂𝒈 𝒈𝒂𝒊𝒏 =
−𝑹𝒇
𝑹𝟏
− 𝟏
Unit voltage gain
with 180𝑜phase
inversion (out-
phased
𝒗𝒐𝒍𝒕𝒂𝒈 𝒈𝒂𝒊𝒏 =
−𝑹𝒇
𝑹𝟏
= −𝒂

27. 1.2.4. Practical Op-amp Circuits
(1) Inverting Amplifier
• The output (𝑽𝒐) is obtained by multiplying the input (𝑽𝟏) by a fixed or
constant gain (𝑨𝒗), set by the input resistor (𝑹𝟏) and feedback resistor
(𝑹𝒇)—this output also being inverted from the input.
• Notice that, the input (𝑽𝟏) will be connected to the (−) of the op-amp as;
Fig. 1.12.invertsing op-
amp
𝑽𝒐 = −
𝑹𝒇
𝑹𝟏
𝑽𝟏

28.  Example-3: If the circuit of Fig. 1.12 has 𝑹𝟏= 100 kΩ & 𝑹𝒇 = 500 kΩ, what
will be the output voltage if 𝑽𝟏 = 2V?
 Solution: use the equation above;
(2) Non-inverting Amplifier
• The input (𝑽𝟏) will be connected to the (+) of the op-amp that works as a
non-inverting amplifier or constant-gain multiplier.
• The output voltage (𝑽𝒐) can be found by using the circuit given in Fig. 1.13
below as;
𝑽𝒐 = −
𝑹𝒇
𝑹𝟏
𝑽𝟏 = −
500 kΩ
100 kΩ
𝟐𝑽 = −𝟏𝟎𝑽

29. • Note that the voltage across 𝑹𝟏 is 𝑽𝟏 since 𝑽𝒊 ≈ 0 V, refer to Fig.1.13. (b).
• Use voltage division rule;
→
Fig. 1.13. Non-inverting constant-gain multiplier
𝑽𝟏 =
𝑹𝒇
𝑹𝟏 + 𝑹𝒇
𝑽𝒐 𝑽𝒐 = (𝟏 +
𝑹𝒇
𝑹𝟏
)𝑽𝟏

30.  Example-3: If the circuit of Fig. 1.13. has 𝑹𝟏= 100 kΩ & 𝑹𝒇 = 500 kΩ, what
will be the output voltage if 𝑽𝟏 = 2V?
 Solution: use the equation above;
(3) Unity Follower
• The circuit provides a gain of unity (1) with no polarity or phase reversal.
• The output (𝑽𝒐) is the same polarity and magnitude as the input (𝑽𝟏).
𝑽𝒐 = 𝟏 +
𝑹𝒇
𝑹𝟏
𝑽𝟏 = 𝟏 +
500 kΩ
100 kΩ
2V = +𝟏𝟐𝑽

31. (4) Summing Amplifier
• The most used of the op-amp circuits is the summing amplifier will have
three-input summing amplifier circuit, which provides a means of
algebraically summing (adding) three voltages, each multiplied by a
constant-gain factor. See Fig. 1.15 in the next slide
Fig. 1.14. Unity Follower
𝑽𝒐 = 𝑽𝟏

32. • The above equation also shows that, each input adds a voltage to the output
multiplied by its separate constant-gain multiplier.
𝑽𝒐 = −(
𝑹𝒇
𝑹𝟏
𝑽𝟏 +
𝑹𝒇
𝑹𝟐
𝑽𝟐 +
𝑹𝒇
𝑹𝟑
𝑽𝟑)
Fig. 1.15. summing Op-amp

33.  Example-4: Calculate the output voltage of an op-amp summing amplifier
for the following sets of voltages and resistors. Use 𝑹𝒇= 1 MΩ in all cases;
(a) 𝑽𝟏 = +1 V, 𝑽𝟐 = +2 V, 𝑽𝟑 = +3 V, 𝑹𝟏 = 500 k Ω , 𝑹𝟐 = 1 M Ω , 𝑹𝟑 = 1 M Ω
(b) 𝑽𝟏 = -2V, 𝑽𝟐 = +3 V, 𝑽𝟑 = +1 V, 𝑹𝟏 = 200 k Ω , 𝑹𝟐 = 500 k Ω , 𝑹𝟑 = 1 M Ω
 Solution: use the equation above;
(a)
(b)
𝑽𝒐 = −
1000 KΩ
500 k Ω
+1 V +
1000 KΩ
1000 KΩ
+2 V +
1000 KΩ
1000 KΩ
+3 V
= −𝟕𝑽
𝑽𝒐 = −(
𝑹𝒇
𝑹𝟏
𝑽𝟏 +
𝑹𝒇
𝑹𝟐
𝑽𝟐 +
𝑹𝒇
𝑹𝟑
𝑽𝟑)
𝑽𝒐 = −
1000 KΩ
200 k Ω
−2 V +
1000 KΩ
500 KΩ
+3 V +
1000 KΩ
1000 KΩ
+1 V
= +𝟑𝑽

