1. Electronic Analog Computerby
Dr. Amin Danial Asham

2. Analog Computers Digital Computers What is Computer:
A computer is a machine which performs mathematical computations.
In general, computers may be classified as:
Analog Computers
Digital Computers

3. A. Analog Computers
An analog computer is a form of computer that uses the continuously changeable aspects of physical quantities such as electrical, mechanical, or hydraulic quantities to model the problem being solved.
In this sort of computer, numbers are represented by continuously-varying quantities.
Since electronic analog computers can be easily constructed and adjusted, they are used to simulate mechanical systems and get all measurements directly with an oscilloscope.

4. A. Analog Computers Heathkit EC-1 educational analog computer

5. A. Analog Computers Polish analog computer ELWAT

6. Polish analog computer AKAT-1
A. Analog Computers
Polish analog computer AKAT-1

7. A. Analog Computers (continue)
Physical quantities such as weight, temperature or area are represented by voltages. Voltage is the electrical analog of the variable being analyzed. Arbitrary scale factors are set up to relate the voltages in the computer to the variables in the problem being solved.
For example, 1 volt equals 5 meters or 10 volts equals 1 kg. The name "analog" comes from the fact that the computer solves by analogy by using physical quantities to represent numbers.

8. B. Digital Computers:
Operate by discrete steps, that is, they actually count.
Numbers are represented by sets of 1's and 0's where 1 and 0 are represented by two different voltages.
Operations are simple logical operations (i.e. AND, OR, etc.) or arithmetic operations (i.e. addition or subtraction) . Calculus-type (differentiation and integration) operations are very complicated to do.
Common Examples of Calculator, Programmable Logic Controller, Personal Computers, Tablets,……etc

9. A. Electronic Analog Computers
Analog computers are especially useful in solving dynamic problems in which the motion can be expressed in the form of a differential equation.
This type of computers can perform addition, multiplication by a constant, integration, and differentiation based on the used elements and the configuration of the circuit.
Since the analog computer can perform these operations, thus it is a convenient device for the solution of differential equations.
The fundamental building block of the electronic analogue computer is the Operational Amplifier.

10. Operational Amplifier (Op-Amp IC)
Op-Amp is a differential DC coupled voltage Amplifier.
Where
𝑉 𝑜𝑢𝑡 is the output.
𝑉 + is the non inverting input.
𝑉 − is the inverting input.
𝑉 𝑆+ is the positive power supply.
𝑉 𝑆− is the negative power supply.

11. Operational Amplifier (Op-Amp) (continue)
An equivalent circuit of an operational amplifier.
𝑉 𝑜𝑢𝑡 = 𝐺 𝑂𝐿 𝑉 𝑖𝑛
Where
𝑉 𝑖𝑛 = 𝑉 + − 𝑉 −
𝐺 𝑂𝐿 is the open loop gain.
𝑍 𝑖𝑛 input impedance.
𝑍 𝑜𝑢𝑡 output impedance.
𝑮 𝑶𝑳 𝑽 𝒊𝒏
𝑽 𝒊𝒏 = 𝑽 + − 𝑽 −
𝑉 +
𝑉 −
𝑉 𝑜𝑢𝑡
𝒁 𝒐𝒖𝒕
𝒁 𝒊𝒏

12. Operational Amplifier (Op-Amp) (continue)
Typical Op-Amp
𝐺 𝑂𝐿 has a typical value greater than
𝑍 𝑖𝑛 is very high impedance which is of order M ohms.
𝑍 𝑜𝑢𝑡 is low impedance which is typically around 50 ohms.
Properties of Op-Amp
𝐼 + and 𝐼 − are considered zero
because of the high input
impedance.
The open loop gain: 𝐺 𝑂𝐿 →∞
Vin
LM324
𝐼 +
𝐼 −
𝑉 𝑠+
𝑉 𝑠−

13. Because of very high open loop gain we get a Narrow Linear Region.
Operational Amplifier (Op-Amp) (continue)
The relation between the 𝑽 𝒐𝒖𝒕 and 𝑽 𝒊𝒏 = 𝑽 + − 𝑽 −
Because of very high open loop gain we get a Narrow Linear Region.
Saturation regions where 𝑽 𝒐𝒖𝒕 is limited by characteristics of the internal circuit and the power supply

14. Operational Amplifier (Op-Amp) (continue)
Because the 𝐺 𝑜𝑙 is very large it is considered infinity.
Since the output voltage 𝑉 𝑜𝑢𝑡 = 𝐺 𝑂𝐿 𝑉 𝑖𝑛
Therefore, as 𝐺 𝑂𝐿 →∞, we get 𝑉 𝑜𝑢𝑡 →∞.
However, 𝑉 𝑜𝑢𝑡 is limited by the internal circuit characteristics of the OpAmp and the power supply.
In Saturation region the relation between 𝑉 𝑜𝑢𝑡 and 𝑉 𝑖𝑛 is non-linear.
The linear region as shown in the Input-Output curve is very narrow and corresponds to very small values of 𝑉 𝑖𝑛 .
In the Linear region:
Since
𝑉 𝑜𝑢𝑡 = 𝐺 𝑜𝑙 𝑉 𝑖𝑛
Therefore,
𝑉 𝑖𝑛 = 𝑉 𝑜𝑢𝑡 𝐺 𝑂𝐿
as 𝑉 𝑜𝑢𝑡 is limited and 𝐺 𝑂𝐿 →∞, therefore
𝑉 𝑖𝑛 ≅0
Therefore
𝑉 + ≅ 𝑉 −

15. Basic Mathematical Operations
To use the Op-Amp as an amplifier in the linear region a negative feedback is used to keep 𝑉 𝑖𝑛 close to zero and hence getting 𝑉 𝑜𝑢𝑡 in the linear region. Mathematical operations are implemented with amplifiers such as:
Multiplication by constant
Addition or Summation
Subtraction
Integration
Differentiation
An open loop Op-Amp circuit is used as a comparator. In this application the output 𝑉 𝑜𝑢𝑡 is either high saturated or low saturated in response to the difference between 𝑉 + and 𝑉 − . Therefore, the Op-Amp produces 1 or 0 as a result of comparing two signals where 1 is high saturated value and 0 is low saturated value or vice versa.

