Electrical Engineering Department
Computer Engineering
Electronic Measurements
24ELEC11C
Academic Year 2024/2025
Y1, Semester Two
Lecture #RVN1
Revision Session
(Read in conjunction with the relevant Lectures and Sheets)
Basic MUST-KNOW rules
• 𝜔 = 2 𝜋 × 𝑓 Radians/sec.
• e.g. 𝜔 = 𝜋 × 100𝐾 𝑅𝑎𝑑/𝑠 ∴ 𝜔 = 2𝜋 × 50𝐾 𝑅𝑎𝑑/𝑠
∴ 𝒇 = 𝟓𝟎 𝑲𝑯𝒛 .
• V units : Volts
Where:
OHMs LAW: 𝑽 = 𝑰 × 𝑹
• I units : Amp (A) or mA = 10−3 A or 𝜇𝐴 = 10−6 A (e.g. 3ma =3 × 10−3)
• R units : OHMs (Ω) or KΩ = 103 Ω
or MΩ = 106 Ω
• Example: 𝑉 = 𝐼 𝑅 = 10𝑚𝑎 × 3𝐾Ω = 30𝑉 (𝑚 × 𝐾 = 10−3 × 103 = 1)
• Voltage Divider:
𝑅2
𝑅1
𝑉𝑅2 = 𝑉 ×
& 𝑉𝑅1 = 𝑉 ×
𝑅1+𝑅2
𝑅1+𝑅2
𝐑𝟏
𝑽
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𝐑𝟐
𝑉𝑅2
2
Errors’ Basics: How to calculate ABS, Relative and % Errors
• If error given in ABS
X = 250 Ω ± 5Ω ≡ 𝑋 ± ∆𝑋
• Express the error in REL
5
X = 250 ±
= 250 ± 0.02
250
• Express the error in REL %
5
X = 250 ±
× 𝟏𝟎𝟎 = 250 ± 2 %
250
• ‫العكس‬:Express the error in ABS
≡
∆𝑋
𝑋 ±
𝑋
X = 250 V ±2% ≡ 𝑋 ± 𝑋 × 2%
= 250V ± 2% × 250𝑉 = 250𝑉 ± (0.02 × 250𝑉)
𝐴𝑝𝑝𝑙𝑖𝑒𝑠 𝑓𝑜𝑟 𝑣, 𝑜𝑟 𝑖 , 𝑜𝑟 Ω
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= 250𝑉 ± 5𝑉
3
Reminder: The Simplified Rules for Error Propagation !!!
• Addition of measured quantities
𝑹=𝑿+𝒀−𝒁
∴ ∆𝑹 = ∆𝑿 + ∆𝒀 + ∆𝒁
• Multiplication with a constant
∴ ∆𝑹 = 𝒄 . 𝚫𝑿
𝑹 = 𝒄. 𝑿
• Multiplication/Division of measured quantities
𝑿. 𝒀
𝑹 = 𝑿. 𝒀. 𝒁
𝒐𝒓 𝑹 =
𝒁
• Polynomial functions
𝑋 2 . 𝑌 −3
𝑅=
𝑍4
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∆𝑹 𝚫𝑿 ∆𝒀 ∆𝒁
∴
=
+
+
𝑹
𝐗
𝒀
𝒁
∆𝑅
Δ𝑋
Δ𝑌
Δ𝑍
∴
= 2.
+ 3.
+ 4.
𝑅
𝑋
𝑌
𝑍
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Add ABS Errors
All added positively
𝐀𝐝𝐝 𝐑𝐞𝐥𝐚𝐭𝐢𝐯𝐞 𝐄𝐫𝐫𝐨𝐬
All added positively
Multiply by the exponent.
Think of the same as
if Partial Derivative is used !!
4
Example (Philosophy of answering)
• 𝑽𝒐 =
𝑹𝟏
× 𝑽𝒊
𝑹𝟐
A
+
𝑹𝟑 × 𝑹𝟒
𝑹𝟐𝟓
× 𝑰𝟐𝒊
B
MULT: ADD Relative (or % errors) positively
MULT: ADD Relative (or % errors) positively
Good Practice
• These are relative errors (e.g. 9% and 64%)
• for 𝑽𝒐 = A + B (ADD problem): you cannot add these Rel % Errors to find error in Vo!!
• 1st: Convert these Rel errors first to ABS errors, then add the ABS errors.