34. (5) Integrator
• The op-amp circuit with the feedback component of a capacitor, as shown
below is called integrator.
• Virtual ground means that we can consider the voltage at the junction of R
and 𝑿𝑪 to be ground (since Vi ≈ 0 V) => short circuit but notice that no
current goes into ground at that point.
Fig. 1.16. Integrator; (a) op-amp circuit, & (b) virtual equivalent
circuit

35. • Notice that the capacitive impedance (𝑿𝑪) can be expressed as;
• Solving for
𝑽𝑶
𝑽𝟏
gives;
• i.e.
• Change the s-domain to t-domain (Invers-Laplace transform);
𝑿𝑪 =
𝟏
𝒋𝝎𝑪
=
𝟏
𝒔𝑪
Where;
𝒔 = 𝒋𝝎 : Laplace
notation
𝑰 =
𝑽𝒊
𝑹
= −
𝑽𝒐
𝑿𝑪
=
−𝑽𝒐
𝟏
𝒔𝑪
= −𝒔𝑪𝑽𝒐
𝑽𝒊
𝑹
= −𝒔𝑪𝑽𝒐 ⇒
𝑽𝒐
𝑽𝒊
=
−𝟏
𝒔𝑪𝑹
=
−𝟏
𝒔(𝑪𝑹)
𝒗𝒐 𝒕 = −
𝟏
𝑪𝑹
𝒗𝒊 𝒕 𝒅𝒕

36. • Notice that the above equation shows that the output is the integral of the
input, with an inversion and scale multiplier of 𝟏
𝐑𝐂.
1.2.5. Offset Currents and Voltages
• In theory the op-amp output should be 0V when the input is 0V, but in
actual operation there is some offset voltage at the output.
• For example, if one connected 0 V to both op-amp inputs and then
measured 26 mV(dc) at the output, this would represent 26 mV of
unwanted voltage generated by the circuit and not by the input signal.
• Thus the manufacturer must specifies an input offset voltage for the op-
amp.
Reading assignment: Differentiator

37. • The output offset voltage is then determined by the input offset voltage and
the gain of the amplifier, as connected by the user.
• The output offset voltage can be shown to be affected by two separate
circuit conditions.
(1) An input offset voltage, 𝑉𝐼𝑂, &
(2) An offset current (𝐼𝐼𝑂) due to the difference in currents resulting at the
plus (+) and minus (-) inputs.
 Input Offset Voltage (𝑽𝑰𝑶)
• The manufacturer’s specification sheet provides a value of VIO for the op-
amp.

38. • To determine the effect of this input voltage on the output, see Fig. 1.17
below
• Recall that 𝑽𝒐 = 𝑨𝑽𝒊, gives;
Fig. 1.17. (a) Operation showing effect of input offset voltage, (b) equivalent circuit
𝑽𝒐 = 𝑨𝑽𝒊 = 𝑨 𝑽𝑰𝑶𝑴 − 𝑽𝑰𝑶𝑪 = 𝑨(𝑽𝑰𝑶 −
𝑹𝟏
𝑹𝟏 + 𝑹𝒇
𝑽𝑶)

39. • Solving for 𝑽𝒐,
⇒
 Example-5: Calculate the output offset voltage of the an op-amp circuit has,
𝑹𝟏= 2 KΩ, 𝑹𝒇 = 150 KΩ & The op-amp spec lists 𝑽𝑰𝑶 = 1.2 mV.
 Solution: use equation above;
𝑽𝒐 =
𝑨
𝟏 + 𝑨
𝑹𝟏
𝑹𝟏 + 𝑹𝒇
𝑽𝑰𝑶 ≈
𝑨
𝑨
𝑹𝟏
𝑹𝟏 + 𝑹𝒇
𝑽𝑰𝑶
𝑽𝒐 𝐨𝐟𝐟𝐬𝐞𝐭 =
𝑹𝟏 + 𝑹𝒇
𝑹𝟏
𝑽𝑰𝑶
𝑽𝒐 𝐨𝐟𝐟𝐬𝐞𝐭 =
𝑹𝟏+𝑹𝒇
𝑹𝟏
𝑽𝑰𝑶 =
2 KΩ+150 KΩ
2 KΩ
(1.2 mV) = 91.2
mV

40.  Output Offset Voltage Due To Input Offset Current (𝑰𝑰𝑶)
• An output offset voltage will also result due to any difference in dc bias
currents at both inputs.
Fig. 1.18. (a) Op-amp connection showing input bias currents, (b) equivalent circuit

41. • The two input resistors are never exactly matched, each will operate at a
slightly different current.
• Using superposition (refer Fig. 1.18 (b)), the output voltage due to input
bias current 𝑰𝑰𝑩
+
, denoted by 𝑽𝒐
+, (resemble non-inverting op-amp) is;
• The output voltage due to only current 𝑰𝑰𝑩
−
, denoted by 𝑽𝒐
−, (resemble
inverting op-amp) is;
• Total voltage;
𝑽𝒐
+ = 𝑰𝑰𝑩
+
𝑹𝑪
𝑹𝟏 + 𝑹𝒇
𝑹𝟏
= 𝑰𝑰𝑩
+
𝑹𝑪 𝟏 +
𝑹𝒇
𝑹𝟏
𝑽𝒐
− = 𝑰𝑰𝑩
−
𝑹𝟏 −
𝑹𝒇
𝑹𝟏
𝑽𝒐 𝐨𝐟𝐟𝐬𝐞𝐭 𝐝𝐮𝐞 𝐭𝐨 𝑰𝑰𝑩
+
& 𝑰𝑰𝑩
+
= 𝑰𝑰𝑩
+
𝑹𝑪 𝟏 +
𝑹𝒇
𝑹𝟏
− 𝑰𝑰𝑩
−
𝑹𝟏
𝑹𝒇
𝑹𝟏

