Basic Electricity
VOLTAGE
 Also known as potential difference, indicates how much energy would be acquired
or lost per unit of charge by a particle as it was moved within an electric field.
 It is also a measure of the potential difference between two points. One particular
potential level is chosen as the reference level and assigned an arbitrary level of
zero. This reference is commonly referred to as ground.
Battery
 The most basic source of voltage is the battery.
1.5 V
1.5 V Battery

A battery has a positive (+) and a (-) terminal. These terminals indicate
how the batteries are to be wired.
Batteries in Series
Et = E1 + E2 + E3
= 1.5 v + 1.5 v + 1.5 v
= 4.5 Volts
E1 +
E2
+
E3
Etotal = 4.5 volts
CURRENT
 Current (I) is the flow of electrons through a wire or other devices in an electric
circuit. The unit of current is the amperes, abbreviated as A.
OHM’S LAW
 According to Ohm’s Law, the voltage ( E ) in a circuit is proportional to the
product of the current I and Resistance R.
E=IxR
Where: E is in volts
I is in amperes
R is ohms
RESISTANCE
 Anything which limits or impedes the current I flowing in a resistor is called a
resistance, This is measured in ohms and symbolized by the capital Greek letter
omega.
POWER
 The amount of energy ( W ) dissipated across a resistance R as current flows
through it is
P = E x I, E2 / R, I2 R
Where:
P is expressed in watts
E is in volts
I is in amperes
R is ohms
SERIES/ PARALLEL CONNECTIONS
A. Resistors
A.1. Series Connected
Resistances of resistors connected in series with each other in an electric
circuit. Resistances are added together to get the total resistance.
R
570 ohms
Rt = R1 + R2 + R3
A.2. Parallel Connected Resistance
Rt =
1
1
1
1
1
+
+
+
R1
R2 R3 R4
B. Capacitors
B.1. Series Connected
The equivalent total capacitance Ct of capacitors in series computed as
Ct =
1
1
1
1
1
+
+
+
C1
C2 C3 C4
B.2. Parallel Connected
The equivalent total capacitance Ct of parallel connected capacitors is computed
as
Ct = C1 + C2 + C3 … Cn
C. INDUCTORS
C.1. Series Connected
The equivalent inductance of series connected inductors is equivalent to the
sum of individual inductances.
Lt = L1 + L2 + L3 … Ln
C.2. Parallel Connected
The equivalent inductance of series connected inductors is equivalent to the
sum of individual inductances.
Lt =
1
1
1
1
1
+
+
+
L1
L2 L3 L4
ALTERNATING CURRENT AND DIRECT CURRENT



Voltage is usually plotted against time. This voltage is commonly called as an
electrical signal.
The graph of this electrical signal against time is its waveform.
Electrical signal that are constant in time is a Direct Current.
ALTERNATING CURRENT

Alternating current is a signal with voltage values
Basic Electronics
Electronics
Semiconductor Diodes
 Learn the characteristics of three semiconductor materials:
Silicon, Germanium, Gallium Arsenide
 Understand conduction of using electron and hole theory
 Be able to describe the difference between n – and p – type material
 Develop the basic operation and characteristics of a diode in the
No – bias, forward bias, and reverse – bias regions.
 Be able to calculate the DC, AC and average resistance of a diode from the
characteristics.
 Be able to understand the impact of an equivalent circuit whether it is ideal or
practical.
 Become familiar with the operation and characteristic of a Zener diode and light
emitting diode.
Important person to note is DR. Gordon E. Moore – predicting every integrated
Circuit produced will double every two years.
Construction of semiconductor devices are manufactured on single semiconductor
wafers also called substrate of the highest quality.
Semiconductors are a special class of elements having a conductivity between that of
a good conductor and that of an insulator.
Semiconductor materials fall into one of two classes:
Single-crystal
 Single-crystal semiconductors such as germanium (Ge) and silicon (Si) have a
repetitive crystal structure,
The three semiconductors used most frequently in the construction of electronic
devices are Ge, Si, and GaAs.
Compound-crystal semiconductors such as gallium arsenide (GaAs), cadmium sulfide
(CdS), gallium nitride (GaN), and gallium arsenide phosphide (GaAsP) are constructed
of two or more semiconductor materials of different atomic structures..
Semiconductor History


discovery of the diode in 1939, transistor in 1947, germanium as the base material
suffered from low levels of reliability due primarily to its sensitivity to changes in
temperature
in 1954 the first silicon transistor, less temperature sensitive.
Semiconductor Theory
Watch animation of semiconductor theory available on You Tube. Copy the link and
paste it your browser’s URL.
https://www.youtube.com/watch?v=ethnHSgVbHs
The pn junctions

