INDUSTRIAL ELECTRONICS N5
CHAP 1: AC THEORY AND RLC CIRCUIT
Capacitor: is a device that has the ability to store electric charge when the voltage is applied.
Inductor called a coil has the ability to generate a magnetic field when the current flows.
The capacitance of capacitor depends on three factors:
- Area of the plate
- Distance between
- Di-electric constant
The rate at which the capacitor charges depends on: its Capacitance and Resistance in
the circuit.
TIME CONSTANT OF RC CIRCUIT
This I is the time taken by the capacitor to charge from 0% to 63,2 % of the total allied voltage
T = RC
RC DIFFERENTIATOR AND RC INTEGRATOR
RC DIFFERENTIATOR Called high pass filter because it allow high frequency (AC) to pass and block Low
frequency (DC). The output is obtained from the resistor.
RC INTEGRATOR Called low pass filter because it allow low frequency (DC) to pass and block high
frequency (AC). The output is obtained from the Capacitor.
CALPACITOR C OUPLING
This circuit is used to block DC signal and while AC signal pass at the output signal of an amplifier.
TIME CONSTANT OF RL CIRCUIT
- This the time taken fo the circuit to reach a value of 63,2% of the total current. T =L / R
-
Inductor known as chock connected in series likecoupling capacitor blocks high frequency signal
(AC) and pass low frequency signal(DC) in a DC power supply or Low pass filter.
Connected in parallel the Capacitorwill pass DC to the load and Inductorwill pass AC signal to
the load.
HARMONICS
-
This is the distortion of sinusoidal signal by waveforms of different frequencies. Or Nonsinusoidal complex waveformsor Combination of squarewaves and sine waves of different
frequncies and amplitudes.
- Sine waveform and square waveform are used for testing of electronic circuit,
- When a signal is sent through reactive components (Capacitor and Inductor) : the square
waveform become distorted but the sine waveform is not distorted.
- Square waves indicate the presence of error.
- Sinewave indicate the existance of an error.
- As square wave is reach in harminc frequencies and phase disturbances can be easily observed.
Shortage of high frequency respopnse can be observed is a leading and trailing edges of a waveform
and low frequency in the upper horizontal level as shown:
SERIES, PARALLEL AND SERIES – PARALLEL CIRCUITS
SERIES CIRCUIT
Q – factoris the voltage amplification where VL= VC
calculated by Q = XL/R (XL = XC) or
Selectivity
1/R√
and bigger than VT
𝑳
𝑪
is the ability of particular circuit to respond to a particular frequency by neglecting all other
frequencies.
Or the fact that higher Q-factor offer a sharper peak neglecting frequencies closed to resonance
frequencies.
Bandwidth is a frequency range in which the current exceed 0,707 of I T
PARALLEL CIRCUITS
RESONANCE IN PERALLEL CIRCUIT
The current through the inductor (IL)is the same as the current through the capacitor (IC)and this value
is larger than the supply current ( IT).This amplification factor is called Q- factor(Q) Or
Q – factor in parallel resonance circuit is the current amplification.
Q-factor
𝐼𝑐
……………Q =𝐼𝑇
Resonance frequency ……fr
Dynamic impedance
=
=
1
2𝜋
𝐼𝐿
𝐼𝑇
√
1
𝐿𝐶
𝑅2
− 𝐿2
but if XL ≫ R …………
…………. ZD = 𝑹𝑪
fr =
𝟏
𝟐𝝅√𝑳𝑪
𝑳
EXERCISES 1.1
CHAP 2 POWER SUPPLIES (PS)
PS is an electronic device used to convert AC to DC .
PS consists of: Transformer, Rectifier, Filter and Voltage regulator.
1. TRANSFORMER
The transformer step down the voltage (e.g. 220 V AC to 12 V AC)
2 RECTIFIERS: Convert AC to pulsating DC there is half wave and full wave.
2.1 HALF WAVE RECTIFIER
2.2 FULL WAVE RECTIFIERS : CENTRE TAP AND BRIDGE
3 FILTER CIRCUIT (Capacitor)
3.1 Simple capacitor filter (One Capacitor)
3.2 𝜋- FILTER : RC 𝜋- π𝜋
− 𝑓𝑙𝑡𝑒𝑟- filter
(Two Capacitors)
VOLTAGE REGULATION (VR)
VR is a degree to which the DC output voltage is kept constant irrespective of changes in AC
mains or load
𝑉 −𝑉
𝑉 −𝑉
VR = 𝑁𝐿𝑉 𝐹𝐿 and percentage voltage regulation is %VR = 𝑁𝐿𝑉 𝐹𝐿 𝑥 100
𝐹𝐿
𝐹𝐿
EXERCISES 2: Q1 to Q13
VOLTAGE DOUBLERS
Voltage multipliers are used as soon as a higher voltage is needed from secondary side of
a transformer
HALF WAVE VOLTAGE DOUBLER
-
OPERATION:
During the”+” half cycle of Vm, D1 is FB, conduct,
charging C1 to Vm, D2 is FB
During the “ -- “ half cycle of Vm, Vc1 and Vm are in
series charging C2 to 2Vm through V2.
FULL WAVE VOLTAGE DOUBLER
OPERATION:
- During the”+” half cycle of Vm, D1 is FB, conduct, charging C1 to Vm, D2 is FB
- During the “ -- “ half cycle of Vm, D2 is FB, conducts, charging C2 to Vm, D1 is RB.
Vc1 and Vc2 are in series, therefore :Vo = Vc1 + Vc2 = Vm + Vm = 2 Vm
VOLTAGE TRIPPLER
OPERATION:
During the”+” half cycle of Vm, D1 is FB, conduct, charging C1 to
Vm, D2 is FB
During the “ -- “ half cycle of Vm, D2 is FB, conducts, charging C2
to Vm, D1 is RB.Vc1 and Vc2 are in series, therefore :
Vo = Vc1 + Vc2 = Vm + Vm = 2 Vm
During the next “+” Vc2 and Vc1 are in series charging C1 to Vm
and C3 to 2 Vm. Vo = Vc1 + Vc3 = Vm+ 2Vm = 3 Vm.
SERIES AND SHUNT REGULATORS
5 Elements making Voltage regulator:
- Reference element: provide a stable reference level
- Feedback network: supply feedback signal,
- Comparator : compares the feedback signal with the reference signal and supply an
error signal.
- Protective element: protect the control element,
- Control element: Reacts when an error signal is supplied keeping Vo constant as the
supply voltage changes.
VOLTAGE REFERENCE SOURCE
- Zener Diode is usually used as a voltage reference source.
- Dz regulate when Vin > Vz
- Rs is needed to absorb the voltage difference.
- Rs(min) is calculated to protect Dz against excessive current because if Vin changes,
VR change, Iz changes but Vz remains constant
Rs (min) =
-
𝑽𝒊𝒏(𝒎𝒂𝒙)− 𝑽𝒁
𝑰𝒛(𝒎𝒂𝒙)
RL causes the Iz to drop and when IZ approaches zero regulation process stops.
RL (min) = 𝑽
𝑽𝒁
𝒊𝒏(𝒎𝒂𝒙)− 𝑽𝒁
EXERCISES 2: Q16
SERIES VOLTAGE REGULATORS
1. SERIES VOLTAGE REGULATORS WITH ADJUSTABLE REFERENCE SOUTCE
To make the output voltage adjustable, a potentiometer is used
as a voltage divider to make reference voltage adjustable.
OPERATION
. P is used to adjust Vr
. Vr is constant across Vz: Vr = Vbe + Vo
. If Vo increases, Vbe decreases, switches off and Vo decreases,
very small current flows.
. If Vo decreases, Vbe increases T1 switches ON more and Vo
increases.
2. SERIES VOLTAGE REGULATORS WITH AN AMPLIFIER
By connecting a feedback network with a comparator or
differential amplifier the stabilizing factor is improved.
- P1, R2 and R3 form an adjustable voltage divider network
supplying Vf from Vo to the differential comparator
- Comparator compares the feedback voltage (Vf) with
reference voltage (Vz).
- This difference in the output of a comparator is an
amplified error signal.
- Any increase in Vo caused by increase in Vin produce
higher Vf, error voltage increases as well.
- T1 is forced to conduct less current, Vo falls therefore
stabilized.
3. OVER CURENT PROTECTION
-
If the load draws too much current, the voltage across P2
increases enough to switch T2 ON, Ib of T1 decreases,
therefore switches OFF decreasing the load current.
- P1 adjusted, T2 is OFF. If there is overload, volt drop at P2
increases, T2 becomes ON, Ib1 ON Ic is limited to a safe
value
4. OVER VOLTAGE PROTECTION
- The value of Dz and R5 determine the maximum
voltage accepted by the load.
- Normally R5 is too low to trigger the SCR.
- If Vo becomes too large, the voltage drop across R5
becomes too large enough to trigger the SCR, Vz
remains constant.
- Once the SCR is triggered, all the current flows through
5. SHUNT VOLTAGE REGULATOR
it and Vo ≅ 0
- If Vo increases, Vbe increases. T switches ON more, Ic
decreases and Vo decreases.
- If Vo decreases, Vbe decreases, the transistor switches
OFF, IL increases and Vo increases while Vz is always
constant.
6. DOUBLE VOLTAGE POWER SUPPLIES
Double voltage power supplies is required for
differential amplifier that need positive and
negative power supplies.
ADJUSTABLE POWER SUPPLY
- Fixed Voltage Power supplies are used to form part of electronic equipment
E.g TV, Radio or measuring instrument
- Adjustable power supplies are used for the testing of electronic circuitry.
7. A REGULATOR WITH VOLTAGE HIGHER THAN FIXED VOLTAGE REGULATOR
-
Vo is always 5 V larger than VG , Vo = VG + 5 V
VG is the volt drop across P1 ( ground voltage)
8. A REGULATOR WITH VOLTAGE LOWER THAN FIXED VOLTAGE REGULATOR
P1 adjust the ground reference to a negative reference -causing
Vo to drop below the normal level
(VG: negative value) Vo = VG + 5 V
9. HIGH STABLE ADJUSTABLE VOLTAGE REGULATOR
-
This regulator is obtained if the feedback
gain is increased with the gain of an
amplifier.
This regulator will produce an adjustable
voltage of+7 to +30 V with a + 5 V
regulator.
-
10. THREE TERMINALS VOLTAGE REGULATOR
These are three terminals Integrated circuit(IC) available in different current and voltage
rating
These regulators are fixed voltage regulator but some can produce an adjustable output or
fixed output,
These output can be positive or negative
Adjustable output (Fixed input)
Positive
Negative
LM317
LM337
Fixed output (fixed input)
Connection of LM7812 regulator in a power supply
e.g. LM7903 ……….Vo = - 3 V and
LM7803………….Vo = 3V
Positive VR
Negative VR
LM 78 XX
LM 79 XX
CHAP 3 : TRANSISTOR AMPLIFIERS
Design of a fixed bias amplifier without Re( unstable)
Ic
Ib
This circuit produce a reasonable gain as amplifier but its bias in unstable
β=
RC =
𝑉𝑐𝑐 − 𝑉𝑐𝑒
Rb =
𝐼𝑐
𝑉𝑐𝑐 − 𝑉𝑏𝑒
𝐼𝑏
β = gain of Transistor
Design of a fixed bias amplifier with Re(stable)
Factors influencing the bias stability in a transistor are:
Inverse saturating current
Base emitter voltage
Current gain.
To prevent the shift of the work point Re is connected.
The function of Re is to limit the Ic during an increase in temperature.
Re increases the bias stability but causes an oscillation due to negative feedback . To keep the
DC stability the capacitor Ce is connected in parallel with Re.
These factors cause a shift on the work point due to change in temperature :
RC =
𝑉𝑐𝑐 − 𝑉𝑐𝑒 −𝑉𝑒
𝐼𝑐
, Rb =
𝑉𝑐𝑐 − 𝑉𝑏𝑒 −𝑉𝑒
𝐼𝑏
, β=
Ic
Ib
,
Ve =
𝑉𝑐𝑐
10
, Ce ≥
10
2𝜋𝑓𝑅𝑒
, Re =
𝑉𝑒
𝐼𝑐
Design of a Voltage divider bias amplifier
As the value of current gain 𝛽 is temperature sensitive it is important to
supply a DC circuit which will not depend on the transistor gain. That
circuit uses a voltage divider network Rb1 and Rb2.
Rb1 =
Ve =
𝑅𝑏2 ( 𝑉𝑐𝑐 −𝑉𝑏 )
𝑉𝑏
𝑉𝑐𝑐
10
Re =
𝑉𝑒
𝐼𝑐
1
Ic
Rb2 = 10 𝛽Re
β=
Vb = Vbe + Ve ,
Vbe = 0,7 V if not given
Ib
DRAWING OF HYBRID PARAMETERS
Design of a fixed bias amplifier without Re(unstable): in H- Parameter
USING PRECICE METHOD
USING PRECISE METHOD
RC = ZL
,
1
Z2 =
(Rs = 0 Ω )
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒
hoe −
ℎ𝑖𝑒 + 𝑅𝑠
Z1 = hie --
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒𝑥𝑍𝐿
1+ℎ𝑜𝑒 . 𝑍𝐿
Current Gain Voltage gain:
ℎ𝑓𝑒
𝑅𝑏
Av = (
Ai = ( 1+ℎ𝑜𝑒.𝑍𝑙)(𝑅𝑏+𝑍1)
Power gain
− ℎ𝑓𝑒 . 𝑍𝐿
ℎ𝑖𝑒+(ℎ𝑖𝑒 .ℎ𝑜𝑒−ℎ𝑓𝑒.ℎ𝑟𝑒)𝑍𝐿
Input impedance
Ap = - Ai . Av
)
Output impedance:
Zi = Rb//Z1
Zo = ZL//Z2
Design of a fixed bias amplifier with Re( stable) :
in H- Parameter
USING APPROXIMATION METHOD
Current Gain: Ai = hfe
Voltage gain: Av
= -
ℎ𝑓𝑒 .𝑍𝐿
ℎ𝑖𝑒
Input impedance: Z1 = hie
-
,
Power gain:
,
Ap =
Output impedance :
ℎ𝑓𝑒 2 . 𝑍𝐿
ℎ𝑖𝑒
Z2 > 1/hoe
Fixed bias amplifier with and without Re use the same formula in HParameter.