42. • For 𝑹𝑪 = 𝑹𝟏, the above equation will be;
 Example-6: Calculate the offset voltage for the circuit of Fig. 1.18 for op-
amp specification listing 𝑰𝑰𝑶 = 100 nA & 𝑹𝒇 = 150 KΩ.
 Solution:
 Total Offset Due To 𝑽𝑰𝑶 & 𝑰𝑰𝑶
• The op-amp output may have an output offset voltage due to both factors
covered above, the total output offset voltage can be expressed as
𝑽𝒐 𝒐𝒇𝒇𝒔𝒆𝒕 𝒅𝒖𝒆 𝒕𝒐 𝑰𝑰𝑶 = 𝑰𝑰𝑶𝑹𝒇
Where;
𝑰𝑰𝑶 = 𝑰𝑰𝑩
+
− 𝑰𝑰𝑩
−
𝑽𝒐 𝒐𝒇𝒇𝒔𝒆𝒕 𝒅𝒖𝒆 𝒕𝒐 𝑰𝑰𝑶 = 𝑰𝑰𝑶𝑹𝒇 = 𝟏𝟎𝟎 𝒏𝑨 × 𝟏𝟓𝟎 𝑲Ω = 𝟏𝟓 𝒎𝑽

43.  Example-7: Calculate the total offset voltage for the circuit having 𝑹𝟏= 5
KΩ, 𝑹𝒇= 500 KΩ, & 𝑹𝑪= 5 KΩ, for an op-amp with specified values of input
offset voltage 𝑽𝑰𝑶 = 4 mV & input offset current 𝑰𝑰𝑶 = 150 nA.
 Solution: The offset due to 𝑽𝑰𝑶 is;
• The offset due to 𝑰𝑰𝑶 is;
𝑽𝒐 𝒐𝒇𝒇𝒔𝒆𝒕 = |𝑽𝒐 𝐨𝐟𝐟𝐬𝐞𝐭 𝐝𝐮𝐞 𝐭𝐨 𝑽𝑰𝑶 | + |𝑽𝒐 𝒐𝒇𝒇𝒔𝒆𝒕 𝒅𝒖𝒆 𝒕𝒐 𝑰𝑰𝑶 |
𝑽𝒐 𝐨𝐟𝐟𝐬𝐞𝐭 𝐝𝐮𝐞 𝐭𝐨 𝑽𝑰𝑶 =
𝑹𝟏 + 𝑹𝒇
𝑹𝟏
𝑽𝑰𝑶 =
5 KΩ + 500 KΩ
5 KΩ
𝟒𝒎𝑽
𝑽𝒐 𝐨𝐟𝐟𝐬𝐞𝐭 𝐝𝐮𝐞 𝐭𝐨 𝑽𝑰𝑶 = 𝟒𝟎𝟒 𝒎𝑽
𝑽𝒐 𝒐𝒇𝒇𝒔𝒆𝒕 𝒅𝒖𝒆 𝒕𝒐 𝑰𝑰𝑶 = 𝑰𝑰𝑶𝑹𝒇
= 𝟏𝟓𝟎 𝒏𝑨 × 𝟓𝟎𝟎 𝑲𝜴
= 𝟕𝟓 𝒎𝑽

44. 1.2.6. Instrumentation Amplifier
• The one & vital application of op-amps are for instrumentation purpose.
• Designed like a circuit providing an output based on the difference between
two inputs (times a scale factor).
• Consists a potentiometer is provided to permit adjusting the scale factor of
the circuit.
• Three op-amps are used, a single-quad op-amp IC & its internal circuitry is
shown in Fig. 1.19.
𝑽𝒐 𝒐𝒇𝒇𝒔𝒆𝒕 = 𝑽𝒐 𝐨𝐟𝐟𝐬𝐞𝐭 𝐝𝐮𝐞 𝐭𝐨 𝑽𝑰𝑶 + 𝑽𝒐 𝒐𝒇𝒇𝒔𝒆𝒕 𝒅𝒖𝒆 𝒕𝒐 𝑰𝑰𝑶
= 𝟒𝟎𝟒 𝐦𝐕 + 𝟕𝟓 𝐦𝐕 = 𝟒𝟕𝟗 𝐦𝐕
Reading assignment: Input Bias Current

45. • Notice that op amps A1 and A2 draw no current, current i flows through
the three resistors as though they were in series.
Fig. 1.19. Instrumentation
amplifier.