The junction is where the p type and type material regions meet and junction
diode another name for a pn crystal. The diode is a contraction of two
electrodes (cathode and anode) where di means 2. This chapter describes how
diodes works and prepares the students for the transistor which is a
combination of two junction diodes.
Unbiased Diode
The figure3 shows an unbiased diode which means that there is no supply
voltage applied to its electrodes.
Figure 3
The depletion layer
The electrons on the n side tend to diffuse or spread in all directions. Some
diffuse across the junction. When an electron diffuse or enters the p region, it becomes
a minority carrier. With so may holes around it, this electron will fall into a hole. When
this happens, the hole disappears and the conduction band electron becomes a
valence electron.
Each time an electron crosses the junction, it creates pair of ions. The figure
above shows a pair of ions on each side of the of the junction. The ions are fixed in
crystal structure because of covalent bonding and cannot move around like conduction
bond electrons or holes
Each pair of positive and negative ions is called a dipole. The creation of a
dipole means that on conduction band electron and one hole have been taken out of
circulation. As the number of dipoles builds up, the region near the junction is emptied
because of movable charges. We call this charge empty region as the depletion layer.
Barrier Potential

In figure 3, having a field between ions in the depletion region is equivalent to a
potential difference, or called the barrier potential, is approximately equal to:
o 0.3 volts for germanium and
o 0.7 volt for the silicon diodes.
Temperature effects






The barrier potential depends on the temperature at the junction.
Higher temperature creates more electron – hole pairs.
As a result, higher temperatures creates more electron - hole pairs.
With this increase in temperature, the drift of minority carriers across the
junction increases. This forces an equilibrium to occur at a slightly lower
barrier potential.
So with this approximation, for changes in barrier potential for either
germanium or silicon diodes, the barrier potential decreases 2.5 mV
for each celcius degree rise.
In equations, the change in barrier potential is
Delta V = - 0.0025 Delta T
or ∆ 𝑉 = −0.0025 ∆𝑇
Example: Calculate the barrier potential of a silicon diode at 75 degree celcius.
Given:
Barrier potential of a silicon diode:
@25 ℃: 0.7 𝑣𝑜𝑙𝑡𝑠
Find:
Barrier potential of a silicon diode:
@75 ℃: ? 𝑣𝑜𝑙𝑡𝑠
Solution:
The amount of decrease in temperature formula is
∆ 𝑉 = −0.0025 ∆𝑇
The amount of decrease is @75 ℃
∆ 𝑉 = −0.0025 (75 − 25) = − 0.125 𝑉
Barrier potential @75 ℃
𝑉 = 0.7 − 0.125 = 0.575 𝑉
Energy Hill
Figure 4: Shows bands before and after diffusion
Forward bias
Figure 4 shows a dc source across a pn junction diode. The negative source
terminal connects to the n – type material
Figure 5: For biased pn junction
Highlights of what happens to an electron in Fig 4.
1. After leaving the negative source terminal, it enters the right end of the
crystal.
2.It travels through the n region as a conduction-band electron.
3. Near the junction it recombines and becomes a valence electron.
4. It travels through the p region as a valence electron.
5. After leaving the left end of the crystal, it flows back into the positive source
terminal.
Energy bands
 Forward bias lowers the energy hill, see figure 6c, because of this,
conduction – band electrons have enough energy to invade the p region.
After entering the p region, they fall into a hole. As a valence electron, it
continues its journey toward the left end of the crystal.
 A conduction – band electron may fall into a hole before it crosses the
junction. A valence electron may cross the junction from right to left ; this
leaves a hole just to the right end of the junction. This hole does not live long
because a conduction – band electron soon falls into it.
 A steady stream of conduction – electrons flows towards the junction and
falls into holes near the junction. The captured electrons because of the
holes are now called valence electrons move left in a steady stream through
the holes in the p region. In this way we get a steady flow of electrons
through the diode.
Fig 6: Energy bands during reverse bias, forward bias and in zero bias.
Retrieved from Research Gate: Joel Jean, etal
Reverse bias
Depletion layer widens


The externally produced field (from the battery) is in the same direction as
the depletion - layer field. Because of this, holes and electrons moves
towards the ends of the crystal (away from the junction). The fleeing
electrons leave positive ions behind, and departing holes leave negative ions
behind; therefore, the depletion layer gets wider. The greater the reverse
bias, the wider the depletion layer becomes.
The depletion layer stops growing when its potential difference equals the
applied reverse voltage.
Transient current

When the depletion layer is growing due to the applied reverse voltage,
therefore a current flows into the external circuit while the depletion layer is
adjusting to its new width. This current is called transient last only in
nanoseconds since it becomes zero after the depletion layer stops growing.
Diode reverse current