VOLTAGE DIVIDER IN H-PARAMETER
-
USING PRECISE METHOD
Rc = ZL
,
Z2 =
Current Gain
ℎ𝑓𝑒
1
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒
hoe −
ℎ𝑖𝑒+𝑅𝑠
Ap = - Ai . Av
Z1 = hie -
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒𝑥𝑍𝐿
1+ℎ𝑜𝑒 . 𝑍𝐿
Voltage gain
𝑅𝑏𝑇
Ai = ( 1+ℎ𝑜𝑒.𝑍𝑙)(𝑅𝑏𝑇+𝑍1)
Power gain
,
Av = (
Input impedance
Zi = RbT //Z1
− ℎ𝑓𝑒 . 𝑍𝐿
ℎ𝑖𝑒+(ℎ𝑖𝑒 .ℎ𝑜𝑒−ℎ𝑓𝑒.ℎ𝑟𝑒)𝑍𝐿
)
Output impedance:
Zo = ZL//Z2
RbT =
𝑅𝑏1 𝑥𝑅𝑏2
𝑅𝑏1+𝑅𝑏2
-
USING APPROXIMATION METHOD
𝑅𝑏2
Current Gain:
Ai = 𝑅𝑏2 +ℎ𝑖𝑒 x hfe
ℎ𝑓𝑒 .𝑍𝐿
Voltage gain:
Av = -
Input impedance:
Z1 = hie ,
ℎ𝑖𝑒
,
Power gain:
Ap =
Output impedance :
ℎ𝑓𝑒 2 . 𝑍𝐿
ℎ𝑖𝑒
= - Ai . Av
Z2 > 1/hoe
COMMON BASE AMPLIFIERS
Common base amplifier in H- Parameter
-
USING PRECISE METHOD
Voltage gain is higher,
Current gain is less than one,
-
For common base calculation all the “e” are replaced by “b” and “b” by ” e”.
For common base:
RBT and Rb are replaced by Re
ZL= Rc//RL =
,
Z2 =
𝑹𝑪𝒙𝑹𝑳
( RC ≠ ZL )
𝑹𝑪+𝑹𝑳
Zi = Re//Z1
Zo = Rc//Z2
1
hob
hie:
hoe:
hfe:
hre:
input impedance of common emitter,
output conductance of common emitter,
forward current gain of common emitter (𝛽 )
reverse voltage gain of common emitter.
ℎ𝑓𝑏 𝑥 ℎ𝑟𝑏 𝑥 𝑍𝐿
ℎ𝑓𝑏𝑥 ℎ𝑟𝑏,
−
ℎ𝑖𝑏+𝑅𝑠
Z1 = hib - 1+ℎ𝑜𝑏 . 𝑍𝐿
Current Gain Voltage gain :
ℎ𝑓𝑏
𝑅𝑒
𝑅𝑐
Av = (
Ai = ( 1+ℎ𝑜𝑏.𝑍𝑙)(𝑅𝑒+𝑍1) (𝑅𝑐+𝑅𝐿 )
Power gain
Ap = - Ai .Av
-
− ℎ𝑓𝑏 . 𝑍𝐿
ℎ𝑖𝑏+(ℎ𝑖𝑏 .ℎ𝑜𝑏 − ℎ𝑓𝑏.ℎ𝑟𝑏)𝑍𝐿
Input impedance
Zi = Re//Z 1
)
Output impedance:
Zo = Rc//Z2
USING APPROXIMATION METHOD
𝑅𝑐
Current Gain:
Ai = hfb(𝑅𝑐 +𝑅𝐿 )
Av = -
Voltage gain:
ℎ𝑓𝑏 .𝑍𝐿
ℎ𝑖𝑏
,
Power gain:
Ap = - Ai . Av
Zo = Rc
3.11 DISTORTION OF COMMON EMITTER AMPLIFIER
Input impedance:
Zi = hib ,
Output impedance :
The output Signal is distorted when the output signal does not adopt the format of the input signal.
Factors responsible for distortion causal
-
- The work point of the circuit is in error,
The load and the circuit output is not adapted ,
- The input characteristics are not linear,
- The amplitude of the input is too big.
Three Types Of Distortions:
- AMPLITUDE DISTORTION
. An increase in input signal will result to increase of an output amplified signal.
. But, if the input amplitude is too big, The amplified signal will be chopped at saturation
or cut-off region as shown:
- NON-LINEAR DISTORTION
The distortion of the output signal is caused by the non-linearity of the input signal. When
Vbe drops below cut-off, the input Ib is distorted therefore Ic and Vce are distorted.
FREQUENCY DISTORTION
The distortion is caused by reactive components (inductor and
capacitor)
CHAP 4 OPERATIONAL AMLIFIERS (op-am)
Op-am is an IC with high voltage gain using an external feedback network for its operation.
It can operate as :
- Amplifier
- Signal generator
- Voltage regulator
Characteristics: High Zin, Low Zo, High Av, Wide bandwidth, Low voltage offset.
Drawing of 741 – Integrated Operational Amplifier In a:
DIFFERENTIAL AMPLIFIER CIRCUIT
Drift is a slow variation of the work voltage or current on the output while the input is
kept constant
Factors causing Drift :
- Temperature variation,
- Changing load condition,
- Aging of component,
- Unstable voltage supplied,
- Parameter variation of component.
FIG2 is circuit diagram of a differential amplifier Using NPN transistor.
- This circuit is designed to reject common mode signal.
-
The difference of the amplifier is expressed by a factor called COMMON MODE REJECTION
RATIO(CMRR).
The large the CMRR the more effective the amplifier becomes.
Common Mode is when the two input signals are equal in polarity, the amplifier’s response
will hardly reflect any difference.
Differential Mode is when the two input signals are opposite in polarity, the amplifier’s
response will reflect the difference between the two input signal..
Disadvantage of this circuit, an unwanted offset called differential input offset (Vio =± 2 mV) is
developed.
differential input offset Is a small deviation of bias potential at input terminal caused by
transistor disorder.
To eliminate this unwanted offset most op-am possesses Null balance facility using a 10 K
potentiometer,
-
Questions
-
How ≠cial amplifier can reject common mode signal?
The two transistors at input are equal and amplify the common mode signal equally.
Because the ≠ce is taken from output the ≠ce will be zero.
-
Briefly describe with aid of a circuit diagram how you can eliminate the ≠cial
input offset in an Op-am with the aid of a potentiometer.
-
Draw a circuit diagram to show how null balance adjustment of the 741 Op-am can
be done by mean of a potentiometer.
1. Inverting amplifier
Vo = Av = -
𝑅𝑓
𝑅𝑖
𝑉𝑜
𝑉𝑖
Vi
=
𝑅𝑓
𝑅𝑖
2 Non-inverting amplifier
Vo = (1 +
Av = -
𝑉𝑜
𝑉𝑖
𝑅𝑓
𝑅𝑖
)Vi
= (1 +
𝑅𝑓
𝑅𝑖
)
Its Function:
3Voltage Follower
This amplifier produce a
gain of unity (Gain = 1)
Vo = Vi
.Acts as buffer between
two stages
. Used for Impedance
matching
. Power amplifier
4 Summing Amplifier
𝑉2
𝑉3
Vo = - Rf( 𝑉1
+ + 𝑅2)
𝑅1 𝑅2
5Integrator
Vo(tb) =1 𝑡𝑏
∫ ±𝑉𝑖
𝑅𝐶 𝑡𝑎
(𝑡𝑏) +
𝑉𝐶(𝑡𝑎)
6 Practical integrator
-
-
Without Rs at low
frequency, the gain
becomes so large,
With Rs at low
frequency Xc≫Rs,
Xc can be neglected
the circuit then
operate as an
-
𝑅1 . 𝑅𝑠
At high frequency
Rs ≫ 𝑋𝐶, Rs can be
neglected and the
circuit operates as
an integrator.
𝑅𝑠
Av = gain = - 𝑅𝑖
1
R2 = 𝑅1+𝑅𝑠 and fs = 2𝜋 𝑅𝑠 .𝐶
inverter.
7 Differentiator
8 Practical Differentiator
In Practical integrator:
In Practical differentiator:
- Rs = shunt
- Rs = series
resistor limits the
resistor limits
low frequency of
the frequency
the circuit and
gain of the
- R2 reduces the
circuit.
DC offset.
-
-
Without Rs at
high frequency,
the gain becomes
so large,
With Rs at high
frequency Xc ≫ 0
the circuit then
operates as an
inverter.
-
At Low frequency
𝑋𝐶 ≫ Rs, Rs can be
neglected and the
circuit operates as a
differentiator.
Av = high frequency gain =
𝑅𝑠
𝑅𝑖
Period: t = Rf . C
9 Comparator
Vo = A (Vr – Vi)
10 Differential Amplifier
𝑅𝑓
Vo = 𝑅1
( V2 – V1)
If R1= R2 = R3 = Rf…..Vo =
V2 – V1
EXERCISE4
Q7, 8, 9, 10, 11, 13
FILTERS
Filter is an appliance which passes electrical current at certain frequency while preventing other frequency
from passing.
ACTIVE FILTER
This is the network that consists of resistors and capacitors around an operational amplifier or a transistor.
Advantages:
- Cost less,
- Can be tuned-in easily without a drastic change in the design,
- Isolate the load from the source,
- No loss because signals are amplified.
Disadvantages: A circuit is required for the supply voltage.
TYPE OF ACTIVE FILTERS
1 Low pass filter Circuit
Gain/frequency response
fo =
1
2𝜋 . 𝑅𝐶
2 High pass filter Circuit
Gain/frequency response
fo =
1
2𝜋 . 𝑅𝐶
3 Band pass filter circuit
Gain/frequency response
fo =
1
2𝜋 . 𝑅𝐶
T-Type RC circuits of low and high pass
filter in parallel are connected in the
feedback line.
4 Band reject filter Circuit
Gain/frequency response
fo =
1
T-Type RC circuits of low and high pass
filter in parallel are connected in the
input line.
2𝜋 . 𝑅𝐶
CHAP 5
HANDLING OF COMPEMENTARY METAL OXYD SEMICONDUCTOR
INTEGRATED CIRCUIT (C – MOS IC)
C-MOS Is The Combination Of N-MOS And P-MOST ogether.
It operates as an inverter; when Q1is ON, Q2 is OFF and vice versa.
Its advantages: Low power consumption and High noise immunity
Its disadvantage is that it is very slow.
C- MOS IC
ZENER DIODE GATE PROTECTION
Oxide layer form a capacitance, the human body
could develops a static charge between 4KV and
15 KVwhile 150 V is enough to destroy a
CMOS – IC when zener diode protection is not used.
Zener Diode gate protection
Example of clamping
CMOS WITHOUT ZENER DIODE GATE PROTECTION
The voltage between gate and source can
not exceed + 15,7 V or – 15,7 V,
this is the volt drop across zener diode
when the core is forward bias and other reverse bias
The following Procedures need to be considered when handling CMOS - IC
Connect all test equipment to earth,
- Switch off the power before working on the circuit,
All unused input must be connected to Vdd or VSS,
- Use earth wristband specially,
Store all unused unit in conductive sponge, conductive container,
- Use earthed point soldering iron as soon as you
Work on the circuit.
-
-
CHAP 6
SINGLE PHASE CONTROL SYSTEM
Summary showing the reaction of diode and thyristor when an AC signal is applied on:
A) Diode
The current flows in Forward Bias
B) SCR
Current flows from + to – but there is a
control gate.
Like the diode, it converts AC signal to pulsating DC except it depends on trigger pulse(
Gate current)
C) TRIAC
Conducts in both direction when SCR is triggered.
D) QUADRAC
BASIC AC AND DC CONTROL CIRCUIT
A) HALF WAVE DC CONTROL
The current flows through the load during positive half cycle when SCR is FB and once the
SCR is triggered.
- Purpose of this circuit is to control the output voltage.
- Advantage of this circuit is that you pay fro energy only when the load is applied
(connected)
B) FULL WAVE DC CONTROLWITH TWO SCR
C) FULL WAVE AC CONTROL WITH TWO SCR
D) FULL WAVE AC CONTROL WITH ONE SCR AND A BRIDGE
TRIGGER CIRCUIT
RC circuit supplies a phase displacement of ac signal,
This displacement form:
Phase angle = trigger angle
Conduction angle
- TRIGGER ANGLE (𝜶): is a period during which a thyristor
is OFF but forward bias because the capacitor is
SCR trigger circuit and waveform
not yet fully charged to a required gate voltage.
- CONDUCTION ANGLE (𝜷):is a period during which
VC is fully charged, the thyristor is ON and forward bias.
The bias drop until the current reaches the holding current.
- The Potentiometer is used to adjust the phase angle to a required value once in use.