46. • Notice that IA is difference amplifier, thus
• Hence,
• But for node a & b:
• Also; 𝒗𝒂 = 𝑽𝟏 & 𝒗𝒃 = 𝑽𝟐, thus;
𝑽𝒐𝟏 − 𝑽𝒐𝟐 = 𝒊(𝑹 + 𝑹𝑷 + 𝑹) = 𝒊(𝑹𝑷 + 𝟐𝑹)
𝒊 =
𝒗𝒂 − 𝒗𝒃
𝑹𝑷
𝒊 =
𝑽𝟏 − 𝑽𝟐
𝑹𝑷
𝑽𝑶 =
𝑹
𝑹
𝑽𝒐𝟏 − 𝑽𝒐𝟐 = (𝑽𝒐𝟏 − 𝑽𝒐𝟐)
𝑽𝑶 = 𝒊(𝑹𝑷 + 𝟐𝑹)
⇒ 𝑽𝑶 = 𝟏 +
𝟐𝑹
𝑹𝑷
𝑽𝟏 − 𝑽𝟐 = 𝒌(𝑽𝟏 − 𝑽𝟐)

47.  Example-8: Calculate the output voltage expression for the circuit of Fig.
1.20 shown below.
Fig. 1.20. for ex-8

48.  Solution: use the above equation;
1.2.7. Filters
• A popular application uses op-amps to build active filter circuits.
• A filter circuit can be constructed using passive components: resistors and
capacitors.
• An active filter additionally uses an amplifier to provide voltage
amplification & signal isolation or buffering.
(1) Passive Filter
• A filter is a passive filter if it consists of only passive elements R, L, and C.
𝑽𝑶 = 𝟏 +
𝟐𝑹
𝑹𝑷
𝑽𝟏 − 𝑽𝟐 = 𝟏 +
𝟐(𝟓𝟎𝟎𝟎)
𝟓𝟎𝟎
𝑽𝟏 − 𝑽𝟐
𝑽𝑶 = 𝟐𝟏 𝐕𝟏 − 𝐕𝟐

49. (2) Active Filter
• It is said to be an active filter if it consists of active elements (such as
transistors and/or op-amps) in addition to passive elements R, L, and C.
• There are four types of filters whether passive or active:
1) Low-pass filter: passes low frequencies and stops high frequencies,
2) High-pass filter: passes high frequencies and rejects low frequencies,
3) Band-pass filter: passes frequencies within a frequency band and blocks
or attenuates frequencies outside the band, and
4) Band-stop filter: passes frequencies outside a frequency band and blocks
or attenuates frequencies within the band.

50. Fig. 1.21. Ideal filter response: (a) low-pass; (b) high-pass; (c) band pass & (d) band stop.

51. 1) Low-pass filter
• A first-order, low-pass filter using a single resistor and capacitor is shown
Fig. 1.22. below, it has a practical slope of = 20 dB per decade.
Fig. 1.22. (a) First-order low-pass active filter & (b) response

52. • Recall that the voltage gain for circuit (non-inverting) shown above is;
• And the above equation will be true for cutoff frequency of;
 Example-9: calculate the cutoff frequency of a first-order low-pass filter for
𝑹𝟏=1.2 KΩ and 𝑪𝟏= 0.02 𝜇F.
 Solution:
𝑨𝒗 = 𝟏 +
𝑹𝒇
𝑹𝟏
𝒇𝑶𝑯 =
𝟏
𝟐𝝅𝑹𝟏𝑪𝟏
𝒇𝑶𝑯 =
𝟏
𝟐𝝅𝑹𝟏𝑪𝟏
=
𝟏
𝟐𝝅(𝟏. 𝟐 × 𝟏𝟎𝟑)(𝟎. 𝟎𝟐 × 𝟏𝟎−𝟔)
= 𝟔. 𝟔𝟑 𝐤𝐇𝐳

53. (2) High-Pass Active Filter
• First- and second-order high-pass active filters can be built as shown in
Fig.;
Fig. 1.22. High-pass filter: (a) first order; (b) second order; (c) response
plot.

54. • The amplifier cutoff frequency is;
 Example-10: calculate the gain & cutoff frequency of a second-order high-
pass filter as in Fig. 1.22 (b) for 𝑹𝒐𝒇= 50 KΩ, 𝑹𝒐𝟏=10 KΩ, 𝑹𝟏= 2.1 KΩ and
𝑪𝟏= 0.05 𝜇F.
 Solution:
(i) Gain:
(ii) cutoff frequency:
𝒇𝑶𝑳 =
𝟏
𝟐𝝅𝑹𝟏𝑪𝟏
Notice that gain
is;
𝑨𝒗 = 𝟏 +
𝑹𝒇
𝑹𝟏
𝑨𝒗 = 𝟏 +
𝑹𝒐𝒇
𝑹𝒐𝟏
= 𝟏 +
50 KΩ
10 KΩ
= 𝟔
𝒇𝑶𝑳 =
𝟏
𝟐𝝅𝑹𝟏𝑪𝟏
=
𝟏
𝟐𝝅(𝟐. 𝟏 × 𝟏𝟎𝟑)(𝟎. 𝟎𝟓 × 𝟏𝟎−𝟔)
= 𝟏. 𝟓 𝐤𝐇𝐳