𝑹𝟏
𝑹𝟑 × 𝑹𝟒
𝟐
• To do that: Calculate first the value of A and B. A =
× 𝑽𝒊 and B =
×
𝑰
𝒊
𝑹𝟐
𝑹𝟐𝟓
∆𝐴
∆𝐵
• Hence get ABS error of each term from 0.09 =
and 0.64 =
hence get ∆𝐴 and ∆𝐵
𝐴
𝐵
Solution: 𝑽 = 𝟑 ± 𝟑 × 𝟎. 𝟎𝟗 + 𝟎. 𝟎𝟎𝟏𝟐 ± (𝟎. 𝟎𝟎𝟏𝟐 × 𝟎. 𝟔𝟒)
𝒐
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5
Answer in Brief!!
A
𝑹𝟏
• 𝑽𝒐 =
× 𝑽𝒊
𝑹𝟐
+
𝑹𝟑 × 𝑹𝟒
𝑹𝟐𝟓
× 𝑰𝟐𝒊
B
=1%
• Multiply: Add Relative (Or % errors)
=10%
MULT: ADD Relative (or % errors) positively
∆𝐴
• Term A:
= 2% + 4% + 3% = 9%
𝐴
=20%
MULT: ADD Relative (or % errors) positively
∆𝐵
• Term B:
= 3% + 1% + (2 × 10%) + (2 × 20%) = 64%
𝐵
Good Practice
• ADD ABS errors:
𝑹𝟏
•A=
× 𝑽𝒊 = 3
𝑹𝟐
• ∆𝐴 = 𝐴 × 0.09 = 0.27
• ∆𝑽𝒐 = ∆𝐴 + ∆𝐵
𝑹𝟑 × 𝑹𝟒
× 𝑰𝟐𝒊 = 0.0012
and
B=
and
∆𝐵 = 𝐵 × 0.64 =0.000768
𝑹𝟐𝟓
Random Error !!!
Solution: 𝑽𝒐 = 𝟑 + 𝟎. 𝟎𝟎𝟏𝟐 𝑽 ± 𝟎. 𝟐𝟕 ± 𝟎. 𝟎𝟎𝟎𝟕𝟔𝟖 𝑽 ≅ 𝟑. 𝟎𝟎 𝑽 ± 𝟎. 𝟐𝟕𝑽
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6
Propagation of Error Detailed Example Solution:
𝑅 𝑅
𝑅3
𝑉𝑜 = 𝐼𝑖 × 1 2 − 0.2 𝑅4 2 ,
𝐼𝑖 = 10 𝑚𝑎 ± 1 𝑚𝑎 , & all R’s have tolerances
Remember: The CTV value is obtained from the equation as is, and also use the variables with its exact sign.
• To find the error, treat each term independently.
• The first alone is a multiplication (MULT) problem,
• The second term alone is a multiplication (MULT) problem.
• Then BOTH terms, will be an Addition/Subtraction (ADD) problem.
• First Find the CTV Value of the function 𝑉𝑜 (to be used later):
𝑹𝟏 . 𝑹𝟐
𝟔𝑲 × 𝟐. 𝟒𝑲
𝑽𝒐𝟏 𝑪𝑻𝑽 = 𝑰 ×
= 𝟏𝟎 𝒎𝑨 ×
= 𝟑𝟎 𝐕
𝑹𝟑
𝟒. 𝟖𝑲
𝑽𝒐𝟐 𝑪𝑻𝑽 = −𝟎. 𝟐 × 𝑅4 2 = −0.2 × 52 = −𝟓 𝑽
∴ 𝑽𝒐 𝑪𝑻𝑽 = 𝟑𝟎 𝑽 − 𝟓 𝑽 = 𝟐𝟓𝐕 This is the Function value (25V), then write the sum of 2 ABS erros in
%
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7
The First Term (Multip: Sum of REL.):
𝑹𝟏 𝑹𝟐
𝑉𝑜 = 𝑰𝒊 ×
− 𝟎. 𝟐 𝑹𝟒 𝟐
𝑹𝟑
2
10𝐾 ×
= 200 Ω
100
Given, as
Relative %
• 𝑅1 = 6K  2% ,
,
12
2.4𝐾 ×
= 288 Ω
100
𝑅2 = 2.4K  12%
and
𝐼𝑖 = 10 𝑚𝑎 ± 1 𝑚𝑎
240 Ω
× 100 = 𝟓%
4.8𝐾
Given, as
ABS