OPEN LOOP AND CLOSED LOOP SYSTEM
OPEN LOOP SYSTEM
OPEN LOOP SYSTEM
is a System with no feedback network, the operator is needed to perform any
necessary adjustment on the input until the load reaches its required value.
APPLICATION OF OPEN LOOP SYSTEM
-
It can be used for speed control of low power series motor and universal motor,
Half wave phase control circuit for DC load (drawing page 6.10, Fig.6.6)
Full wave phase control circuit for DC load (drawing page 6.11, Fig. 6.7)
It can be used for light dimmer.(Page 6.11, Fig. 6.8)
CLOSED LOOP SYSTEM
CLOSED LOOP SYSTEM
is a system that uses a feedback network which feedback a sample of the load
condition to the comparator.
The comparator compares Vr and Vf:
- If Vr = Vf the load is at a required value,
- If Vr ≠ Vf due to the load deviation, this difference is an error signal and is used by the control
circuit to bring the load back to normal.
- When the load is back to normal, the error signal decreases until becomes zero.
- The load will be kept constant without human operation.
Examples of closed loop system
-
Self balancing Potentiometer,
- Synchronous motor system,
- Tacho-generator feedback.
CHAP 7 :TESTING EQUIPMENTS
Analogue signal
Digital signal
Difference between Analogue and Digital signal
is a signal with infinite number of value between two ultimate values.
e.g. 0,1,2,3,4,5,6,7,8,9, Sine wave, Radio wave
is a signal that has only two ultimate values. High and Low or 1 and 0
ANALOGUE TO DIGITALN CONVERTER is a
voltmeter that measures an analogue signal (voltage or current) and convert it to
digital signal.
A) STAIR CASE VOLTMETER
Operation:
At start the control circuit Reset the counter to zero. Clock pulse generator send pulse which are counted by the
counter. These counts are converted to a stair case type Vr. When Vr = Vu, the clock pulse stop, counter stop counting
and the digital output Vu appears on the display unit. This process is repeated several time very fast as it can not be
seen by human eye.
B) DUAL SLOPE VOLTMETER
This A/D converter generates an integrating waveform having two slopes and two time interval proportional to Vu.
During T1 Vu is integrated, the capacitor charge, the gate is opened and clock pulses pass to the counter, as soon
as the counter has counted through its full capacity, this duration can be determined.
During T2a transfer pulse is transferred to the flip-flop to switch Vr, the capacitor discharges to zero, the
comparator close the gate, the counter stop counting and Vu appears on the display unit.
C) VOLTAGE TO FREQUENCY CONVERTER
It operates as dual slope except that its input voltage is converted to frequency.
-
SUCCESSIVE APPROXIMATION A/D CONVERTER
The code register is converted to analogue Vr.
Comparator enable the logic programmer to program the register with certain code.
The large counting binary number is set to 1 while the other remains to zero.
Vr is compared to Vu.
If Vr < Vu : 1 is kept
If Vr > Vu I is reset to zero
If Vr = Vu The process stop and Vu appears on the display unit.
FREQUNCY COUNTER
convert analogue input signal(Vu) to digital signal.
Time base generator produce pulse which open the gate for a fixed time.
The counter counts the number of pulse allowed through the gate and stop counting as soon as the gate closes. These counts
values appear on the display unit.
Schmitt trigger
CHAP8 OSCILLATORS
OSCILLATOR is a circuit that generates waveforms and repeat it to a fixed rate.
Or Unstable amplifier using positive feedback.
Waveforms generated are: Sine wave, Square wave, Saw-tooth wave.
Two types of oscillators:
- High frequency oscillator
- Low frequency oscillator
High frequency oscillator is used to develop and carry wave for telecommunication.
Low frequency oscillator is used for testing of electronic equipment in some digital circuit.
BASIC TUNED CIRCUIT
-
-
When the switch S is on point A, the capacitor charges.
When the switch is on point B, the capacitor discharges through the coil setting up a
magnetic field.
When the current drop to zero, the magnetic field collapses, induces an EMF in the coil
which then charge the capacitor in opposite polarity.
The capacitor discharges through the coil again but in opposite direction. The above
process generates an oscillation. Due to the Energy loss the oscillation will have a dampin
effect. To eliminate this damping effect:
The energy must be in phase and at the same frequency therefore positive feedback is
required.
Resonance frequency and feedback signal must be in phase,
Power supplies must be used.
L,C and Centre tap form a resonant circuit.
Centre tap cause a mutual induction,
L1 controls the amplitude of oscillation,
L2 controls the frequency of oscillation.
Capacitor block DC from tank circuit.
This frequency must be stable, and factors that are causing instability are:
Variation of transistor parameter,
Change in load condition,
-
Unstable supply voltage.
HARTLEY OSCILLATORS
-
fo =
1
2𝜋 𝑥 √𝐿𝑇 .𝐶
COLLPITS OSCILLATORS
-
This circuit is similar to Hartley except that it has a single inductor
and two capacitors.
C1 controls the amplitude of oscillation,
C2 controls the frequency of oscillation.
1
𝐶1 𝑥 𝐶2
fo = 2𝜋 𝑥
CT = 𝐶1+𝐶2
.𝐿
√𝐶𝑇
-
RC PHASE SHIFT OSCILLATORS
-
-
.
This oscillator consists of three RC sections, each section causes a
phase shift of 60 o making a total of 180 o.
As the transistor supply itself a phase shift of 180 o, the total phase
shift will be 360 o, the input and output are in phase.
Hfe must be more than 60 to keep the oscillation.fo =
1
2𝜋 𝑥 𝑅𝐶√6
CRISTAL CONTROLLED OSCILLATOR
-
This oscillator uses Quartz crystal which is the most efficient method to stabilize the frequency of
oscillator.
The most important property of this crystal is the piezo-electric effect.
-
Piezo-electric effect: as soon as mechanical stress is applied upon two opposite surfaces of a
-
crystal a potential difference is developed across this surface.
Action of a crystal can be expressed by an equivalent electrical circuit which resonates:
This crystal has two resonant circuits:
Series Resonant takes place at low frequency when XL and XC are equa
Z= R, impedance has the minimum value.
Parallel Resonance takes place at high frequency f2 when the reactan
of RLC circuit is equal to reactance of parallel capacitor Cm (melting capac
The total impedance has the maximum value.
UNIJUNCTION TRANSISTOR OSCILLATOR
-
This one is a relaxation oscillator because the relaxation
(discharge) takes place after a period of activity.
R and C determine the rate of oscillation.
It is used in digital circuit to trigger a SCR.
-
Its Oscillation frequency is calculated by: fo =
-
MULTI-VIBRATORS
1,5
𝑅𝐶
Multi-vibrator is an electronic circuit which uses positive feedback. It is a relaxation oscillator because it develops
a non-sine waveform with the aid of resistor and capacitor.
Multi-vibrator can be self generating or use trigger pulse.
1) A-Stable Multi-vibrator
A-stable = unstable multi-vibrator : it has two semi stable conditions.
This circuit does not need trigger circuit therefore called free running oscillator.
It generates square pulses and it has two output which are in anti-phaseWhen the voltage is applied, both transistors tend to switch ON, however it is impossible
both of them to be on at the same time due to the feedback capacitor.
- When T1 switches ON ,T2 switches Off, when T2 switches ON T1 switches Off.
- The period during which each transistor is ON is determine by the time consta
- T1 = R2C1 and T2 = R1C2,
1
-
The oscillation frequency = fo =
-
(Exercise 8 : Question 13.1 & 13.2 A - stable Multi-vibrator)
For approximation values the time taken by the capacitor to charge or
𝑇1+𝑇2
discharge up to 50 % of it maximum value is 70 % of the total time therefore
0,7 R2C1 and T2 = 0,7 R1C2 so
-
fo =
1
1,4 𝑅𝐶
In order to adjust T1 and T2 , R1 and R2 can be replaced by a potentiometer. FIG K
The change of time constant R1C2 and R2C1 causes the rate of collector voltage to
increase, as result no increase in time of square wave. The method used to
Improve the increase in time is called Rozner modification,
- Diode in RB isolate the charge across the capacitor from collector and
- The capacitor charge through R4 not R3
A-stable operation of 555-timer
When the voltage is applied to Vcc the capacitor charges through RA and RB until reaches
2/3 of VCc. Vo is low, T switches ON. When The capacitor discharges Vc drops to 1/3 Vcc ,
Vo is high to complete the full cycle and the circuit oscillate as the process repeat itself.
tH = 0,693(RA + RB)C
tL = 0,693RBC
tT = TLow + THigh
fo =
1,443
(𝑅𝐴+2𝑅𝐵)𝐶
(Exercise 8 : Question 20 A - stable oscillator )
Mono-stable multi-vibrator
This circuit has one stable and one temporary stable state that is triggered by an
appropriate circuit.
The period during which one state is temporary stable can be calculate by T = 1.1 RC
Bi-stable multi-vibrator
This circuit has two stable conditions with two separated inputs called set and rese
- When the voltage is applied any of two transistors switches ON.
- If T1 is ON, T2 is Off, Vc1 becomes low, Vc2 high,
- A positive trigger pulse at reset inputmust be applied to switch T2 ON, and V
becomes low, T1 switches Off.
- To switch T1 On again a positive trigger pulse must be applied on the set inpu
Application:
-
As flip-flop, it is used in a Digital circuit to store data.
It is used in Double slope A/D convrter
It is in 555-Precision timer.
SCHMITT TRIGGER
Schmitt trigger is a Bistable Used for:
- Pulse tracing
- View level discrimination,
- To form square wave pulses from other wace forms.
-
555 PRECISION TIMER
Input and output waveform of Schmitt trigger
-
Pin 1: connect the circuit to the ground,
Pin 2: Send trigger pulse to bottom comparator,
Pin 3: Provide output pulse to other circuit,
Pin 4: Supply circuit to the circuit both with Pin 8,
Pin 5: Connect with the capacitor to the ground to reduce the noise (not in use)
Pin 6: Threshold voltage to the upper comparator,
Pin 7: Discharge the external capacitor,
Pin8: Supply positive voltage to the circuit both with pin 4. This voltage is 5 V to18 V.
-
555 TIMER AS DC-DC CONVERTER
-
Used for low power application,
The output voltage will be the double of the input.
CHAP 9 TRANSDUCERS
Def . Transducer is a device that converts one form of energy (input) to another form of energy (output).
Input energy can be:
- Electrical,
- Mechanical,
- Solar,
- Chemical,
- Sound,
- Optical
There is two category of transducer: Passive and Active Transducer
a)
b)
Passive transducers need
electrical power to operate.
E.g. Resistive,
Inductive,
Capacitive,
Hall Effect
Photo emissive cell.
Active Transducers : Called
again self-generating Transducers because they do not need
electrical power to operate.
…..These transducers generate an EMF as soon as they are stimulate by some physical form of energy.
e.g. Photo voltaic cell
Thermo coupler,
Voltage and current transducer.
According to their operating principles , they are :
1)
RESISTIVE
2) INDUCTIVE
3) VOLTAGE AND CURRENT TRANSDUCER
Strain Gauge
LVDT
Potentiometer
Tacho-generator,
Thermistor
Synchronous motor
Hall Effect Transducer, Photo electric
Photo voltaic cell, Photo diode, LASCR
Photo transistor, Infra-red detector, optic fiber.
Question to be asked before selecting a transducer:
What is the physical quantity that should be measured: Type and Range of measurement?
Which converting principle is the most suitable: The input—output characteristic must be adapted to the
rest of the system.
Accuracy required: factors influencing the accuracy are:
-
-
Transducer Parameter,
- Physical condition,
- Transducer Ratio,
Environmental condition,
Adaptability of equipment.
1)
RESISTVE TRANSDUCER
-
Although the resistance does not require electrical power to determine it resistance, The resistive
transducer need electrical power to convert the input form of energy to electrical current or voltage.
A resistive transducer uses the change in resistance to convert a form of energy to another.
e.g. displacement, heat, light, power humility convert to current or voltage.
1.1)
STAIN GAUGE
A strain gauge is a passive transducer which converts a mechanical displacement into change in resistance
STRAIN is a force that pushes, pulls or stretches something in such a way that it may damage it.
A GAUGE is a device that measures an amount or quantity and display the measurement.
……The change in resistance (∆𝑅 ) is proportional to the applied voltage and measured by a Wheatstone
bridge:
Gauge factor (K) is the sensitivity of the strain gauge which is the unit change in resistance
K =
∆𝑅/𝑅
∆𝑙/𝑙
per unit change in
length.
∆𝑅 : change in Gauge resistance and ∆𝑙: change in length.
Stress
is a strong physical pressure applied to an object.
Stress
=
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
tress can cause stain
Stress
=
∆𝑙
𝑙
= 𝜎
so
K =
∆𝑅/𝑅
𝜎
and
𝑙
R = 𝜌 𝐴 where
A =
𝜋𝑑2
4
Hooke’ law: This law determine the relationship between strain and stress in terms of modulus of
elasticity of material under stress.
𝜎 =
𝑆
𝐸
where
S …… Stain
in Kg/cm2
and
E …… young’s modulus in Kg/cm
2
𝝈
stress
(no unit)
Example :
A distribution gauge with gauge factor of 3 is pasted to a steel beam under strain of 1600 Kg/cm 2.
The modulus of elasticity for steel is approximately 2,1 Kg/cm2. Calculate the percentage change in
resistance of the distortion gauge because of strain.
1.2 ) POTENTIOMETER
1.2.1
LINEAR VARIABLE POTENTIOMETER
This transducer is used to measure the linear displacement and can be connected in a Wheatstone
bridge as shown:
This bridge is in balance when the slider moves to A or B, V1 and V
differ, the output is the amount of displacement and the polarity
indicate the direction of displacement.