16. Basic Mathematical Operations (continue)
Note: For simplicity the power supply terminals 𝑽 𝐬+ 𝐚𝐧𝐝 𝑽 𝒔− will be omitted.
Basic Rules for Closed Loop Op-Amp Circuits as Amplifiers:
𝑽 + = 𝑽 − because of 𝑮 𝑶𝑳 →∞
𝑰 + =𝟎 and 𝑰 − =𝟎 because of the high input impedance.
Output Impedance is small compared to the input impedance of load and hence negligible.
V +
V −
I +
I −

17. Basic Mathematical Operations (continue)
Multiplications by Constant
This operation can be implemented by inverting and non-inverting amplifiers.
Inverting Amplifier
From the circuit diagram and applying the basic rules we get the following relations:
Since I − =0, therefore the same current is passing through 𝑅 𝑖 and 𝑅 𝑓
Since V + = V − , we get that V − =0, hence
𝐼= 𝑉 𝑖 𝑅 𝑖 = 0− 𝑉 𝑜 𝑅 𝑓
∴ 𝑉 𝑖 𝑅 𝑖 =− 𝑉 𝑜 𝑅 𝑓
∴ 𝑽 𝒐 =− 𝑹 𝒇 𝑹 𝒊 𝑽 𝒊
Inverting Amplifier
𝑉 𝑖
𝑉 𝑜 I
𝑺𝒂𝒎𝒆 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 I
𝑽 − =𝟎

18. Basic Mathematical Operations (continue)
Multiplication by constant (continue)
Inverting Amplifier (continue)
The gain of the inverting amplifier is − 𝑹 𝒇 𝑹 𝒊
The input voltage 𝑽 𝒊 is multiplied by − 𝑹 𝒇 𝑹 𝒊
The input impedance of the amplifier is 𝑹 𝒊

19. Example Solution Since Design an inverting amplifier of gain -4
𝑉 𝑜 =− 𝑅 𝑓 𝑅 𝑖 𝑉 𝑖
Therefore,
𝑅 𝑓 𝑅 𝑖 =4
If we choose 𝑅 𝑓 =1𝑀𝑜ℎ𝑚
𝑅 𝑖 = 𝑅 𝑓 4 =250𝑘𝑜ℎ𝑚
Inverting Amplifier
𝑉 𝑖
𝑉 𝑜 I I

20. Non-Inverting Amplifier
Basic Mathematical Operations (continue)
Multiplication by constant (continue)
Non-Inverting Amplifier
Using the basic rules we get:
From the first rule we get
𝑉 − = 𝑉 𝑖
From the second rule, the same current is passing through 𝑅 𝑖 and 𝑅 𝑓
∴ 𝐼= 𝑉 𝑖 𝑅 𝑖 = 𝑉 0 − 𝑉 𝑖 𝑅 𝑓
∴ 𝑉 𝑖 𝑅 𝑖 + 𝑉 𝑖 𝑅 𝑓 = 𝑉 𝑜 𝑅 𝑓
∴ 𝑽 𝒐 = 𝑽 𝒊 (𝟏+ 𝑹 𝒇 𝑹 𝒊 )
Notes:
Non-Inverting Amplifier
𝑽 𝒊 𝑰
The effective input resistance of the non- inverting circuit is at least as high as that of the operational amplifier itself. 𝑰
𝑉 𝑖
𝑉 𝑜
The Gain =(𝟏+ 𝑹 𝒇 𝑹 𝒊 ) is positive and higher than one.

21. Non-Inverting Amplifier
Example
Design an non-inverting amplifier of gain 3
Solution
Since
𝑉 𝑜 = 1+ 𝑅 𝑓 𝑅 𝑖 𝑉 𝑖
Therefore,
1+ 𝑅 𝑓 𝑅 𝑖 =3
∴ 𝑅 𝑓 𝑅 𝑖 =2
If we choose 𝑅 𝑓 =1𝑀𝑜ℎ𝑚
∴ 𝑅 𝑖 = 𝑅 𝑓 2 =500𝑘𝑜ℎ𝑚
Non-Inverting Amplifier
𝑉 𝑖
𝑉 𝑜

22. Voltage Follower Amplifier
Basic Mathematical Operations (continue)
Multiplication by constant (continue)
Non-Inverting Amplifier (continue)
As a special case of non-inverting amplifier is voltage follower amplifier.
This amplifier can be derived from Non-inverting amplifier by putting R f =0 or R i =∞ i.e. removed.
The gain is Unity.
Voltage Follower Amplifier
𝑉 𝑖
𝑉 𝑜
The effective input resistance of the non-inverting circuit is at least as high as that of the operational amplifier itself.

23. Basic Mathematical Operations (continue)
Multiplication by constant (continue)
Sometimes it is needed to multiply by a fractional constant, which can be obtained by using an inverting amplifier with gain less than unity , i.e. 𝑅 𝑓 𝑅 𝑖 .
However, amplifiers with gain less than unity may cause unstable operation. Therefore this
method is undesirable.
It is more desirable to use a potential
divider as shown, where
𝑉 𝑜 = 𝑅 𝑜 𝑅 𝑖 𝑉 𝑖
𝑅 𝑖
𝑅 𝑜
𝑉 𝑖
𝑉 𝑜

24. Thanks