𝑅3 = 4.8K  240
∆𝑽𝟏
∆𝑰
∆𝑹𝟏
∆𝑹𝟐
∆𝑹𝟑
𝟏𝒎
•
= +
+
+
=
+ 𝟎. 𝟎𝟐 + 𝟎. 𝟏𝟐 + 𝟎. 𝟎𝟓 = 𝟎. 𝟐𝟗 ≡ ± 29%
𝑽𝟏
𝑰
𝑹𝟏
𝑹𝟐
𝑹𝟑
𝟏𝟎𝒎
• Or directly and simply:
∆𝑽𝟏
= 𝟐% + 𝟏𝟐% + 𝟓% + 𝟏𝟎% = 𝟐𝟗%
𝑽𝟏
• 𝑮𝒆𝒕 𝒕𝒉𝒆 𝑨𝑩𝑺 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒕𝒉𝒆 𝑬𝒓𝒓𝒐𝒓:
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∴ ∆𝑽𝟏 = 𝟑𝟎𝑽 × 𝟎. 𝟐𝟗 = ±𝟖. 𝟕𝑽 (ABS)
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8
The Second Term (Multip: SUM of REL):
𝑹𝟏 𝑹𝟐
𝑉𝑜 = 𝑰𝒊 ×
− 𝟎. 𝟐 𝑹𝟒 𝟐
𝑹𝟑
,
𝐼𝑖 = 10 𝑚𝑎 ± 1 𝑚𝑎
Given, as
Relative %
• 𝑅4 = 5  5% ,
• Applying the error propagation rule for Multiplication & Exponents on the 2nd term:
∆𝑽𝟐
∆𝑹𝟒
•
= ±𝟎. 𝟐 × 𝟐
= ±𝟎. 𝟐 × 𝟐 × 𝟓% = ± 𝟐%
𝑽𝟐
𝑹𝟒
• 𝑮𝒆𝒕 𝒕𝒉𝒆 𝑨𝑩𝑺 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒆𝒓𝒓𝒐𝒓: ∴ ∆𝑽𝟐 = 𝟓𝑽 × 𝟎. 𝟎𝟐 = 𝟎. 𝟏𝑽 (ABS)
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9
The Overall Function Vo (Add: Sum of ABS’s):
𝑹𝟏 𝑹𝟐
𝑉𝑜 = 𝑰𝒊 ×
− 𝟎. 𝟐 𝑹𝟒 𝟐
𝑹𝟑
,
𝐼𝑖 = 10 𝑚𝑎 ± 1 𝑚𝑎
• The total error = +ve SUM of ABS errors (bcz it is now an Addition case):
ABS
ABS
∆𝑉𝑜 = ∆𝑽𝟏 + ∆𝑽𝟐 = 𝟖. 𝟕𝑽 + 𝟎. 𝟏𝑽 = ±𝟖. 𝟖𝑽
• And to express the net result, we get first the CTV value of the function:
𝑉𝑜 = 𝟑𝟎 𝑽 − 𝟓𝑽 = 𝟐𝟓 𝑽
• Then add the error part:
ABS
∴
%
𝟖. 𝟖
𝑉𝑜 = 𝟐𝟓 𝑽 ± 𝟖. 𝟖 𝑽 = 𝟐𝟓 𝑽 ±
× 𝟏𝟎𝟎 = 𝟐𝟓 𝑽 ± 𝟑𝟓. 𝟐%
𝟐𝟓
𝑽𝟎 = {𝟏𝟔. 𝟐𝑽 𝒕𝒐 𝟑𝟑. 𝟖𝑽}
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10
Design of a Voltmeter (for a specific Range)
• Design a voltameter of Decade Range to measure a
voltage of 6.5 Volts. Given PMMC 50𝜇𝑎 , 600Ω
• Select a FS Decade Range =10V (Very Important Step)
S=
𝟏
ImFS
+
+
ImFS
VFS
rm
(The Range)
𝑰𝒎𝑭𝑺
vmFS
-
Rmult
Vmult
-
𝑹𝑽𝒐𝒍𝒕𝒎𝒆𝒕𝒆𝒓 = 𝑺 × 𝑽𝑭𝑺
Should be in the order of 10’s KΩ to 100’s KΩ - if much low, check your solution !!!
𝑹𝒎𝒖𝒍𝒕 = 𝑺 × 𝑽𝑭𝑺 − 𝒓𝒎
This 𝑹𝑽𝒐𝒍𝒕𝒎𝒆𝒕𝒆𝒓 would be used
in the loading effect Calc’s
This is the Voltmeter Design result
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11
Example: PMMC 𝟓𝟎𝟎 𝝁𝒂 , 600Ω
𝟏
𝟏
S=
=
𝑰𝒎𝑭𝑺
𝑰𝝁
e.g.