1.2.2
ROTATING POTENTIOMETER
This transducer is used to convert the angular position of an axle or any other object to a voltage.
The connection in the Wheatstone bridge is the same as linear one.
A single phase potentiometer can rotate almost ≅ 300 o of angle and has a Low resolution.
RESOLUTION of a potentiometer is the relation of the voltage drop across two adjacent turns to the
total volt drop.
Resolution
=
𝑽𝒐𝒍𝒕 𝒅𝒓𝒐𝒑 𝒂𝒄𝒓𝒐𝒔𝒔 𝒕𝒘𝒐 𝒂𝒅𝒋𝒖𝒄𝒆𝒏𝒕 𝒕𝒖𝒓𝒏𝒔
𝑻𝒐𝒕𝒂𝒍 𝑽𝒐𝒍𝒕 𝒅𝒓𝒐𝒑
=
𝟏
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑻𝒖𝒓𝒏𝒔
Example: Calculate the %tage Resolution of a potentiometer with 1000 Turns
To get a better resolution, a potentiometer with more revolution to a gear box must be used:
1.3
-
Pf , feedback potentiometer sends a feedback signal to the comparato
after being mechanically adjusted by an axle of the canon.
The comparator compares the feedback signal with reference signal w
is adjusted manually. Their difference is an error signal which is applied
the motor. The motor rotates, the gear box as well as canon rotates th
axle of Pf in return.
As soon as the canon reaches its required position, the error signal is z
and the motor stop.
THERMISTOR
This transducer measure and control the temperature (𝑻 𝒐 ).
It has a large resistance change per degree change in temperature.
A temperature control circuit with the aid of thermistor using AC is shown:
The bridge is balance as soon as R1 = R2 = R3 = Rt
If the temperature drops, Rt increases and the bridge unbalance.
Rt and 𝑻 𝒐 are inversely proportional, therefore the thermistor has a NTC.
𝜷
Rt = A𝒆𝑻 A and 𝛽 are constant , T is the temperature in degree Kevin (oC + 273) 𝒐 K
-
Example: Exercises 9, Question 9 and 10 ( fig. 9.13)
2) INDUCTIVE TRANSDUCERS
This transducer uses the change of inductive reactance to convert other form of energy to electrical signal.
2.1 LINEAR VARIABLE DIFFERENTIAL TRANSFORMER (LVDT)
OPERATION: Initially the core is in
the center, the two secondary voltages are equal
therefore Vo = o. As soon as external voltage is applied the core moves to A or B.
if it ‘s A , more magnetic field is developed that side than B side, VA > VB so Vo
becomes the difference between VA and VB
Application: The output of LVDT serve as a closed loop servo-system
2.2 TACHO-GENERATOR
The Tacho-generators are used in closed loop motor control system.
Operation: feedback voltage (Vf) invoked by the anchor connected to the axle of the
motor is proportional to the speed of the motor.
This Vf is compared with reference voltage (Vr) and give an error signal (Ver) at output
of comparator. This Ver is used to adapt the speed of the motor again.
2.3 SYNCHRONOUS MOTOR
This transducer is a rotating inductive one used in pair to synchronous a transmitter
and a receiver. They have three stators winding wrapped electrically 120o and a rotor
winding connected to two collector rings.
Operation: Initially the Tx (transmitter) and Rx (receiver) are in balance. As soon as the
rotors of Rx produces an error signal to turn the motor. The antenna also turns the roto
of Rx at the same angle so that the error signal becomes zero.
3) VOLTAGE AND CURRENT TRANSDUCERS
These transducers convert a form of energy to current or voltage when they are stimulated by some
physical form of energy. They are :
Passive: Hall effect and photo emitting cell
Active (self generating): photo voltaic cell, thermo coupler.
3.1 HALL EFFECT TRANSDUCERS
hall effect displacement transducers
-
This transducer is a recording head which serves to measur
magnetic field and convert that measurement into voltage
Hall Voltage.
-
3
It measures the power in DC and AC circuit as well. Vhall is
proportional to the magnetic field
PHTO-ELECRIC TRANSDUCER
These transducers convert light energy to electrical energy.
3.1 PHOTO – VOLTAIC CELL (PVC)
This transducer is a solar silicon cell which converts the energy of the sun
to electric power,
It operates in the principle of current and voltage principle.
It has a very thin P-type silicon crystal, thin layer n-type diffusion layer.
The light shines through the P-type and reaches PN-junction because P-type is
so thin. An EMF is developed across this junction as soon as the light shines on
Application:
3.2 PHOTO DIODE (PD)
-
Satellite generator,
Solar heat devices,
Battery less calculator.
PD is a semiconductor PN-junction component which conducts in reverse bias region
once light fall on it.
In this reverse bias region the inverse current increases with the increase in light,
and this is an advantage to use this region. This current must be amplified first
before it is used as it is too small for direct application.
Application:
- Infra-red detector
- devises using optique fibre, etc.
3.3 PHOTO TRANSISTOR (PT)
-
Application:
3.4 LIGHT ACTIVATED SCR (LASCR)
-
3.5 LIGHT DEPENDANT RESISTOR (LDR)
(Photo conductive cell)
Drawing of Schmitt trigger circuit which include a photo transistor
The relay and the transistor T2 are in series.
D2 protects the transistor T2
T2 will switch ON as soon as the light intensity drops, the relay will be
ON as well.
The potentiometer adjusts the bias of T1 to compensate for temperature
surrounding light.
LASCR is used in a circuit to convert AC to pulsating DC,
To eliminate the half wave vibration, the capacitor is connected in parallel
with the coil of the relay
Used in AND function where two LASCR are connected in series.
Both two LASCR must receive the light to pass the current to the load.
OR function where two LASCR are connected in parallel.
Only one LASCR needs to receive light to pass the current to the load
It resistance decrease when the light intensity increases
(normally from 100 kΩ to 100 Ω).
Operation:
R1 and R2 form a voltage divider to supply Vr to the comparator,
P adjusts Vin for different light intensity.
When the light intensity drops, R of LDR increases and Vin increases.
When Vi < Vr, Vo is Low , T1 switches ON as Vbe increases across R3,
The current flows, the relay is energized and closed the contact.
By changing LDR and P the circuit will react reversely .
Application:
LDR is used in precision light switch to control the street light
4.6 INFRA-RED DETECTOR
Infra-red beam is an invisible beam which can be invoked by an infra-red emitting diode.
Operation:
C1 passes pulses and block the slow variation of surrounding light.
C2 faces the pulse to a smooth DC keeping T1 ON and letting the current
to flow through the relay. Photo diode conveys the pulse to the amplifier.
As soon as emission of an object or person is interrupting, no signal is sent
to the transistor therefore switches OFF.
The relay switches back to it normal condition so that the contact short circuited
and activates the Alarm.
Application:
-
Alarm system, Doorbell, Automatic door opener,
-
Light switch, Detection of human movement
4.7 OPTIC FIBER
Optic fiber is
a flexible, transparent fiber made of glass (silica) or plastic slightly thicker than human
hair used to conduct light between two ends of the fiber most specially in communication system
and medical camera.
IN MEDECINE
If a fiber is placed on the one end, it can be seen directly on the other end. That is why is suitable
for medical camera with which a specialist can examine a human stomach.
IN COMMUNICATION
A Transmitter and a Receiver are connected to both side of optic fiber.
The transmitter sends light pulses and the receiver observes and invokes these pulses
with the aid of a photo electric transducer.
INDUSTRIAL ELECTRONICS N5
nnnnnvvvvnnN5N5
CHAP 1
EXERCISES 1 Page 1.33 To 1.38
QUESTION 11
VT = 200 V
C = 10− 4 f (100 µf)
L = 0, 15 H
f= 50 Hz
R = 10 Ω
1
1
XC = 2πfC = 2π .50.0,0001 = 31.831 Ω
XL = 2πfL = 2π. 50.0.15 = 47,12 Ω
(a) Total impedance
ZT = R + j (XL – XC)
ZT = 10 + j15.3 Ω
in Pol: 18, 27 [56.8o Ω
(b) Total current
𝑉
250
IT = 𝑍𝑇 = 𝟏𝟖,𝟐𝟕 [56,8
𝑇
= 13, 68 [- 56,8o A
(c) Q- factor
Q=
1
R
√
𝐿
1
=
C
10
√
0,15
10− 4
= 3, 87
QUESTION 12
VT = 10 V
R = 12 Ω
L = 150 µH
f= 50 KHz
(a) The capacitor value to let the circuit resonate at 50 Khz
1
fr =
2π√𝐿 𝑥 𝐶
1
50 x 10 =
2π√𝟏𝟓𝟎 𝐱 10−6 𝑥 𝐶
= 67,55 nf
(b) The total current through the circuit:
𝑉
10
IT = 𝑇 =
0,83 A
3
𝑍𝑇
𝟏𝟐
(at resonance R = Z)
(C – d) Total Reactance (XL, XC) and Volt drop across R and X
XL = 2π x fL (2π x 𝟓𝟎 𝐱 103 𝑥 𝟏𝟓𝟎 𝐱 10−6 )
XL = XC = 15𝜋 Ω
VL = 0,83 x 47,12 = 39,3 90 o V
VC = 0, 83 x 47,12 = 39,3 – 90 o V
VR = 0, 83 x 12 = 10 V
(e) ZT = R = 12 Ω
(f) Q – factor
Q=
1
R
√
𝐿
C
=
Q=
1
12
𝑋𝐿
R
√
=
𝟏𝟓𝟎 𝐱 10−6
67,55 𝑥 10− 9
47,12
12
= 3, 93
= 3,93
QUESTION 13
R = 5 , XL = j40, Xc = - j80, Vt = 30 V
ZT = R + j (XL – XC)
ZT = 15 + j (40 – 80) Ω
= 5 – j 40 Ω
in Pol: 40,3 [- 82,875 o Ω
𝑉
IT = 𝑍𝑇 =
𝑇
30 [ 𝟎
𝟒𝟎,𝟑 [− 𝟖𝟐 ,𝟖𝟕𝟓
= 0,744 [𝟖𝟐 , 𝟖𝟕𝟓 A
VR = IT x R = 0,744 [𝟖𝟐 , 𝟖𝟕𝟓 x 5 [ 𝟎 = 3,72 [𝟖𝟐 , 𝟖𝟕𝟓 V
VL = IT x XL = 0,744 [𝟖𝟐 , 𝟖𝟕𝟓 x 40 [𝟗𝟎
= 29, 76 [𝟏𝟕𝟐 , 𝟖𝟕𝟓 𝐕
VC = IT x XC
= 0,744 [𝟖𝟐 , 𝟖𝟕𝟓 x 80 [−𝟗𝟎
= 59,52 [−𝟕, 𝟏𝟐𝟓 𝐕
c ) Phasor diagram of impedance
QUESTION 14
1
fr = 2π√𝐿 𝑥 𝐶
3 x 10 3 =
1
2π√𝟓 𝐱 10−3 𝑥 𝐶
C= 56 pf
QUESTION 15
R =20 L = j10 mH, C = 10 µf,
VT = 250 V, f = 50 Hz
XL = 2πfL = 2π x 50 x 0,01 = 3,142 Ω
ZT = R + j (XL – XC)
ZT = 20 + j (3,142 – 318, 31) Ω
= 20 – j40 (Pol: 315, 8 [- 86 ,37 o Ω)
𝑉
250 [ 𝟎
IT = 𝑍𝑇 = 𝟑𝟏𝟓,𝟖 [− 𝟖𝟔 ,𝟑𝟕 = 0,792 [𝟖𝟔, 𝟑𝟕 A
𝑇
V R = IT x R
= 0,792 [𝟖𝟔 , 𝟑𝟕 x 20 [ 𝟎 = 15, 833 [𝟖𝟔 , 𝟑𝟕 V
VL = IT x XL = 0,792 [−𝟖𝟐 , 𝟖𝟕𝟓 x 3,142 [𝟗𝟎 = 2, 4 [𝟏𝟕𝟔 , = 𝟑𝟕 𝐕
VC = IT x XC = 0,792 [−𝟖𝟐 , 𝟖𝟕𝟓 x 318, 31 [−𝟗𝟎 = 251, 987 [−𝟑, 𝟔𝟑 𝐕
The capacitor value at which the circuit will resonate
1
fr = 2π√𝐿 𝑥 𝐶
50 =
1
2π√𝟏𝟎 𝐱 10−3 𝑥 𝐶
= 1013,2 µf
(c) Phasor diagram of impedance
QUESTION 16
𝑉𝑇
110
XC = 2π = 0,25 = 440 Ω
XC =
440 =
1
1
2πfC
= 31.831 Ω
C = 6, 03 µf
2π x 60.x C`
QUESTION 17
XL = 2pi x f L = 2pi x 50 x 1,2 = 376, 99
ZT = 15 + j 376,99 = 377, 29[ 𝟖𝟕, 𝟕
𝑉𝑇
250
IT = ZT = 377,29[ 𝟖𝟕,𝟕
= 0,663[ − 𝟖𝟕, 𝟕 A
QUESTION 18
XC =
XC =
1
1
2πfC
2π x 2,5 𝑥 106 x`10− 4
= 636, 62 µΩ
ZT = 1200 + j 636, 62 𝑥 10− 6 = 1200 [−𝟑
𝑉𝑇
100
IT = ZT = 1200 [−𝟑 = 83, 3[ 𝟑 mA
VR = IT x R
= 83, 3[ 𝟑 mA x 1, 2 = 100 V
VL = IT x XL
= 83, 3[ 𝟑 mA x 636, 62 µΩ = 53 µV
OFF-BOOK QUESTIONS
QUESTION 1
Bandwidth = 2 Khz,
fr = 10 Khz, Calculate L and Vc
𝑓
10 000
Q = 𝑓 𝑟 = 2000 = 5
Q=
𝑋𝐿
𝐻−𝑓𝐿
, XL = Q x R =10[90
XL = 2 𝜋fL
𝑋𝐿
10
L = 2 𝜋f = 2 𝜋.1000 = 0, 159 mH
IT =
𝑉𝑇
𝑅
20 30 o
(R = Z) = 2
= 10 [30
VC = IT x XC
= 10 [30 x 10[−90 = 100 [−60
VL = IT x XL
= 10[30 x 10[90 = 100 [120
R
QUESTION 2
IT = 2 [42 Ao
L= 20 mA, C = 120 µ f., R = 5 Ω
1
1
XC =
=
= 26, 53 Ω
2πfC
2π .50.0,00012
XL = 2πfL = 2π x 50 x 0, 02 = 47,12 Ω
ZT = 5 + j (6,28 – 26,53 ) = 20, 85 [− 76,86
VT = IT x ZT = 2 [42 x 20, 85 [− 76,86
= 41, 7 [− 34,86
QUESTION 3
R= 4 Ω , XL = 20 [80 Ω, L = 18 H
𝑓𝑟
10 000
Q =
=
=5
𝑓𝐻− 𝑓𝐿
2000
BW = 𝑓𝐻− 𝑓𝐿
𝑓
𝑓
𝑓
BW = Q 𝑟 and fr = 𝐻 +2 𝐿
𝑋
𝑋
fr = 2𝜋𝐿𝐿
BW =
{
20
Q = 𝑅𝐿 = 5 = 5
XL = 2𝜋fL …..