55. (3) Band-pass Filter
• A band-pass filter using two stages, the first a high-pass filter and the
second a low-pass filter, the combined operation being the desired band-
pass response.
Fig. 1.23. Band-pass active
filter.
Response

56.  Example-11: calculate the cutoff frequencies of the band-pass filter circuit of
Fig. 1.23 with 𝑹𝟏= 𝑹𝟐= 2.1 KΩ, 𝑪𝟏= 0.1 𝜇F and 𝑪𝟐= 0.002 𝜇F.
 Solution:
• For low frequency;
• For high frequency;
𝒇𝑶𝑳 =
𝟏
𝟐𝝅𝑹𝟏𝑪𝟏
=
𝟏
𝟐𝝅(𝟏𝟎 × 𝟏𝟎𝟑)(𝟎. 𝟏 × 𝟏𝟎−𝟔)
= 𝟏𝟓𝟗. 𝟏𝟓 𝐇𝐳
𝒇𝑶𝑯 =
𝟏
𝟐𝝅𝑹𝟐𝑪𝟐
=
𝟏
𝟐𝝅(𝟏𝟎 × 𝟏𝟎𝟑)(𝟎. 𝟎𝟎𝟐 × 𝟏𝟎−𝟔)
= 𝟕. 𝟗𝟔 𝐤𝐇𝐳

57. 1.2.8. Converters
(1) Current-to-Voltage Converter (CVC)
• CVC converts a variable input current to a proportional output voltage.
Fig. 1.24. Current-to-voltage converter.

58. • Because of the virtual ground input current (𝑰𝒊 ) will flow through the
feedback path, & the voltage dropped across 𝑹𝒇 is 𝑹𝒇𝑰𝒇.
• The output voltage equals the voltage across 𝑹𝒇, which is proportional to 𝑰𝒊.
⇒
(2) Voltage-to-Current Converter (VCC)
• VCC is a circuit that is equivalent to a controlled current source (voltage
controlled current source). The input voltage controls the current.
• The current is independent of the load resistance (𝑹𝑳 ), but by input
resistance (𝑹𝟏).
• The output (load) current that is controlled by an input voltage.
𝑽𝒐𝒖𝒕 = 𝑹𝒇𝑰𝒇

59. • By neglecting the input offset voltage, both inverting and non-inverting
input terminals of the op-amp will be at the same voltage, 𝑽𝒊𝒏. => the
voltage across 𝑹𝟏 equals 𝑽𝒊𝒏. Refer Fig. 1.25.
𝑰𝑳 =
𝑽𝒊𝒏
𝑹𝟏
Fig. 1.25. Voltage-to-current converter.

60.  Example-12: what is the load current in Fig. 1.26? The load power? What
happens if the load resistance changes to 4Ω?
 Solution:
• This 2A current will flow through 2Ω, producing:
, if for 4Ω
𝒊𝒐𝒖𝒕 =
𝑽𝒊𝒏
𝑹𝟏
=
𝟐 𝑽
1Ω
= 𝟐 𝐀
𝑷𝑳 = 𝑰𝟐𝑹𝑳 = 𝟐𝑨 𝟐(2Ω) = 𝟖𝐖 𝑷𝑳 = 𝑰𝟐𝑹𝑳 = 𝟐𝑨 𝟐(4Ω) = 𝟏𝟔𝐖
Fig. 1.26. For Example-
12.

61. • An amplifier (large-signal amplifier) receives a signal from some pickup
transducer or other input source and provides a larger version of the signal
to some output device or to another amplifier stage. Egg: to drive a speaker
• The main features of a large-signal amplifier are the circuit’s power
efficiency, the maximum amount of power that the circuit is capable of
handling, and the impedance matching to the output device.
• One method used to categorize amplifiers is by class.
• Basically, amplifier classes represent the amount the output signal varies
over one cycle of operation for a full cycle of input signal.

62. 1.3.1. Class-A large signal amplifier
• Class A: The output signal varies for a full 360° of the cycle.
• Amplifier Efficiency: of an amplifier, defined as the ratio of power output to
power input, improves (gets higher) going from class A to class D.
Fig. 1.26. Class-A
amplifier response .

63.  Series-fed Class-A Amplifier
• It is the type of class-A power amplifier, this circuit is not the best to use as
a large-signal amplifier because of its poor power efficiency.
• Unlike operational amplifiers, power amplifiers uses a vital active circuit
element, i.e. transistors (power transistors).
Fig. 1.27. Series-fed class
A large-signal amplifier.
beta of a power
transistor

64.  DC Bias Operation
• The dc bias set by 𝑉𝐶𝐶 and 𝑅𝐵 fixes the dc base-bias current & apply
Kirchhoff’s voltage law;
• The collector current will be;
• And for collector-emitter path (apply Kirchhoff’s voltage law);
𝑰𝑩 =
𝑽𝑪𝑪 − 𝑽𝑩𝑬
𝑹𝑩
=
𝑽𝑪𝑪 − 𝟎. 𝟕𝑽
𝑹𝑩
𝑰𝑪 = 𝜷𝑰𝑩
𝑽𝑪𝑬 = 𝑽𝑪𝑪 − 𝑰𝑪𝑹𝑪

65.  AC Operation
• Transistor characteristic showing load line and Q-point; allows us to know
the input & output relation along with the DC-bias for power amplifier’s
transistor.
Fig. 1.28. Transistor characteristic showing load line and Q-point.