• V to be measured = 7.5V
𝟏
𝟏
𝟑
S=
=
=
𝟐
×
𝟏𝟎
= 𝟐𝑲
−𝟔
𝟓𝟎𝟎 𝝁𝒂
𝟓𝟎𝟎×𝟏𝟎
Select a Decade Range = 10 V
𝑹𝑽𝒐𝒍𝒕𝒎𝒆𝒕𝒆𝒓 = 𝑺 × 𝑽𝑭𝑺 = 𝟐𝑲 × 𝟏𝟎 = 𝟐𝟎𝟎𝑲Ω
& 𝑹𝑴𝒖𝒍𝒕𝒊𝒑 = 𝟐𝟎𝟎𝑲𝛀 − 𝟔𝟎𝟎𝜴
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12
Design an Ammeter (for a specific Range)
• Design an Ammeter of integer Range out of 1, 2, 5 to measure a current of
15.8 ma. Given PMMC 50𝜇𝑎 , 500Ω
IFS = 20 mA
• Select a FS Integer Range 20ma (not 1 or 2 or % Amp )
• The Impedance seen by the circuit is: 𝑹𝒔𝒉 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒓𝒎
• Calculate the Shunt resitter:
•
𝐼𝑚𝑒𝑡𝑒𝑟 𝜇 × 𝑟𝑚𝑒𝑡𝑒𝑟
𝑅𝑠ℎ =
𝐼𝑅𝑎𝑛𝑔𝑒 𝑏𝑖𝑔 − 𝐼𝑚𝑒𝑡𝑒𝑟 𝜇
ImFS
?
≅ 𝑅𝐴𝑚𝑚𝑒𝑡𝑒𝑟 (low value)
• To improve your calculation, consider parallel combination of 𝑟𝑚 and Rsh:
𝑹𝑺𝒉 × 𝒓𝒎
𝑹𝑨𝒎𝒎𝒆𝒕𝒆𝒓 =
𝑹𝑺𝒉 + 𝒓𝒎
Value should be in order of 10’s Ω to 100’s Ω. If much high, recheck your solution !!!!!
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13
Example:
• 𝑃𝑀𝑀𝐶 ∶ 𝐼𝑚𝑒𝑡𝑒𝑟 𝜇 = 500𝜇𝑎 𝑟𝑚𝑒𝑡𝑒𝑟 = 600Ω
• 𝐼 𝑡𝑜 𝑏𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑠 15.6𝑚𝑎
𝑅𝑎𝑛𝑔𝑒 = 𝟐𝟎𝒎𝒂 ≥ 15.6 𝑚𝑎 !)
• 20ma selected bcz it was requested among an integer numbers: 1, 2, 5, ….etc.
𝐼𝑚𝑒𝑡𝑒𝑟 𝜇 × 𝑟𝑚𝑒𝑡𝑒𝑟
𝑅𝑠ℎ =
𝐼𝑅𝑎𝑛𝑔𝑒 𝑏𝑖𝑔 − 𝐼𝑚𝑒𝑡𝑒𝑟 𝜇
(𝟓𝟎𝟎 × 𝟏𝟎−𝟔 ) × 𝟔𝟎𝟎
(𝟓𝟎𝟎 × 𝟏𝟎−𝟔 ) × 𝟔𝟎𝟎
0.3
𝑅𝑠ℎ =
=
=
= 𝟏𝟓. 𝟑𝟖 Ω
−𝟔
−𝟑
−𝟔
𝟐𝟎𝒎 − (𝟓𝟎𝟎 × 𝟏𝟎 ) 𝟐𝟎 × 𝟏𝟎 − (𝟓𝟎𝟎 × 𝟏𝟎 ) 0.0195
𝑹𝑺𝒉 × 𝒓𝒎
𝟏𝟓. 𝟑𝟖 × 𝟔𝟎𝟎
• For more precision : 𝑹𝑨𝒎𝒎𝒆𝒕𝒆𝒓 =
=
= 𝟏𝟒. 𝟗𝟗𝟔 ≅ 𝟏𝟓Ω
𝑹𝑺𝒉 + 𝒓𝒎
𝟏𝟓. 𝟑𝟖 + 𝟔𝟎𝟎
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14
Basic Simplified Design of a Transistor Amplifier:
• Always use the Simplified Design (not Thevenin’s). i.e. 𝐼𝐵 ≅ 0
• Always 𝑽𝑩 > 𝑽𝑬 by 0.7 V
𝒊. 𝒆.