20
= 2𝜋 𝑥 18.10−6 = 176, 84 Khz
𝑓𝑟
Q
= 176,84 Khz
=
5
= 35, 57 Khz
𝑓𝐻− 𝑓𝐿
𝑓
𝑓
176, 84 Khz = 𝐻−2 𝐿
{
BW = 𝑓𝐻− 𝑓𝐿
35, 57 Khz = 𝑓𝐻− 𝑓𝐿
353, 68 = 𝑓𝐻 + 𝑓𝐿
{
35,37 = 𝑓𝐻 − 𝑓𝐿
fr =
2
353, 68 − 𝑓𝐻 = 𝑓𝐿 (1)
− 35,37 + 𝑓𝐻 = 𝑓𝐿 (2)
(1) = (2)
353, 68 − 𝑓𝐻= − 35,37 + 𝑓𝐻
2 x fH = 353,68 + 35,37 = 389,05
fH = 194, 53 Khz
in (1) or (2) fL = - 35,37 + fH = -35,37 + 194, 53
= 159, 15 Khz
fL = 159, 15 Khz
QUESTION 4
VR
R =
6 25
=
= 2 ( R = Z)
3 25
BW = 𝑓𝐻− 𝑓𝐿
Vc
90 −60
XC = IT = 3 25 = 30 - 85
IT
(i)
Q=
Xc =
(ii)
𝑋𝐿
𝑅
1
=
30
2
= 15
2𝜋𝑓𝑐
1
1
fr = 2𝜋.𝑋𝐶 = 2𝜋.30.11.0−6 = 482,3 Hz
Bandwidth
𝑓
𝑓
𝑓
(iii)
BW = Q 𝑟 and fr = 𝐻 +2 𝐿
482,3
Bw =
= 241 15 Hz
15
Upper half power frequency and lower half power frequency
𝑓
BW = Q 𝑟
Q(fH − fL) = 𝑓𝑟
{
{
2 x fr = 𝑓𝐻+ 𝑓𝐿
2fr = 𝑓𝐻+ 𝑓𝐿
𝑓𝐻 = 498, 4 𝐾ℎ𝑧
15 (𝑓𝐻 − 𝑓𝐿) = 482,3
{
{
𝑓𝐿 = 466, 2 𝐾ℎ𝑧
996, 8 = 𝑓𝐻 + 𝑓𝐿
QUESTION 5
XL = XC
1
2𝜋𝑓𝑙 =
2𝜋𝑓𝑐
1
2𝜋 𝑥 1000 𝑥 0,0015 = 2𝜋 𝑥 1000 𝑥 𝐶
(2𝜋 𝑥 1000)2 𝑥 0,015 𝑥 𝐶 = 1
1
(i)
C = (2𝜋 𝑥 1000)2𝑥 0,015
= 1,688 µf
XL = 2𝜋𝑓𝐿
= 2𝜋𝑥 1000 𝑥 0,015 = 94,25 Ω
VT
20
IT = R = 10 = 2 A
(ii)
VL = IT x XL = 2 x 94, 25 [90 V
= 188,5[90 V
(iii)
VC = IT x XC = 188, 5[−90 V
XL 94,25
(iv)
𝑄 = R = 10 = 9,4
(v)
BW =
fr
Q
=
1000
9,4
= 106,4 Hz
QUESTION 6
VR
10 60
IR = IT = R = 2 0 = 5[60 A
XL = 2𝜋𝑓𝐿 = 2𝜋 𝑥 1000 𝑥 0,955 = 6[90 Ω
VL = IT x XL
= 5[60 x 6[90 = 30[150 V
V T = VR + VL + V C
V C = VT – V R – V L
VC = 15, 94[76,50 - 10[60 - 30[150
( 3,72 + j15,4996) – (5 + j8,66) – (-25, 98 + j15)
= 3, 72 + j15, 4996 – 5 – j8,66 + 25, 98 – j15
= 24,7 – j8,16 (Pol :26, 01[−18,28)
VC = XC x IT
Vc
26,01[−18,28
Xc =
=
IT
5[60
= 5,202 [−78,28 Ω
1
1
C = 2𝜋𝑓 𝑋𝑐 = 2𝜋 𝑥 1000 𝑥 5,202 = 30,6 µf
QUESTION 7: PARALLEL
Xc =
1
2𝜋𝑓𝐶
=
1
2𝜋106 𝑥 53,05 10− 9
= 3⌊− 90 Ω
IT = IR + IL +IC
IC = IT - IR - IL
= 7,68 ⌊17,42 - (6 ⌊− 60 + 5 ⌊− 150 )
= 7,33 + j2,3 – [(3 – j5,2) + (- 4,33 – j2,5)]
=7,33 + j2,3 – 3 + j5,2 + 4,33 + j2,5
= (7,33 - 3 + 4,33) + (j2,3 + j5,1) + j2,5)
IC = 8,66 + j9,9 A (rect.), pol: 13,23 ⌊49,1 A
Vc = Xc x Ic = 3 ⌊− 90 x 13,23 ⌊49,1
= 39,69⌊− 40,9 V
In parallel VT = VR = VL = VC = 39,69 ⌊− 40,9 V
𝑉
39,69 ⌊− 40,9
VL = XL x IL so XL = 𝐼 𝐿 = 5 ⌊− 150 = 7,94 ⌊109 Ω
𝐿
XL = 2𝜋fL so
L=
𝑋𝐿
2𝜋f
7,94
= 2𝜋 x 106 = 1, 26 µf
QUESTION 19 PARALEL
Z1 = 3 + j10 (10,4 ⌊73,2) and
Z2 = 5 - j6 (16,76 ⌊−72,6)
a) Series
b) Parallel
𝑍 𝑥𝑍
ZT = Z1 + Z2
ZT = 𝑍1+ 𝑍 2
1
= 3 +j10+5–j16
= 8 – j6
ZT =
2
10,4 ⌊73,2) 𝑥 16,76 ⌊−72,6
10 ⌊− 36,87
17,4 ⌊ 37,47 Ω
Pol: 10 ⌊− 36,87 Ω
QUESTION 20
IR =
𝑉𝑇
IL = 𝑋𝐿 =
𝑉𝑇
=
20
= 1A
𝑅
20
20 ⌊0
= 1,33⌊−
15 ⌊90
90 A
IT = IR + IL = 1 – j1,33 =1,667 ⌊− 53 A
OR
1
1
1
= −𝑗
𝑍𝑇
𝑅
𝑋𝐿
1
1
1
=
−𝑗
𝑍𝑇
20
15
1
= 0,05 − 𝑗 0,0666667
𝑍𝑇
1
= 0,083 ⌊− 53 A
𝑍𝑇
ZT = (0,083)−1 ⌊0 − (− 53) A
= 12 ⌊ 53 Ω
𝑉𝑇
20 ⌊0
IT = =
= 1,667 ⌊− 53 A
12 ⌊ 53
𝑍𝑇
QUESTION 21
Replace j 15 by – j15
IR =
𝑉𝑇
Ic 𝑋𝑐 =
𝑉𝑇
=
20
𝑅
20
20 ⌊0
=
15 ⌊− 90
= 1A
1,33⌊90 A
IT = IR + Ic = 1 + j1,33 =1,667 ⌊ 53 A
OR
1
1
1
= +𝑗
𝑍𝑇
𝑅
𝑋𝑐
1
1
1
=
+𝑗
𝑍𝑇
20
15
1
= 0,05 + 𝑗 0,0666667
𝑍𝑇
1
= 0,083 ⌊53 A
𝑍𝑇
ZT = (0,083)−1 ⌊0 − (53) A
= 12 ⌊53 Ω
𝑉𝑇
20 ⌊0
IT =
=
= 1,667 ⌊ 53 A
𝑍𝑇
12 ⌊− 53
QUESTION 22
XL = 6,28 Ω and Xc = 677, 2 Ω
IR =
𝑉𝑇
𝑅
=
220
200
= 1,1 A
220 ⌊0
𝑉𝑇
IL =
= 6,28 ⌊ 90 = 35, 03⌊−90 A
𝑋𝐿
𝑉𝑇
Ic =
=
𝑋𝑐
220 ⌊0
677,2 ⌊−90
= 0,325⌊90 A
IT = IR - j (IL - Ic)
= 1,1 - j (35,03 - 0.325)
or 1,1 + 35,03⌊−90+0,325⌊90
IT = 1,1 – j34,7 (Pol = 34, 7 ⌊− 88,18 A)
OR
1
1
1
1
= −( − )
𝑍𝑇
𝑅
𝑋𝐿 𝑋𝑐
1
1
1
1
= 200 − 𝑗( 6,28 - 677,2)
𝑍
𝑇
1
= 0,005 − 𝑗(0,159236 − 0,00014767)
𝑍𝑇
1
= 0,005 − 𝑗(0,15766
𝑍𝑇
1
= 0,15784 ⌊−88,18 A
𝑍𝑇
ZT = (0,15784)−1 ⌊0 − (−88,18) A
= 6,33 ⌊88,18 Ω
𝑉𝑇
220 ⌊0
IT =
=
= 34,7 ⌊−88,18 A
6,33 ⌊88,18
𝑍𝑇
QUESTION 23
a) Resonant frequency
1
= 2𝜋 0,02𝑥 0,00001 = 355,88 Hz
1
f = 2𝜋√𝐿 𝑥 𝐶
√
b) Dynamic impedance
𝐿
0,02
Zd = 𝑅 𝑥 𝐶 = 5 𝑥 0,00001 = 400 Ω
c) Q-factor
𝑋𝐿
2𝜋𝑓𝐿
2𝜋 𝑥 355,88 𝑥 0,002
Q= 𝑅 = 5 =
= 8,94
5
d) What are the values IT, I1 and I2,
If VT = 40 V
ZT = Zd + Zs = 400 + 5000 = 5,4 KΩ
𝑉𝑇
40
IT = 𝑍𝑇 = 4500 = 7,41 mA
V// = IT x Zd = 7,42 mA x 400 = 2,973 V
XL = 2𝜋𝑓𝐿 = 2𝜋 𝑥 355,88 𝑥 0.02 = 44,72
Z1 = 5 + j44,72 (Pol: 45⌊83,53
Z2 = 44,727 ⌊−90
I1 =
I2 =
𝑉//
𝑍1
𝑉//
𝑍2
=
=
2,973
45
2,973
44,73
= 66 mA
= 66 mA
QUESTION 24
a) The value of capacitor at resonance
At resonance XL = XC = 3,14 Ω
1
1
XC =
=
2πfC
C=
𝑓
2πf x Xc
=
2π .50.0,0001
1
2π x 500 x 31,4
= 10, 14 µf
b) Current through each branch
ZRL = 2 + j 3,4 = 31,46⌊86,35
𝑉
100
IRL = 𝑍 = 31,46⌊86,35 =3,18⌊−86,35 A
𝑅𝐿
XL = Xc = 31,4 so Zc = 31,4⌊−90 Ω
𝑉
100
Ic = =
= 3,18⌊90 A
Zc
31,4⌊−90
c) Dynamic impedance
𝐿
Zd =
𝑋𝐿
RC
31,4
L = 2πf = 2π x 500 = 9,995 . 10 - 3 H
9,995 x 10−3
𝐿
Zd = R.C = 2 x 10,14 x 10−6 = 493 Ω
d) Q - factor
31,4
Q = R = 2 = 15,7
𝑋𝐿
QUESTION 25
XL = 471,2 Ω and Xc = 14,5 Ω
IR =
IL =
Ic =
𝑉𝑇
𝑋𝐿
𝑉𝑇
𝑋𝑐
=
=
𝑉𝑇
=
250
𝑅
10
250 ⌊0
=
471,2 ⌊ 90
250 ⌊0
14,5 ⌊−90
= 25 A
0,5306⌊−90 A
= 17,27876⌊90 A
IT = IR - j (IL - Ic)
= 25 - j (0.5306 – 17,27876)
or 25 + 0,5306 ⌊−90 + 17,27876 ⌊90
IT = 25 + j16,71082 (Pol = 30,07 ⌊ 33,76 A)
OR
1
1
1
1
= −( − )
𝑍𝑇
𝑅
𝑋𝐿 𝑋𝑐