66. • An ac load line is drawn using the values of 𝑉𝐶𝐶 and 𝑅𝐶. The intersection of
the dc bias value of 𝐼𝐵 with the dc load line then determines the operating
point (Q-point) for the circuit.
• When an input ac signal is applied to the amplifier of Fig. 1.27, the output
will vary from its dc bias operating voltage and current.
• If a small input signal, as shown in Fig.1.29 (a), will cause the base current
to vary above and below the dc bias point, which will then cause the
collector current (output) to vary from the dc bias point set as well as the
collector–emitter voltage to vary around its dc bias value.

67. Fig. 1.29. Amplifier input and output signal variation; (a) small & (b) large
signal

68. • If the input signal is made larger, the output will vary further around the
established dc bias point until either the current or the voltage reaches a
limiting condition as shown in Fig.1.29 (b).
• For the current this limiting condition is either zero current at the low end
or 𝑉𝐶𝐶/𝑅𝐶 at the high end of its swing.
• For the collector–emitter voltage, the limit is either 0V or the supply
voltage, 𝑉𝐶𝐶.
 Power Considerations
• The power into an amplifier is provided by the supply.

69. • With no input signal, the dc current drawn is the collector bias current, 𝐼𝐶𝑄.
The power then drawn from the supply is:
• Even with an ac signal applied, the average current drawn from the supply
remains the same, so that the equation above represents the input power
supplied to the class A series-fed amplifier.
 Output Power
• The output voltage and current varying around the bias point provide ac
power to the load. => delivered to the load, 𝑹𝑪.
𝑷𝒊(𝒅𝒄) = 𝑽𝑪𝑪𝑰𝑪𝑸

70. • Using rms signals: The ac power delivered to the load (RC) expressed using;
• Thus; or
• Using peak signals: The ac power delivered to the load can also be
expressed using;
• Thus;
or
𝑷𝒐(𝒂𝒄) = 𝑽𝑪𝑬(𝒓𝒎𝒔)𝑰𝑪(𝒓𝒎𝒔)
But;
𝑽𝑪𝑬 𝒓𝒎𝒔 = 𝑰𝑪(𝒓𝒎𝒔)𝑹𝑪
𝑷𝒐(𝒂𝒄) = 𝑰𝑪
𝟐
(𝒓𝒎𝒔)𝑹𝑪 𝑷𝒐(𝒂𝒄) =
𝑽𝑪𝑬
𝟐
(𝒓𝒎𝒔)
𝑹𝑪
𝑷𝒐(𝒂𝒄) =
𝑽𝑪𝑬(𝒑)𝑰𝑪(𝒑)
𝟐
But;
𝑽𝑪𝑬 𝒑 = 𝑰𝑪(𝒑)𝑹𝑪
𝑷𝒐(𝒂𝒄) =
𝑰𝑪(𝒑)𝟐𝑹𝑪
𝟐
𝑷𝒐(𝒂𝒄) =
𝑽𝑪𝑬(𝒑)𝟐
𝟐𝑹𝑪

71.  Efficiency
• The efficiency of an amplifier represents the amount of ac power delivered
(transferred) from the dc source. The efficiency of the amplifier is calculated
using;
 Example-13: for class-A series-fed circuit shown in Fig. 1.30 next slide, an
input voltage that results in a base current of 10 mA peak. Find the
efficiency of the power amplifier.
Reading assignment: Maximum Efficiency
% ղ =
𝑷𝒐(𝒂𝒄)
𝑷𝒊(𝒂𝒄)
× 𝟏𝟎𝟎%

72. Fig. 1.30. Operation of a series-fed circuit for Example-
13
Marked using;
𝑰𝑪𝑸 ≅
𝟒𝟖𝟎 𝒎𝑨 &
𝑽𝑪𝑬𝑸 = 𝟏𝟎. 𝟒 𝑽

73.  Solution: the Q-point can be determined to be;
• And current;
• The voltage;
• This bias point is marked on the transistor collector characteristic of Fig.
1.30b.
• For ac-power; when the input ac base current increases from its dc bias
level, the collector current rises by;
𝑰𝑪𝑸 = 𝜷𝑰𝑩 = 𝟐𝟓 × 𝟏𝟗. 𝟑 × 𝟏𝟎−𝟑 = 𝟎. 𝟒𝟖 𝑨
𝑽𝑪𝑬𝑸 = 𝑽𝑪𝑪 − 𝑰𝑪𝑹𝑪 = 𝟐𝟎 𝑽 − 𝟎. 𝟒𝟖 𝑨 × 𝟐𝟎𝒌Ω = 𝟏𝟎. 𝟒 𝑽
𝑰𝑩 =
𝑽𝑪𝑪 − 𝟎. 𝟕𝑽
𝑹𝑩
=
𝟐𝟎 𝑽 − 𝟎. 𝟕 𝑽
𝟏𝒌Ω
= 𝟏𝟗. 𝟑 𝒎𝑨
𝑰𝑪 𝒑 = 𝜷𝑰𝑩 𝒑 = 𝟐𝟓 𝟏𝟎 𝒎𝑨 𝒑𝒆𝒂𝒌 = 𝟐𝟓𝟎 𝒎𝑨 𝒑𝒆𝒂𝒌