𝑽𝑩 − 𝑽𝑬 = 𝟎. 𝟕 𝑽
• The Generic Circuit Design Procedure:
• 𝑽𝑩 = 𝑽𝒄𝒄 ×
𝑹𝟐
= 3.428 V
𝑹𝟏 +𝑹𝟐
(a Voltage Divider Eqn.)
• ∴ 𝑽𝑬 = 𝑽𝑩 − 𝟎. 𝟕 = 𝟑. 𝟒𝟐𝟖 − 𝟎. 𝟕 = 𝟐. 𝟕𝟐𝟖𝑽
𝑽𝑩
B
VBE = 0.7V
𝑽𝑬
(emitter V = 2.728V less than base V = 3.428V)
• 𝑭𝒊𝒏𝒂𝒍𝒍𝒚 ∶
𝑽𝑬
𝟐.𝟕𝟐𝟖
𝑰𝑬 =
=
= 𝟐. 𝟕𝟖 𝒎𝒂
𝑹𝑬
𝟏𝑲
𝑰𝑬
• Possible alteration in the next slides ➔
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15
Ammeter and Voltmeter Loading effect when measuring
transistor circuits:
• Four Steps:
1. Design the basic Transistor circuits to get : 𝑽𝑩 , 𝑽𝑬 , 𝑰𝑬
2. Knowing the voltages and currents:
Design your Ammeter and Voltameter with decade or integer
range as asked in the question, FS just ≥ the measured value.
e.g. 𝑰𝑬 = 7.5 ma , FS = 10 ma or if 𝑰𝑬 = 10 ma , FS = 10 ma ….etc.
3. Connect the Ammeter in series, or the Voltmeter in
parallel.
Hence, Find the new element value after connection
of the instrument either in series or in parallel.
𝑽𝑩
B
VBE = 0.7V
𝑰𝑬
4. Redesign the circuit to find any of 𝑽′𝑩 , 𝑽′𝑬 , 𝑰′𝑬 as
needed. Hence calculate their loading error.
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16
Ammeter and Voltmeter Loading effect when measuring
transistor circuits:
• Four Steps:
1. Connect the Ammeter in series, OR the Voltmeter in parallel.V
2.
1𝐾×20𝐾
′
e.g. 20KΩ Voltmeter // RE ∴ 𝑅𝐸 =
= 0.952 𝐾Ω
1𝐾+20𝐾
3. e.g. 80 Ω Ammeter series with RE ∴ 𝑅𝐸′ = 1𝐾 + 15 Ω = 1015 V
4. Redesign the circuit (Slide 15) to find any of 𝑽′𝑩 , 𝑽′𝑬 , 𝑰′𝑬 as
needed. Hence calculate the loading error.
5. % error =
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𝑁𝑒𝑤 𝑉𝑎𝑙𝑢𝑒 − 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
Systematic Error !!!
× 100
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V
A
Any one of these
possible connections,
Or a combination ,
e.g. A voltmeter on
the base and ground,
and an ammeter in
series with RE!!
17
Other examples of Instruments Loading Effect
OR
OR
𝑉𝑜 = 𝑰𝒊 ×
𝑹𝟏 𝑹𝟐
𝑹𝟑
In a mathematical Relation,
e.g. we get 𝑅2′ due t a
Voltmeter in Parallel
• Solve the Circuit with no Loading (CTV values).
• Any Element can be loaded by a Voltmeter or an Ammeter.
• The Element will Change its value.
• Repeat the Circuit Solution.
Systematic Error !!!
• Then find the error (difference between Loaded and Unloaded).
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Ohmmeter
Design an Ohmmeter with a half scale reading of 4KΩ
at its Midscale, given a PMMC of 1.5 mA and 10  coil
resistance.
The Ohmmeter is driven by a battery of voltage E = 6Vdc
Hint: find 𝑰𝑭𝑺 and 𝑹𝑺 + 𝑹𝑴 .