1
1
1
1
= 10 − 𝑗( 471,2 - 14.5)
𝑍𝑇
1
= 0,1 − 𝑗(0,0021222 − 0,0689655)
𝑍𝑇
1
= 0,1 + 𝑗0,06684332
𝑍𝑇
1
= 0,12028 ⌊33,76A
𝑍𝑇
ZT = (0,12028)−1 ⌊0 − (33,76) A
= 8,314 ⌊−33,76 Ω
𝑉𝑇
250 ⌊0
IT =
=
= 30,07⌊33,76 A
𝑍𝑇
8,314 ⌊−33,76
QUESTION 26
a) Total impedance of the circuit
Z1 = 4 + j3 – j6 = 4 – j3 (Pol: 5 ⌊−36,87 Ω
Z2 = 5 + j8 (Pol: 9,434 ⌊57,995 Ω
Z3 = 3 – j2 (Pol: 3,61 ⌊−33,69 Ω
Z1 + Z2 = 9 + j5 (Pol: 10,2956 ⌊29,055 Ω
Z// =
𝑍1 𝑥 𝑍2
𝑍1 +𝑍2
=
5 ⌊−36,87 𝑥 9,434 ⌊57,9
10,2956 ⌊29,055
= 4,5816 ⌊−8.025
Z// = 4,5816 ⌊−8.025 Ω
(Rect : 4,537 – j0,6396 Ω )
𝑍 𝑥𝑍
ZT = 1 2 + Z3
𝑍1 +𝑍2
= Z// + Z3
= 4,537 – j 0,6396 + 3 - j2
= 7,537 – j2,6396
= 7,986 ⌊−19,3 Ω
b) Total current of the circuit
𝑉𝑇
18⌊30
IT =
=
= 2,254 ⌊49,3 A
7,986 ⌊−19,3
𝑍𝑇
c) Current flowing through each branch
V// = Z// x IT
= 4,5816 ⌊−8.025 x 2,254 ⌊49,3
= 10,33⌊41,3 V
𝑉//
10,33⌊41,3
I1 =
=
= 2,066 ⌊78,17 A
I2 =
𝑉//
5 ⌊−36,87
10,33⌊41,3
9,434 ⌊57,995
𝑍1
𝑍2
=
= 1,096 ⌊−16,695 A
d) Volt drop across Z3 or Series voltage
Vs = VT – V//
= 18⌊30 - 10,33⌊41,3
= 15,588457 + j 9 - 7,76 – j 6.8178
= 7,828 + j2,1822
= 8,13⌊15,6 V
Or
Vs = IT x Z3
= 2,254 ⌊49,3 x 3,61 ⌊−33,69
= 8,13⌊15,6 V
Check your answer: VT = Vs + V//
= 8,13⌊15,6 + 10,33⌊41,3
= 7,8305+j2,1863 + 7.76056 + j6,8178
= 15,591 + j9 =18⌊30 V
CHAP 2
EXERCISES 2: POWER SUPPLIES (2.32 to 2.34)
QUESTION (note)
𝑉1 𝑁1 𝐼2 220 11
𝑉2
+
𝑁2
+
𝐼1
=
40
𝑉2
+
2
……. V2 = 40 V
Vrms = 2 = 20 V (Centre tap)
Vrms = 0,707 Vm
𝑉
Vm = 𝑟𝑚𝑠 = 28, 28 V
0,707
Vdc = 0,637 Vm = 18 V
Vr(rms) = 0,305 x Vm = 8,6 V
PIV = 2 Vm = 56,58 V
𝑉𝑟(𝑟𝑚𝑠)
8,63
r=
=
= 0,48
𝑉𝑑𝑐
18
% r = 0,48 x 100 = 48 %
𝑉𝑑𝑐
18
Idc = 𝑅𝐿
= 2000 = 9 mA
QUESTION 1
C1 = 470 µf, C2 = 47 µf, RL = 1,5 KΩ,
R = 10 Ω, Vdc = 1 V, Vr(rms) = 1,8 V
f = 50 Hz
1.1 voltage across the load
𝑅
/
𝑉𝑑𝑐 = 𝑅 +𝐿 𝑅 x Vdc
𝐿
1500
/
𝑉𝑑𝑐 = 1500+ 10
x 16 = 15,89 V
1.2 Ripple voltage at second capacitor C2:
𝑋𝐶
/
𝑉𝑟(𝑟𝑚𝑠)=
x Vr(rms),
√𝑅 2+𝑋𝑐2
XC =
1
4πfC
= 33,86 Ω
33,86
/
𝑉𝑟(𝑟𝑚𝑠) =
x 1,8 = 1,73 V.
√102+33,862
1.3 – 1.4 ripple factors
/
𝑟/=
𝑟=
𝑉𝑟(𝑟𝑚𝑠)
/
𝑉𝑑𝑐
𝑉𝑟(𝑟𝑚𝑠)
1,73
=
𝑉𝑑𝑐
15.89
1,8
=
16
= 10,8 %
= 11,25 %
QUESTION 2
1.1 voltage across the load
𝑅
/
𝑉𝑑𝑐 = 𝑅 +𝐿 𝑅 x Vdc
𝐿
2000
/
𝑉𝑑𝑐 =
2000+ 12
x 12 = 11,93 V
1.2 Ripple voltage at second capacitor C2:
1
/
𝑉𝑟(𝑟𝑚𝑠)=
x Vr(rms)
2
(4𝜋𝑓) 𝐿𝐶
1
/
𝑉𝑟(𝑟𝑚𝑠)=
x
(4𝜋𝑓)25.0,0001
1,5 = 0,076 V
1.2 Ripple factor across C2
/
𝑟/=
𝑉𝑟(𝑟𝑚𝑠)
/
𝑉𝑑𝑐
0,076
=
11.93
= 0,006
QUESTION 3
RL = 120
Ω, Vr(rms) = 0,2 V
Vdc = 15 V,f = 50 Hz
𝑉
Vr(rms) = 4√3 𝑥 𝑓𝑥𝑑𝑐𝐶 𝑥 𝑅
0,2 =
𝐿
15
4√3 𝑥 50𝑥 𝐶 𝑥120
C= 1804 µf
QUESTION 4
r = 3% = 0,03
RL = 500 Ω and f = 50 Hz
r = 4√3
C=
C=
1
𝑥 𝑓𝑥 𝐶 𝑥 𝑅𝐿
1
4√3 𝑥 𝑓𝑥 𝑟 𝑥 𝑅𝐿
1
4√3 𝑥 50𝑥 0,03 𝑥 500
C = 192 µf
QUESTION 5
Vm = 15 V, C = 470 µf, Idc = 10 mA
𝑉𝑑𝑐 = 𝑉𝑚 −
𝐼𝑑𝑐
( Vdc = Idc x RL)
4 𝑥 𝑓𝑥 𝐶
Idc x RL = 𝑉𝑚 −
RL =
15
=
0,01
−
𝑉𝑚
𝐼𝑑𝑐
𝐼𝑑𝑐
4 𝑥 𝑓𝑥 𝐶
−
1
2𝑓𝑐
1
2𝑥50𝑥0,00047
= 1,478 KΩ
QUESTION 6
After full wave rectification the frequency is divided by 2
f = 100/2 = 50 Hz:
𝐼𝑑𝑐
4 𝑥 𝑓𝑥 𝐶
0,01
𝑉𝑑𝑐 = 16 − 4 𝑥 50 𝑥 0,001 = 15,95 V
𝑉𝑑𝑐 = 𝑉𝑚 −
r=
𝐼𝑑𝑐
=
4√3 𝑥 𝑓𝑥 𝐶 𝑥 𝑉𝑑𝑐
0,01
4√3 𝑥 50𝑥 0,001 𝑥 15,95
= 0,00018
%r = 0,00018 x 100 = 0,18 %
QUESTION 7
Before full wave rectification the frequency remains the same:
𝑋𝐶
/
𝑉𝑟(𝑟𝑚𝑠)=
x Vr(rms)
√𝑅 2+𝑋𝑐2
/
2
2
( 𝑉𝑟(𝑟𝑚𝑠) 𝑥√𝑅 + 𝑋𝑐 )2 = [𝑋𝐶 𝑥Vr(rms)]2
2
/
𝑉𝑟(𝑟𝑚𝑠) 𝑥 (𝑅2 + 𝑋2𝑐 ) = 𝑋𝐶 2 x Vr(rms)2
2
/
2
/
2
2
( 𝑉𝑟(𝑟𝑚𝑠) 𝑥 𝑅 ) + ( 𝑉𝑟(𝑟𝑚𝑠) 𝑥 𝑋𝐶 ) =[ 𝑋𝐶 2 x Vr(rms)2 ]
2
/
2
/
𝑉𝑟(𝑟𝑚𝑠) 𝑥 𝑅2 ) = 𝑋𝐶 2 x Vr(rms)2 - 𝑉𝑟(𝑟𝑚𝑠) 𝑥 𝑋𝐶 2
/
2
/
2
2
(𝑉𝑟(𝑟𝑚𝑠) 𝑥 𝑅 ) = 𝑋𝐶 2 (Vr(rms)2 - 𝑉𝑟(𝑟𝑚𝑠) )
𝑋𝐶
2
/
2
2
(𝑉𝑟(𝑟𝑚𝑠) 𝑥 𝑅 )
=
2
/
(Vr(rms)2 − 𝑉𝑟(𝑟𝑚𝑠) )
2
/
(𝑉𝑟(𝑟𝑚𝑠) 𝑥 𝑅 2 )
𝑋𝐶 = √
/
2
(Vr(rms)2 − 𝑉𝑟(𝑟𝑚𝑠) )
𝑋𝐶 = √
(0,82 𝑥 202 )
(1,82 − 0,82 )
1
XC =
C=
1
4πf 𝑋𝑐
= 9,923 Ω
4πfC
1
= 4π x 50 x 9,923
C = 160 µf
QUESTION 8
/
𝑉𝑑𝑐 =
𝑅𝐿
𝑅𝐿 + 𝑅
1000
/
x Vdc
𝑉𝑑𝑐 = 1000+ 10 x 14 = 13,86 V
/
𝑟/=
𝑉𝑟(𝑟𝑚𝑠)
/
𝑉𝑑𝑐
0,8
= 13,86 = 0,0577
QUESTION 9
/
𝑉𝑑𝑐 =
/
200
𝑅𝐿
𝑅𝐿 + 𝑅
x Vdc
𝑉𝑑𝑐 = 200+ 5 x 18 = 17,56 V
QUESTION 10
1
/
𝑉𝑟(𝑟𝑚𝑠)= (4𝜋𝑓)2𝐿𝐶 x Vr(rms)
1
0,8 = (4𝜋 𝑥 50)2 𝑥 0,0001 𝑥 𝐿 x 2,2
L = 69,66 mH
QUESTION 11
r = 0,02 and Vr(rms) = 0,4 V
𝑉𝑟(𝑟𝑚𝑠)
VNL = Vdc = 𝑟 = 20 V
𝑉𝑁𝐿 + 𝑉𝐹𝐿
% 𝑉𝑅 =
𝑥 100
𝑉𝐹𝐿
20 + 11,5
% 𝑉𝑅 =
𝑥 100
11,5
= 73,9 %
QUESTION 12
VFL = 11,9 V and %VR = 3,2 %
𝑉𝑁𝐿 − 𝑉𝐹𝐿
% 𝑉𝑅 =
𝑥 100
𝑉𝐹𝐿
𝑉𝑁𝐿 − 11,9
0,032 =
11,9
𝑉𝑁𝐿 = 12,28 V
QUESTION 13
VFL = 12 V and %VR = 3 %
The fall of Vo will be 3 % of 12 = 0,36
Or
𝑉𝑁𝐿 − 𝑉𝐹𝐿
% 𝑉𝑅 =
𝑥 100
𝑉𝐹𝐿
𝑉𝑁𝐿 − 12
0,03 =
12
𝑉𝑁𝐿 = 12,36 V
The fall will be:12,36 – 12 = 0,36
QUESTION 16:
ZENER AS VOLTAGE REGULATOR
Pz = 500 mW
Vz = 12 V
Vs = 18±1 = 19 V (Vin max)
𝑝
0,5
Iz = 𝑍 = = 0,0416 A
𝑣𝑍
12
Rs (min) =
𝑽𝒊𝒏(𝒎𝒂𝒙)− 𝑽𝒁
𝑰𝒛(𝒎𝒂𝒙)
𝟏𝟗−𝟏𝟐
Rs (min) = 𝟎,𝟎𝟒𝟏𝟔 =168 Ω
RL (min) = 𝑽
𝑽𝒁
x Rs
𝒊𝒏(𝒎𝒂𝒙)− 𝑽𝒁
𝟏𝟐
RL (min) = 𝟏𝟕−𝟏𝟐 x 100
= 240 Ω
CHAP3
QUESTION 2
Fixed bias amplifier without Re
Vcc = 10 V, Vce = 8 V, Ic = 12 mA, Vbe = 0,6 V, β = 280
Calculate Rb and Rc
Ic
0,012
β = I ……… 280 = I ……Ib = 42,8 µA
b
b
𝑉𝑐𝑐 − 𝑉𝑏𝑒
10 − 0,6
Rb =
=
= 219,63 KΩ
𝐼𝑏
42,8 𝑥 10−6
𝑉𝑐𝑐 − 𝑉𝑐𝑒
10− 8
Rc =
=
= 166,67 Ω
𝐼𝑐
0,012
Question 3
Fixed bias amplifier with Re