74. • Now find powers;
• And for dc;
• The amplifier’s power efficiency can then be calculated;
𝑷𝒐 𝒂𝒄 =
𝑰𝑪 𝒑 𝟐𝑹𝑪
𝟐
=
𝟐𝟓𝟎 × 𝟏𝟎−𝟑𝑨
𝟐
(𝟐𝟎 Ω)
𝟐
= 𝟎. 𝟔𝟐𝟓 𝑾
𝑷𝒊 𝒅𝒄 = 𝑽𝑪𝑪𝑰𝑪𝑸 = 𝟐𝟎 𝑽 × 𝟎. 𝟒𝟖 𝑨 = 𝟗. 𝟔 𝑾
% ղ =
𝑷𝒐(𝒂𝒄)
𝑷𝒊(𝒂𝒄)
× 𝟏𝟎𝟎% =
𝟎. 𝟔𝟐𝟓 𝑾
𝟗. 𝟔 𝑾
× 𝟏𝟎𝟎% = 𝟔. 𝟓%
Reading assignment: Transformer-coupled
Class-A Amplifier

75. 1.3.2. Class B Amplifier Operation
• A class-B circuit provides an output signal varying over one-half the input
signal cycle, or for 180° of signal
• Two class-B operations => one to provide output on the positive output half-
cycle and another to provide operation on the negative-output half-cycle are
necessary.
• The combined half-cycles then provide an output for a full 360° of operation.
• This type of connection is referred to as push-pull operation.
• Note that class-B operation by itself creates a very distorted output signal
since reproduction of the input takes place for only 180° of the output signal
swing.

76. • Since one part of the circuit pushes the signal to high during one half-cycle
and the other part pulls the signal to low during the other half-cycle, the
circuit is referred to as a push-pull circuit.
Fig. 1.31. (a) representation of push-pull operation (b) using
two voltage supplies; (c) using one voltage supply

77.  Input (DC) Power
• The power supplied to the load by an amplifier is drawn from the power
supply (or power supplies) & that provides the input or dc power;
• Where 𝑰𝒅𝒄 is the average or dc current drawn from the power supplies.
• In class-B operation, the current drawn from a single power supply has the
form of a full-wave rectified signal, while that drawn from two power
supplies has the form of a half-wave rectified signal from each supply.
𝑷𝒊(𝒅𝒄) = 𝑽𝑪𝑪𝑰𝒅𝒄
𝑰𝒅𝒄 =
𝟐
𝝅
𝑰(𝒑)
Where;
𝑰 𝒑 =
𝑽𝑳
𝑹𝑳
:
peak value of the output current

78. • Thus;
 Output (AC) Power
• The power delivered to the load (𝑹𝑳), in rms form:-
• Again in the peak, or peak-to-peak form:-
𝑷𝒊(𝒅𝒄) = 𝑽𝑪𝑪
2
𝝅
𝑰𝑪 𝒑
𝑷𝒐(𝒂𝒄) =
𝑽𝑳
𝟐
(𝒓𝒎𝒔)
𝑹𝑳
𝑷𝒐 𝒂𝒄 =
𝑽𝑳
𝟐
𝒑−𝒑
𝟖𝑹𝑳
=
𝑽𝑳
𝟐
(𝒑)
𝟐𝑹𝑳

79.  Efficiency
• The efficiency of the class B amplifier:-
 Power Dissipated by Output Transistors
• The power dissipated (as heat) by the output power transistors is the
difference between the input power delivered by the supplies and the
output power delivered to the load; i.e. for all transistors =>
• For each transistor is:
% ղ =
𝑷𝒐(𝒂𝒄)
𝑷𝒊(𝒅𝒄)
× 𝟏𝟎𝟎%
𝑷𝟐𝑸 = 𝑷𝒊 𝒅𝒄 − 𝑷𝒐(𝒂𝒄)
𝑷𝑸 =
𝑷𝟐𝑸
𝟐

80.  Example-14: For a class-B amplifier providing a 20-V peak signal to a 16-Ω
load (speaker) and a power supply of 𝑽𝑪𝑪 = 30 V, determine the input power,
output power, and circuit efficiency?
 Solution: the load voltage 20-V peak signal will be across a 16- Ω load
provides a peak load current of:
• And the dc value of the current drawn from the power supply is:
𝑰𝑳 𝒑 =
𝑽𝑳 𝒑
𝑹𝑳
=
𝟐𝟎 𝑽
𝟏𝟔 𝜴
= 𝟏. 𝟐𝟓 𝑨
𝑰𝒅𝒄 =
𝟐
𝝅
𝑰𝑳 𝒑 =
𝟐
𝝅
𝟏. 𝟐𝟓 𝑨 = 𝟎. 𝟕𝟗𝟔 𝑨