𝑰𝑭𝑺
𝟐
𝟔𝑽
=
𝟔𝑽
𝑹𝑺 + 𝑹𝑴 + 𝑹𝒙 𝒎𝒊𝒅 𝒔𝒄𝒂𝒍𝒆
=
𝑹𝑺 + 𝑹𝑴 + 𝟒 𝑲
(1)
𝟔𝑽
𝑰𝑭𝑺 =
𝑹𝑺 + 𝑹𝑴 + 𝟎
(𝟐)
Equate both equations to find (𝑹𝑺 + 𝑹𝑴 ).
Then Substitute in one of them to get 𝑰𝑭𝑺 .
Hence select a PMMC with 𝑰𝑭𝑺 and 𝑅𝑀 (should be given).
Then get 𝑹𝑺 , because (𝑹𝑺 + 𝑹𝑴 ) has been calculated.
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Ohmmeter Example
Design an Ohmmeter with a half scale reading of 4KΩ at its Midscale, given a PMMC
of 1.5 mA and 100  coil resistance. Battery of voltage E = 6Vdc
Hint: find 𝑰𝑭𝑺 and 𝑹𝑺 + 𝑹𝑴 .
𝑰𝑭𝑺
𝟐
𝟔𝑽
=
To design an Ammeter, we
need to find the value of 𝑅𝑠
and a suitable PMMC (even
if it was given !!!).
𝟔𝑽
𝑹𝑺 + 𝑹𝑴 + 𝑹𝒙 𝒎𝒊𝒅 𝒔𝒄𝒂𝒍𝒆
=
𝑹𝑺 + 𝑹𝑴 + 𝟒 𝑲
(1)
𝟔𝑽
𝑰𝑭𝑺 =
Equate both equations:
(𝟐)
𝑹𝑺 + 𝑹𝑴 + 𝟎
𝑰𝑭𝑺 =
𝟐×𝟔𝑽
𝟔𝑽
=
𝑹𝑺 +𝑹𝑴 +𝟒𝑲
𝑹𝑺 +𝑹𝑴 +𝟎
∴ (𝐑 𝐒 + 𝐑 𝐌 ) = 𝟒𝑲
Then Substitute in one of them (2) to get:
𝑰𝑭𝑺 =
𝟔𝑽
= 𝟏. 𝟓 𝒎𝒂𝒂𝒂.
𝟒𝑲+ 𝟎
Since the PMMS is 1.5 ma, hence it can be used directly.
Then get 𝑹𝑺 𝒇𝒓𝒐𝒎 ∶ (𝐑 𝐒 + 𝐑 𝐌 ) = (𝐑 𝐒 + 𝟏𝟎𝟎) = 𝟒𝑲
Hence 𝑹𝑺 = 𝟒𝑲 − 𝟏𝟎𝟎 = 𝟑. 𝟗𝑲Ω
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If IFS ≠ 𝑰𝑴𝑭𝑺, 𝒕𝒉𝒆𝒏 𝒅𝒆𝒔𝒊𝒈𝒏 𝒂𝒏 𝑨𝒎𝒎𝒆𝒕𝒆𝒓 𝒘𝒊𝒕𝒉 𝑹𝒂𝒏𝒈𝒆 = 𝑰𝑭𝑺 (𝟏. 𝟓𝒎𝒂)
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Oscilloscopes
• How to measure Amplitude NOTE:
• DC shift (Must take care of the Ground Level).
• AC Coupling or DC Coupling
• V/div value
• Vpp or Vp or VRMS
• The screen is vertically ± 4 Division from the Center.
• Use the Golden Rule 𝑉𝑃𝑒𝑎𝑘 = 𝑁 × (𝑉/𝑑𝑖𝑣)
• How To measure Time or Frequency:
• Note the Time per Division: s/div or ms/div or µs/div or ns/div.