β = 280, Vcc = 15 V, Ic = 6 mA, Vce = 5 V, f = 50 Hz, Vbe = 0,7 V
Calculate Rc, Rb, Re ,Ce
0,006
Ic
β = I …… 280 =
…… Ib = 21,43 µA
b
Ib
Vcc
Ve = 10 = 1,5 V
RC =
𝑉𝑐𝑐 − 𝑉𝑐𝑒 −𝑉𝑒
𝐼𝑐
Rb =
=
10− 5−1,5
0,006
𝑉𝑐𝑐 – 𝑉𝑏𝑒 – 𝑉𝑒
𝐼𝑏
15− 0,7−1,5
=
21,43 𝑥 10−6
Re =
𝑉𝑒
𝐼𝑐
=
Ce ≥
≥
= 1,4 KΩ
= 597, 3 KΩ
1,5
0,006
10
= 250
2𝜋𝑓𝑅𝑒
10
2𝜋 𝑥 50 𝑥 250
= 127 µf
QUESTION 4
VOLTAGE DIVIDER
Vcc = 12 V, Vce = 5 V, Vbe = 0,7 V, Ic = 15 Ma, f(min)=100 Hz, β = 300
Ve =
𝑉𝑐𝑐
12
= 10 = 1,2 V
10
Vb = Vbe + Ve = 0,7 + 1,2 = 1,9 V
Re =
𝑉𝑒
𝐼𝑐
=
Rb2 =
=
1
10
1,2
0,015
1
10
= 80 Ω
𝛽Re
300 x 80 = 2,4 KΩ
( 𝑉 −𝑉 )
𝑅
Rb1 = 𝑏2 𝑉𝑐𝑐 𝑏
𝑏
2400 ( 12 − 1,9)
=
= 12,76 𝐾Ω
1,9
𝑉 −𝑉 −𝑉
RC = 𝑐𝑐 𝐼𝑐𝑒 𝑒
𝑐
=
RbT =
12− 5−1,2
0,015
𝑅𝑏1 𝑥 𝑅𝑏2
𝑅𝑏1+𝑅𝑏2
=
= 386,67 Ω
12,76 𝑥 2,4
12,76+2,4
= 2,02 KΩ
HYBRID PARAMETERS
COMMON EMITTER CALCULATIONS
QUESTION 6 (a1)
If hie = 1,3 KΩ, hoe = 30 µA/V, hre 2,5 . 10 – 4 , hfe = 𝛽 = 280
Rb = 219,63 KΩ, Rc = 166,67 Ω
Calculate current gain, voltage gain, Power gain, input and output impedances using precise and
approximation method
RC = ZL
Z1 = hie -
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒𝑥𝑍𝐿
1+ℎ𝑜𝑒 . 𝑍𝐿
= 1300 Z2 =
=
280 𝑥 2,5.10−4 𝑥 166,67
1+ 30.10−6 𝑥 166,67
1
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒
hoe − ℎ𝑖𝑒+𝑅𝑠
1
280 𝑥 2,5.10−4
30.10−6 −
1300+0
= 41,9 KΩ
= 1288,3 Ω
PRECISE METHOD
Current gain (Ai):
ℎ𝑓𝑒
𝑅𝑏
Ai = ( 1+ℎ𝑜𝑒.𝑍𝑙)(𝑅𝑏+𝑍1)
280
219,63
= ( 1+30.10−6 .166,67 )(219,63+1,288 )
= 277
Voltage Gain (Av):
Av = (
=(
− ℎ𝑓𝑒 . 𝑍𝐿
ℎ𝑖𝑒+(ℎ𝑖𝑒 .ℎ𝑜𝑒−ℎ𝑓𝑒.ℎ𝑟𝑒)𝑍𝐿
− 280 𝑥 166,67
)
)
1300+(1300𝑥30.10−6 −280𝑥2,5.10−4 ).166.67
= -36
Power gain
Ap = - Ai x Av = - 277 x (-36 ) = 9973
Input and Output impedances
𝑅 𝑥𝑍
Zi = Rb//Z1 Zi = 𝑅 𝑏+𝑍1
1
𝑏
219,63 𝑥 1,288
= 219,63 + 1,288 = 1,28 K
𝑍 𝑥𝑍
Zo = ZL//Z2 Zo = 𝑍𝐿+𝑍2
𝐿
=
166,67 𝑥 41935
166,67 + 41935
2
= 166 Ω
QUESTION 6 (a2)
APPROXIMATION METHOD
ℎ𝑓𝑒
Ai = 1 +ℎ𝑜𝑒 𝑥 𝑍𝐿 =
Av = -
Ap =
ℎ𝑓𝑒 .𝑍𝐿
ℎ𝑖𝑒
ℎ𝑓𝑒 2 . 𝑍𝐿
ℎ𝑖𝑒
280
1 +30.10−6 𝑥 166,67
= -
280 𝑥 166,67
= Ap =
= - 35,8
1300
2802 𝑥 166,67
1300
= 279
= 1005
Z1 = hie = 1300 Ω
Zo = Rc = 167 Ω
Question 6 (b1)
If hie = 1,3 KΩ, hoe = 30 µA/V
hre 2,5 . 10 – 4 , hfe = 𝛽 = 280
Rb = 597 KΩ
Rc = 1,42 Ω
Calculate current gain, voltage gain, Power gain, input and output impedances using precision and
approximation method
RC = ZL
Z1 = hie -
= 1300 -
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒𝑥𝑍𝐿
1+ℎ𝑜𝑒 . 𝑍𝐿
280 𝑥 2,5.10−4 𝑥 1416,6
1+ 30.10−6 𝑥 1416,6
= 1205,56 Ω
1
Z2 =
=
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒
hoe − ℎ𝑖𝑒+𝑅𝑠
1
30.10−6 −
280 𝑥 2,5.10−4
1300+0
= 41,9 KΩ
PRECISION METHOD
Current gain (Ai):
ℎ𝑓𝑒
𝑅𝑏
Ai = ( 1+ℎ𝑜𝑒.𝑍𝑙)(𝑅𝑏+𝑍1)
280
597
= ( 1+30.10−6 .1416,6)(597+1,2055)
= 268
Voltage Gain (Av):
Av = (
=(
− ℎ𝑓𝑒 . 𝑍𝐿
ℎ𝑖𝑒+(ℎ𝑖𝑒 .ℎ𝑜𝑒−ℎ𝑓𝑒.ℎ𝑟𝑒)𝑍𝐿
− 280 𝑥 1416,6
)
)
1300+(1300𝑥30.10−6 −280𝑥2,5.10−4 ).1416,6
= -315
Power gain
Ap = - Ai x Av = - 268 x (-315 ) = 84629
Input and Output impedances
𝑅 𝑥𝑍
Zi = Rb//Z1 Zi = 𝑅 𝑏+𝑍1
𝑏
1
519 𝑥 1,2055
= 519 + 1,2055 = 1,2 KΩ
Zo = ZL//Z2 Zo =
𝑍𝐿 𝑥𝑍2
𝑍𝐿 +𝑍2
1,42 𝑥 41,9
= 1,42 + 41,9 = 1,37 KΩ
QUESTION6 (b2)
APPROXIMATION METHOD
ℎ𝑓𝑒
Ai = 1 +ℎ𝑜𝑒 𝑥 𝑍𝐿 =
Av = -
Ap =
ℎ𝑓𝑒 .𝑍𝐿
ℎ𝑖𝑒
ℎ𝑓𝑒 2 . 𝑍𝐿
ℎ𝑖𝑒
280
= 268,58 (*)
1 +30.10−6 𝑥 1416,6
= -
= Ap =
280 𝑥 1416,6
= -305
1300
2802 𝑥 1416,6
1300
= 85431,8
Z1 = hie = 1300 Ω (*)
Zo = Rc = 1416,6 (*)
QUESTION 6 (C 1)
If hie = 1,3 KΩ, hoe = 30 µA/V, hre 2,5 . 10 – 4 , hfe = 𝛽 = 300
Rb 1 = 12,76 KΩ, Rb2 = 2,4 K and RbT = 2,02 KΩ
Rc = 386,76 Ω
Calculate current gain, voltage gain, Power gain, input and output impedances using precision and
approximation method
RC = ZL
Z1 = hie -
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒𝑥𝑍𝐿
= 1300 -
1+ℎ𝑜𝑒 . 𝑍𝐿
Z2 =
=
280 𝑥 2,5.10−4 𝑥 386,65
1+ 30.10−6 𝑥 386,65
1
ℎ𝑓𝑒𝑥 ℎ𝑟𝑒
hoe −
ℎ𝑖𝑒+𝑅𝑠
1
30.10−6 −
= 1,27133 KΩ
300 𝑥 2,5.10−4
1300+0
= 36,111 KΩ
PRECISION METHOD
Current gain (Ai):
ℎ𝑓𝑒
𝑅𝑏
Ai = ( 1+ℎ𝑜𝑒.𝑍𝑙)(𝑅𝑏+𝑍1)
300
2020
= ( 1+30.10−6 .386,67 )(2020+1271,33)
= 182
Voltage Gain (Av):
Av = (
− ℎ𝑓𝑒 . 𝑍𝐿
ℎ𝑖𝑒+(ℎ𝑖𝑒 .ℎ𝑜𝑒−ℎ𝑓𝑒.ℎ𝑟𝑒)𝑍𝐿
)
− 300 𝑥 386,87
1300 + (1300𝑥30. 10−6 − 300 𝑥 2,5. 10−4 ). 386,87
= - 90
Power gain
Ap = - Ai x Av = - 182 x (-90 ) = 16380
Input and Output impedances
𝑅 𝑥𝑍
Zi = RbT//Z1 Zi = 𝑅 𝑏𝑇+𝑍1
1
𝑏𝑇
2020 𝑥 1271,33
= 2020 + 1271,33 = 780 Ω
𝑍 𝑥𝑍
Zo = ZL//Z2 Zo = 𝑍𝐿+𝑍2
𝐿
2
386,67 𝑥 36111,11
= 386,67 + 36111,11 = 383 KΩ
QUESTION 6 (C2)
APPROXIMATION METHOD
𝑅𝑏2
Current Gain: Ai = hfe X𝑅𝑏2 +ℎ𝑖𝑒
2,4
= 300 X 2,4 + 1,3 = 195
Voltage Gain: AV = - =-
300 𝑥 386,67
1300
ℎ𝑓𝑒 .𝑍𝐿
ℎ𝑖𝑒
= - 89,2
Power Gain : AP = - AV X AI
= - (- 89,2) X 195 = 17357,8
Input and Output impedances
Z1= hie = 1300 Ω and
1
1
Z2 > ℎ𝑜𝑒 = 30 𝑥 10−6 = 33,33 K Ω
𝑅
𝑥𝑍
Zi = RBT//Z1 = 𝑅 𝑏𝑇+𝑍1
1
𝑏𝑇
2020 𝑥 1300
= 2020 + 1300 = 791 Ω
𝑍 𝑥𝑍
Zo = ZL//Z2 = 𝑍𝐿+𝑍2
𝐿
=
386,67 𝑥 33333,33
386,67 + 33333,33
2
= 382,2 Ω
II COMMON BASE CALCULATIONS
( RC ≠ ZL )
Given : Re = 1.2 KΩ , Rc = 8.2 KΩ ,
RL = 15 K Ω, hib = 42 Ω, hrb = 200.10−6 , hob = 0,47 µA/V, hfb = - 0,95
For common base calculation all the “e” are replaced by “b” and “b” by ” e”.