81. • Thus the input power delivered by the supply voltage is:
• Also the output power delivered to the load is:
• for a resulting efficiency of;
𝑷𝒊 𝒅𝒄 = 𝑽𝑪𝑪𝑰𝒅𝒄 = 𝟑𝟎 𝑽 × 𝟎. 𝟕𝟗𝟔 𝑨 = 𝟐𝟑. 𝟗 𝑾
𝑷𝒐 𝒂𝒄 =
𝑽𝑳
𝟐
(𝒑)
𝟐𝑹𝑳
=
(𝟐𝟎 𝑽)𝟐
𝟐(𝟏𝟔 𝜴)
= 𝟏𝟐. 𝟓 𝑾
% ղ =
𝑷𝒐(𝒂𝒄)
𝑷𝒊(𝒂𝒄)
× 𝟏𝟎𝟎% =
𝟏𝟐. 𝟓 𝑾
𝟐𝟑. 𝟗 𝑾
× 𝟏𝟎𝟎% = 𝟓𝟐. 𝟑 %

82.  Maximum Power Considerations
• For class B operation, the maximum output power is delivered to the load
when, 𝑽𝑪𝑪 = 𝑽𝑳 i.e.
• The corresponding peak ac current 𝑰 𝒑 :
• Thus the maximum value of average current from the power supply is
𝑴𝒂𝒙 𝑷𝒐 𝒂𝒄 =
𝑽𝑳
𝟐
(𝒑)
𝟐𝑹𝑳
=
𝑽𝑪𝑪
𝟐
(𝒑)
𝟐𝑹𝑳
𝑰(𝒑) =
𝑽𝑪𝑪
𝑹𝑳
𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝑰𝒅𝒄 =
𝟐
𝝅
𝑰 𝒑 =
𝟐𝑽𝑪𝑪
𝝅𝑹𝑳

83. • Then the maximum value of input power:-
• The maximum circuit efficiency for class B operation is then;
𝑴𝒂𝒙 𝑷𝒊 𝒅𝒄 = 𝑽𝑪𝑪 𝑴𝒂𝒙 𝑰𝒅𝒄 = 𝑽𝑪𝑪
2
𝝅
𝑰𝑪 𝒑
= 𝑽𝑪𝑪
𝟐𝑽𝑪𝑪
𝝅𝑹𝑳
=
𝟐𝑽𝑪𝑪
𝟐
(𝒑)
𝝅𝑹𝑳
𝑴𝒂𝒙 % ղ =
𝑴𝒂𝒙 𝑷𝒐(𝒂𝒄)
𝑴𝒂𝒙 𝑷𝒊(𝒅𝒄)
× 𝟏𝟎𝟎% =
𝑽𝑪𝑪
𝟐
(𝒑)
𝟐𝑹𝑳
𝟐𝑽𝑪𝑪
𝟐
(𝒑)
𝝅𝑹𝑳
× 𝟏𝟎𝟎% =
𝝅
𝟒
× 𝟏𝟎𝟎%
= 𝟕𝟖. 𝟓𝟒%

84. • And the maximum transistor power dissipation is:
 Example-15: workout Egg.-14 for maximum efficacy & the maximum power
dissipated by each transistor ? (Quiz#1)
𝑴𝒂𝒙 𝑷𝟐𝑸 =
𝟐𝑽𝑪𝑪
𝟐
𝝅𝟐𝑹𝑳

85.  Solution:
Step-1: find maximum output power;
• Step-2: maximum input voltage;
• Step-3: efficacy;
𝑴𝒂𝒙 𝑷𝒐 𝒂𝒄 =
𝑽𝑪𝑪
𝟐
(𝒑)
𝟐𝑹𝑳
=
(𝟑𝟎 𝑽)𝟐
𝟐(𝟏𝟔 𝜴)
= 𝟐𝟖. 𝟏𝟐𝟓 𝑾
𝑴𝒂𝒙 𝑷𝒊 𝒅𝒄 =
𝟐𝑽𝑪𝑪
𝟐
(𝒑)
𝝅𝑹𝑳
=
𝟐(𝟑𝟎 𝑽)𝟐
𝝅 𝟏𝟔 𝜴
= 𝟑𝟓. 𝟖𝟏 𝑾
𝑴𝒂𝒙 % ղ =
𝑴𝒂𝒙 𝑷𝒐(𝒂𝒄)
𝑴𝒂𝒙 𝑷𝒊(𝒂𝒄)
× 𝟏𝟎𝟎% =
𝟐𝟖. 𝟏𝟐𝟓 𝑾
𝟑𝟓. 𝟖𝟏 𝑾
× 𝟏𝟎𝟎% = 𝟕𝟖. 𝟓𝟒%

86. Step-4: find dissipate power for each transistor;
𝑴𝒂𝒙 𝑷𝟐 =
𝑴𝒂𝒙 𝑷𝟐𝑸
𝟐
=
𝑽𝑪𝑪
𝟐
𝝅𝟐𝑹𝑳
=
(𝟑𝟎 𝑽)𝟐
𝝅𝟐(𝟏𝟔 𝜴)
= 𝟓. 𝟕 𝑾
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