• The TB is per only one division (ms/div and not per the total screen width 10 div)
• Frequency =
1
𝑃𝑒𝑟𝑜𝑑𝑖𝑐 𝑇𝑖𝑚𝑒𝜏
• Use the Golden Rule
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𝜏 = 𝑁 × (𝜇𝑠/𝑑𝑖𝑣)
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where 𝜏 =
𝑓
21
The Vertical Amplifier Golden Rule
𝑉𝑃𝑒𝑎𝑘 = 𝑁 × 𝑉𝑜𝑙𝑡𝑠/𝑑𝑖𝑣
𝑉𝑃 = 𝑁 × 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐺𝑎𝑖𝑛
𝑉𝑃 = 𝑁 × 𝐺𝑎𝑖𝑛𝑣/𝑑𝑖𝑣
𝑉𝑃 = 𝑁 × (𝑉/𝑑𝑖𝑣)
𝑽𝒑 = 𝟑 × 𝟓𝟎𝒎𝒗/𝒅𝒊𝒗 = 𝟏𝟓𝟎 𝒎𝑽
• If the input is DC, then logically:
MAR - 2025
𝑉𝐷𝐶 = 𝑁 × (𝑉/𝑑𝑖𝑣)
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The TB Golden Rule
𝜇𝑠 , 𝑜𝑟 𝑚𝑠 , 𝑜𝑟 𝑠
𝜏 = 𝑁 × 𝜇𝑠/𝑑𝑖𝑣
𝜏 = 𝑁 × 𝑇𝐵
𝜏 = 𝑁 × (𝜇𝑠/𝑑𝑖𝑣)
1
𝑓=
𝜏
𝟏
𝟏
𝝉 = 𝟒 × 𝟓𝟎𝝁𝒔/𝒅𝒊𝒗 = 𝟐𝟎𝟎𝝁𝒔 ∴ 𝒇 = =
= 𝟓𝑲𝑯𝒛
𝝉 𝟐𝟎𝟎𝝁
𝜆
OR
1
𝜏=
𝑓
f is the frequency of the input signal
𝜏
• This applies for a complete sine wave. If part of the sinewave (e.g. 2 ), then note that the time will be 2 , and so on.
1
• If you want to display one sine wave per full screen, then N = 10, and 𝜏 =
𝑓
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If using an X10 probe:
• The scope impedance (Resistance) alone is 1 MΩ
• If using an X10 probe, its resistance will be 10MΩ and 𝑉𝑃 = 𝟏𝟎 × 𝑁 × (𝑉/𝑑𝑖𝑣)
• The loading effect of a scope on a transistor circuit will be exactly similar to the
Voltmeter case,
• The overall resistance will be the element resistance, in parallel with: either the
1MΩ, or the 10MΩ (if an X10 probe is used) of the scope.
• This will be the loading effect of the scope on the circuit.
• Then calculate the loading error as usual.
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Example on Scopes (CRO)
Sinewave with
DC Shift
• CH-Y1:
• GAIN = V/div = 20 𝑚𝑣
With DC Shift
• DC coupling
+ve Slope Trigger
• DC shift.
• Trigger at 0 level
–ve Slope Trigger
• +ve Slope trigger.
✓
✓
• What is the Vin? 𝑽𝒊𝒏 = 𝑵 × (𝑽/𝒅𝒊𝒗)
Thick upper Red Trace:
• hence Vin = 𝟑 × 𝟐𝟎𝒎/𝒅𝒊𝒗 = 𝟔𝟎 𝒎𝒗
𝑽 = 𝟒𝟎𝒎𝒗 𝐬𝐢𝐧 𝟐𝝅𝒇𝒕
• CH-Y2
Thin middle Orange Trace:
𝑽 = 𝟔𝟎𝒎𝒗 𝐬𝐢𝐧 𝟐𝝅𝒇𝒕
+ 𝟒𝟎 𝒎𝒗
Lower Trace:
𝑽 = 𝟐𝟎𝒎𝒗 𝐬𝐢𝐧 𝟐𝝅𝒇𝒕 − 𝟔𝟎 𝒎𝒗
• CH-Y2 has gain of 3 V/div
• It shows a coinciding trace over CH-Y1 trace !!!! What is the value of the Signal?
• 𝑽𝒊𝒏 = 𝑵 × (𝑽/𝒅𝒊𝒗) = 𝟑 × 3 V/div = 𝟗 𝑽
Thin middle Orange Trace assume it has 30mV DC :
• If the signal frequency = 200KHz, what is the TB value ? 𝑽 = 𝟔𝟎𝒎𝒗 𝐬𝐢𝐧 𝟐𝝅𝒇𝒕 + 𝟑𝟎𝒎𝒗 𝒘𝒊𝒕𝒉 𝑨𝑪 𝑪𝒐𝒖𝒑
𝟏
𝟏
𝟏
• 𝝉 = 𝑵 × (𝝁𝒔/𝒅𝒊𝒗) = =
= 𝟒 × 𝑻𝑩 ∴ 𝑻𝑩 =
= 𝟏. 𝟐𝟓𝝁𝒔
𝒇
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𝟐𝟎𝟎𝑲
𝟖𝟎𝟎 𝑲
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Example on Scopes (CRO)
• CH-Y1:
• Vin = 60 𝑚𝑣
• DC coupling
• No DC shift.
• Trigger at 0 level
• +ve Slope trigger.