ZL= Rc//RL =
𝑹𝑪𝒙𝑹𝑳
𝑹𝑪+𝑹𝑳
Zi = Re//Z1
Zo = Rc//Z2
USING PRECISION METHOD
ZL = Rc//RL =
=
𝟖,𝟐 𝒙 𝟏𝟓
𝟖,𝟐+𝟏𝟓
Z1 = hib -
= 42 -
𝑹𝑪𝒙𝑹𝑳
𝑹𝑪+𝑹𝑳
= 5301,71 Ω
ℎ𝑓𝑏 𝑥 ℎ𝑟𝑏 𝑥 𝑍𝐿
1+ℎ𝑜𝑏 . 𝑍𝐿
−0,95 𝑥 200.10−6 𝑥 5301,72
1+ 0,47.10−6 . 5301,72
= 43,0053 Ω
Z2 =
1
hob
ℎ𝑓𝑏𝑥 ℎ𝑟𝑏 =
−
ℎ𝑖𝑏+𝑅𝑠
1
0,47.10−6 −
− 0,95 𝑥 200.10−6
42 + 0
= 200247,926 Ω
Current Gain:
ℎ𝑓𝑏
𝑅𝑒
𝑅𝑐
− 0,95
Ai = ( 1+ℎ𝑜𝑏.𝑍𝑙)(𝑅𝑒+𝑍1) (𝑅𝑐+𝑅𝐿 ) = (
1200
8200
)(1200+43,0053) (8200+1500 )
1+0,47.10−6. 5301,72
= - 0,323
Voltage gain :
Av = (
− ℎ𝑓𝑏 . 𝑍𝐿
ℎ𝑖𝑏+(ℎ𝑖𝑏 .ℎ𝑜𝑏 − ℎ𝑓𝑏.ℎ𝑟𝑏)𝑍𝐿
− (−0,95) . 5301,72
)=(
42 + (42 . 0,47.10−6 −(−0,95) . 200.10−6 ) 5301,72
= 116,8
Input and Output impedances:
Zi = Re//Z1 =
𝑹𝒆 𝒙 𝒁𝟏
𝑹𝒆 + 𝒁𝟏
=
𝟏𝟐𝟎𝟎𝒙 𝟒𝟑,𝟎𝟎𝟓𝟑
𝟏𝟐𝟎𝟎 + 𝟒𝟑,𝟎𝟎𝟓𝟑
Zo = Rc//Z2 =
=
𝑹𝒄 𝒙 𝒁𝟐
𝑹𝒄 + 𝒁𝟐
𝟖𝟐𝟎𝟎 𝒙 𝟐𝟎𝟎𝟐𝟒𝟔,𝟗𝟐𝟔
= 7,88 KΩ
𝟖𝟐𝟎𝟎 + 𝟐𝟎𝟎𝟐𝟒𝟔,𝟗𝟐𝟔
𝒁𝑳 𝒙 𝒁𝟐
Zo = ZL//Z2 =
=
= 41,53 Ω
𝒁𝑳 + 𝒁𝟐
𝟓𝟑𝟎𝟏,𝟕𝟐 𝒙 𝟐𝟎𝟎𝟐𝟒𝟔,𝟗𝟐𝟔
= 5,165 KΩ
𝟓𝟑𝟎𝟏,𝟕𝟐 + 𝟐𝟎𝟎𝟐𝟒𝟔,𝟗𝟐𝟔
(wrong one in the book)
USING APPROXIMATION METHOD
𝑅𝑐
Current Gain: Ai = hfb (𝑅𝑐 +𝑅𝐿 )
8200
= - 0,95 (8200 +15000 ) = 0,3357
Voltage gain: Av = - -
ℎ𝑓𝑏 .𝑍𝐿
=−
ℎ𝑖𝑏
− 0,95 . 5301,72
42
Input impedance: Zi = hib = 42
Output impedance : Zo = Rc = 8,2K
CHAP 4
OPERATIONAL AMPLIFIERS
EXERCISE 4
QUESTION 7
Vcc =Vee =± 12 V
[ Vcc = +12 V, Vee = - 12 V]
Re = 2 KΩ, Vce = 8 V, hie = 900 Ω
𝑉 −𝑉
a) RL1 = 𝑐𝑐𝐼 𝑐𝑒1 and
𝑐1
RL2 =
𝑉𝑐𝑐 −𝑉𝑐𝑒2
𝐼𝑐2
𝑉𝑒
𝐼𝑒
Ie = 𝑅 and Ic1 = Ic2 =
𝑒
2
12
Ie = 2000 = 6 mA
Ic1 = Ic2 =
RL1 = RL2 =
6 𝑚𝐴
2
12−8
0,003
= 3 mA
= 1,333 KΩ
b) Current gain(hfe), if hre = 100
ℎ𝑓
Voltage gain gmRL = ℎ𝑖 𝑒 x RL
ℎ𝑓𝑒
𝑒
100 =
x 1,333
900
ℎ𝑓𝑒 = 67,5
QUESTION 8
f= 1 KHz ,t =1/f =1/1000 = 0,001 s
= 119,9
to = 0 s , t1 = 0,0005 s, t2 = 0,001 s
Vin = ±2 V
R = 1 MΩ
C = 10 µf
Area between 0 s and 0,0005 s
1
Vo(ta) = Vo(t1) = -
𝑡
𝑏
∫𝑡 ±Vin 𝑑𝑡 − 𝑉𝑐(𝑡𝑎)
RC
1
𝑎
106 𝑥 10−5
0,0005
∫0
+2𝑑𝑡 + 0
= - 0,1 𝑥 [2𝑡]0,0005
−0
0
= - 0,1 𝑥 [2𝑡]0,0005 − 0
0
= – 0,1 x [(2 x 0,0005) – 2 x 0)] - 0
= - 0,1 mV
Area between 0,0005 s and 0,001 s
Vo(tb) = -
1
0,001
RC
∫0,0005 −Vin 𝑑𝑡 − 𝑉𝑐(𝑡𝑎)
0,001
1
Vo(t2) = - 2 x 106 𝑥 10−5 ∫0,0005 −2𝑑𝑡 − 0,0001
0,001
= - 0,1 𝑥 [−2 𝑡]
− 0,0001
0,0005
= - 0,1 [(- 2 x 0,001 – (-2) x 0,0005)] – 0,0001
=
0,0001 - 0,0001
= 0V
Input – output waveform
QUESTION 9
Gain = 10
Rin = 1 K and C = 10 nf
𝑅 𝑥𝑅
R2 = 𝑅1+ 𝑅𝑆
1
𝑅
𝑆
AV = 𝑅𝑆 ……. Rs = Av x R1
1
Rs = 1000 x 10 = 10 KΩ
1 𝑥 10
R2 = 1+ 10 = 0,909 Ω
1
fr = 2𝜋 𝑅
𝑆
1
𝑥𝐶
= 2𝜋 104 𝑥10−8 = 1,591 KHz
QUESTION 10
Gain = 10,
f = 1 Khz,
C = 1 µf
1
t = Rf X C,
1
t = 𝑓 = 1000 = 0,001
𝑡
𝑅𝑓 =
𝐶
=
0,001
10−6
= 1 KΩ
At high frequency, gain =
1000
10 =
𝑅𝑠
Rs = 100 Ω
𝑅𝑓
𝑅𝑠
QUESTION 13
f = 100 KHz and C = 10 nf
fr =
105
=
1
2𝜋 𝐶 𝑅
1
2𝜋 10−8 𝑅
R = 159,15 Ω
CHAP 7 TESTING EQUIPEMENT
Demonstrate by making use of a 4-bit binary code how a successive approximation A/D converter can be used
to measure an initial unknown of 3V, 7V, 9V,11V,12V, 13V, 15V
Solution: The largest counting binary number is set to 1, while the other bits remain 0
Vu is compared to Vr
If Vr < Vu: 1 is kept
If Vr > Vu: 1 is reset to 0
If Vr = Vu: the process stop Vu appears on the display unit.
a) Vu = 3 V (unknown voltage)
8 4 2 1
Binary code
Start 0 0 0 0
Step 2 1 0 0 0 (8) > (3): 𝑟𝑒𝑠𝑒𝑡 1 𝑡𝑜 0
Step 3 0 1 0 0 (4) > (3): 𝑟𝑒𝑠𝑒𝑡 1 𝑡𝑜 0
Step 4 0 0 1 0
(2) < (3): 𝑘𝑒𝑒𝑝 1
Step 5 0 0 1 1
(3) = (3): 𝑘𝑒𝑒𝑝 1
0 0 1 1 appears on the display unit
b) Vu = 7 V (unknown voltage)
8 4 2 1
Binary code
Start 0 0 0 0
Step 1 1 0 0 0 (8) > (7): 𝑟𝑒𝑠𝑒𝑡 1 𝑡𝑜 0
Step 2 0 1 0 0
(4) < (7): 𝑘𝑒𝑒𝑝 1
Step 3 0 1 1 0
(6) < (7): 𝑘𝑒𝑒𝑝 1
Step 4 0 1 1 1
(7) = (7): 𝑘𝑒𝑒𝑝 1
0 1 1 1 appears on the display unit
a) Vu = 9 V (unknown voltage)
Start
Step 1
Step 2
Step 3
Step 4
8
0
1
1
1
1
4
0
0
1
0
0
2
0
0
0
1
0
0
1
Binary code
0
0
(8) < (9): 𝑘𝑒𝑒𝑝 1
0 (12) > (9): 𝑟𝑒𝑠𝑒𝑡 1𝑡𝑜 0
0 (10) > (9): 𝑟𝑒𝑠𝑒𝑡 1 𝑡𝑜 0
1
(9) = (9): 𝑘𝑒𝑒𝑝 1
0 0 1 appears on the display unit
b) Vu = 11 V (unknown voltage)
8
4
2
1
Binary code
Start
Step 1
Step 2
Step 3
Step 4
0
1
1
1
1
0
0
1
0
0
0
0
0
1
1
0
0
0
0
1
1
c) Vu = 13 V
8
Start 0
Step 1 1
Step 2 1
Step 3 1
Step 4 1
1101
(8) < (11): 𝑘𝑒𝑒𝑝 1
(12) > (11): 𝑟𝑒𝑠𝑒𝑡 1𝑡𝑜 0
(10) < (11): 𝑘𝑒𝑒𝑝 1
(9) = (11): 𝑘𝑒𝑒𝑝 1
0 1 1 appears on the display unit
(unknown voltage)
4 2 1
Binary code
0 0 0
0 0 0
(8) < (13): 𝑘𝑒𝑒𝑝 1
1 0 0
(12) < (13): 𝑘𝑒𝑒𝑝 1
1 1 0 (14) > (13): 𝑟𝑒𝑠𝑒𝑡 1 𝑡𝑜 0
1 0 1
(13) = (13): 𝑘𝑒𝑒𝑝 1
appears on the display unit
CHAP 8: OSCILLATORS
QUESTION 4
LT = 0,5 H and C = 10 nf
1
1
fo = 2𝜋√𝐿 .𝐶 =
= 2,25 KHZ
−8
𝑇 1
2𝜋√0,5 . 10
QUESTION 11
R = 10 KΩ, C= 10 µf
1,5
1,5
fo = 𝑅 . 𝐶 = 104 𝑥 10−5 = 15 HZ
QUESTION 13:
Calculate the oscillation frequency of A-stable multivibrator:
13.1
R1 = 10 KΩ, C2= 10 µf and R2 = 1 KΩ, C1= 4,7 µf
T1 = 0,7 x R2 x C1 = 0,7 x 10 3 x 4,7 . 10 -6 = 3,29 mS
T2 = 0,7 x R1 x C2 = 0,7 x 10 4 x 10 -5 = 0,07 S
1
1
fo = 𝑡 + 𝑡 = 0,00329+ 0,07 = 13, 6 Hz
1
2
13.2
R1 = R2 = 10 KΩ and C1 = C2 = 10 µf
1
1
fo = 1,4 𝑥 𝑅𝐶 = 1,4 𝑥 104 𝑥 47.10−6 = 1,519 𝐻𝑧
QUESTION 20:
Calculate the oscillation frequency of A-stable OSCILLATOR:
fo = (𝑅
RA = 1 KΩ, RB = 220 Ω, C = 47 µf
1,443
=
= 21,32 Hz
+ 2𝑅 ) 𝑥 𝐶 (103 +2 𝑥 220 ) 𝑥 47 𝑥 10−6
1,443
𝐴
𝐵
Or
tLow = 0,693 x RB x C
= 0,693 x 220 x 47x10 - 6 = 7,16 mS
tHigh = 0,693 (RA + RB) C
= 0,6932 (1000+220) 47x10 - 6 =39,74 Ms
1
1
fo = 𝑡 + 𝑡
= 0,00716+ 0,03974 = 21,32, 6 Hz
𝐿𝑜𝑤
𝐻𝑖𝑔ℎ
CHAP 9: TRANSDUCERS
QUESTION 5
6
K = 3, E =2,1.10 Kg/Cm2 and S = 1600 Kg/Cm2
𝑆
1600
𝜎 = 𝐸 = 2,1.10−6 = 0,00076
K =
∆𝑅/𝑅
𝜎
……. 3 =
∆𝑅/𝑅 = 0,00228
% ∆𝑅/𝑅 = 0,228 %
QUESTION 7
1
% Resolution = 1000 x 100 = 0,1 %
QUESTION 9
𝑇1𝑜 = 10 oC + 273 = 283 oK
𝑇2𝑜 = 30 oC + 273 = 303 oK
𝐵
𝑅 𝑇1𝑜 = 12.103 = A𝑒 283 (1)
𝐵
𝑅 𝑇2𝑜 = 2,4.103 = A𝑒 303 (2)
𝐵
𝑅𝑇 𝑜
1
𝑅𝑇 𝑜
=
12000
2
2400
𝐵
5 = 𝑒 303
𝐵
=
A𝑒 283
𝐵
A𝑒 303
−
Ln (5) =( 303 −
𝐵
303
𝐵
) Lne
303
303 𝐵−283 𝐵
Ln 5 =
85749
85749 x Ln5 = 20 B
B = 6900
In (1) or (2)
𝐵
6900
12.103 = A𝑒 283 …. 12.103 = A𝑒 283
A = 3,0927.10 – 7
b) 𝑅 𝑇 𝑜 at 20 oC [ 𝑇=𝑜 20 +273 = 293 oK]
𝐵
𝑅 𝑇 𝑜 = A𝑒 283
∆𝑅/𝑅
0,00076
6900
= 3,0927.10 – 7 𝑥 𝑒 293
= 5,22 KΩ,
QUESTION 10
Vt = 10 V , B = 3600, Rt = 8 KΩ. To = 25 oC (298 oK), Vo = 12 V
At 25 oC ,
𝛽
3600
Rt = A𝑒 𝑇 ….. 8000 = A𝑒 298
A = 0,04535
At 35 oC ,
𝛽
3600
Rt = A𝑒 𝑇 = 0,04535𝑒 308 = 5,404 KΩ
8
VA =
x 10 = 5 V
8+8
5,404
VB = 5,404+8 x 10 = 4,0316 V
VAB = Vin = VA – VB
= 5 – 4,03163 = 0,9684 V
𝑉𝑜
12
The gain = AV = 𝑉𝑖𝑛 = 0,9684
= 12, 39