✓
✓
• What is the CH-Y1 Gain? 𝑽𝒊𝒏 = 𝟔𝟎𝒎𝒗 = 𝑵 × (𝑽/𝒅𝒊𝒗)
𝟔𝟎𝒎𝒗
• hence V/d =
= 𝟐𝟎𝒎𝒗/𝒅𝒊𝒗
• CH-Y2
𝟑 𝒅𝒊𝒗
• CH-Y2 has gain of 3 V/div
• It shows a coinciding trace over CH-Y1 trace !!!! What is the value of the Signal?
• 𝑽𝒊𝒏 = 𝑵 × (𝑽/𝒅𝒊𝒗) = 𝟑 × 3 V/div = 𝟗 𝑽
• If the signal frequency = 200KHz, what is the TB value ?
𝟏
𝟏
𝟏
• 𝝉 = 𝑵 × (𝝁𝒔/𝒅𝒊𝒗) = =
= 𝟒 × 𝑻𝑩 ∴ 𝑻𝑩 =
= 𝟏. 𝟐𝟓𝝁𝒔
𝒇
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𝟐𝟎𝟎𝑲
𝟖𝟎𝟎 𝑲
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Examples
With DC Shift
The red are the unknowns
𝝉𝟐 =
𝟏
𝝉𝟏 =
𝒇𝟏
𝟏
𝒇𝟐 CH1-IN
𝑉𝑃 = 𝑁 × (𝑉/𝑑𝑖𝑣)
✓
✓
𝑉𝑃 = 10𝑚𝑣 = 1 × 10𝑚𝑉/𝑑𝑖𝑣)
✓
✓
𝑉𝑃 = 10𝑉 = 1 × 10𝑉/𝑑𝑖𝑣)
CH2-OUT
• If the frequency = 1KHz, what the TB?
• Find the frequency of each signal if the TB = 20µs/div
• Find first a complete sinewave: N=10
✓
✓
• 𝜏1 = 10 Div × 20𝑢𝑠/𝑑𝑖𝑣 = 200us 𝜏 = 𝑁 × 𝑇𝐵(𝜇𝑠/𝑑𝑖𝑣)
1
𝟏
• 𝑓 of CH1 = 200𝜇𝑠 = 5 KHz
𝝉=
𝒇
• 𝜏2 = 5 Div × 20𝑢𝑠/𝑑𝑖𝑣 = 100us
1
• 𝑓 of CH2 = 100𝜇𝑠 = 10 KHz
TB is the same !!!!!!!!!
• The circuit is a frequency doubler.
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𝟏
𝒇
• 𝝉= =
𝟏
= 1 𝑚𝑠
𝟏𝑲
✓
✓
• 𝜏 = 𝑁 × 𝑇𝐵 = 1𝑚𝑠 = 5 × 𝑇𝐵
• What is the CH2 DC shift, if the V/div =
20mv? = +2 Div x 20mv/div = 40mv
• What is the V/div if Vp of CH1 = 5V?
5𝑉
• Vp =
= 5 V/div
1 𝑑𝑖𝑣
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Example, Let’s Solve it together !!
• Notes:
• The amplitude of the input signal = 20 V
• It has +10 V DC shift
• USE the Golden 𝑽𝒊𝒏 = 𝑵 × (𝑽/𝒅𝒊𝒗)
for the Vi , and for the DC Shift !
• Same for CH-Y2. There will be Different DC shift
and Different AC Amplitude: (by the Divider Ratio!)
if any of Y1 ot Y2 in AC Mode: WoW. No DC Shift !!
• 𝜔 is not given, then there must be TB given
• Or if frequency is given, then get the TB
𝟏
• 𝝉 = 𝑵 × (𝝁𝒔/𝒅𝒊𝒗) =
𝒇
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X-Y Mode
Blocking
CAP
Blocking
CAP to
get the Lissajous centered
C
෡ 𝐬𝐢𝐧(𝝎𝒕 + 𝝓)
𝒗𝒚 = 𝑬
𝑣𝑦
+TB
Trigger
CRO
Sweep
C
෡ 𝐬𝐢𝐧(𝝎𝒕)
𝒗𝒙 = 𝑬
Out of phase (180 Deg phase shift
In-Phase (e.g. Resistive Circuits)
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Good Luck ISA
Next Exam:
Final Sit EXAM
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MAR - 2025
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Dr. Sherif Guinena
Sherif.Guinena@BUE.edu.eg
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