In reverse biased condition, a small amount of reverse diode current exist
which can be attributed to these two factors:
o Saturation current ( Minority carrier current) – caused by caused by
thermal energy creating pairs of hole and free electrons
o Surface leakage current (skin effect) – free electrons can cross to the
other junction through the skin of the material due to surface
impurities and imperfection on the crystal structure.
o Generally current is zero in a reversed - biased diode.
Reverse Breakdown Voltage






Increasing the reverse voltage and eventually reaching the reverse voltage.
For rectifier diodes, diodes that conducts only one way, the breakdown
voltage is usually greater than 50 volts.
Once the breakdown voltage is reached, a large number of minority carriers
appear in the depletion layer and the diode conducts heavily
Most diodes are not allowed to breakdown.
The reverse voltage across a diode is kept less than the breakdown voltage.
A reversed biased diode conducts poorly.
A reversed biased diode acts as a open switch whereas a forward biased
diode acts as a closed switch and conducts easily.
Bipolar and unipolar devices




Junction diode is a bipolar device
Bipolar device needs holes and conduction – band electrons to work properly.
Some devices are unipolar. They need holes or electrons to operate normally.
A carbon resistor is a unipolar device. It uses only conduction band electrons. A
resistor can be made out of a p -type semiconductor material
The Diode Equation

The Shockley Ideal Diode equation gives the I – V voltage characteristic of an
ideal diode in either forward, reverse or no bias condition.
𝐼𝑑 = 𝐼𝑠(𝑒 𝑞𝑉𝑑/𝑛𝑘𝑇𝑘 − 1 )
where: Id = Diode current
Is = reverse saturation current
q = electron charge (1.6 x 10-19 Joules/Volt or 1.6 x 10-19 coulombs)
Vd = diode voltage
Ŋ = empirical constant, below the knee voltage = 1 for Ge. 2 for Silicon
= above the knee voltage = 1 for Ge and Si
k = Boltzmann’s constant (1.38 x10-23 J/Kelvin
Tk = absolute temperature in kelvin
DC or Static Resistance of a Diode
 The DC resistance of a diode at the operating point can be found by finding the
corresponding levels of Vd and Id.
Figure 7: DC or Static Resistance
Dynamic Resistance or AC Forward Resistance


The resistance offered by the diode when AC is applied to the diode is called
AC resistance or dynamic resistance. The current flows in both the direction
when AC voltage is applied.
Looking at Figure 8:
𝑟𝑑 =
∆𝑉𝑑
∆𝐼𝑑
Where: ∆ signifies a finite change in the quantity
@ room temperature,
𝑟𝑑 =
26 𝑚𝑉
𝐼𝑑
;
Considering the effect of the junction resistance (rB, usually at 1 ohm)
𝑟′𝑑 =
26 𝑚𝑉
+ 𝑟𝐵
𝐼𝑑
Figure 8: How to determine the AC Resistance
Average Ac Resistance

It is determined by the straight line that is drawn linking the intersection of the minimum
and maximum values of external input voltage.
𝑟𝑎𝑣𝑒 =
∆𝑉𝑑
| 𝑝𝑡 𝑡𝑜 𝑝𝑡
∆𝐼𝑑
Figure 9
Peak Inverse Voltage
 The maximum value of reverse voltage designated as peak inverse voltage
(PIV) occurs when the diode is reverse biased.
 PIV = Vin , if Vin is 12V then PIV is 12V
Figure 10
Diode approximations
 Using these approximations, diode analysis is much simpler and easy to
understand and will guide the students in some mathematical analysis of
transistors. Below is the schematic symbol of a diode.
o The Ideal Diode ,1st Approximation
Figure 11
o Practical Diode Model, Second Approximation
Conventional
a
b
Figure 12 a. Second Approximation b. Switch – battery equivalent Circuit
o Complex Diode Model, 3rd Approximation
The complete model of a diode consist of a diode of the barrier potential,
the small forward dynamic resistance rb, and the large internal reverse
resistance rr
Rr = Vr/ Ir
where: Ir is the reverse current and Vr is the reverse voltage
Rectifier Diodes




Figure 12a, the ac sources pushes electrons up the resistor during the
positive half cycle, and down the resistor during negative half cycle.
A rectifier diode is forward biased on the first positive half cycle of the of the
input voltage.
In figure 8a, a diode was inserted into the circuit. The diode conducts at the
first positive half cycle and on the negative half cycle, the diode is reversed
biased. See output waveform across the resistor figure 8b.
When analyzing diode circuits, one of the things to decide is whether they
forward biased or reverse biased.
Figure 12a: Schematic Diagram without diode device in a circuit
Figure 12b: Monitoring the applied
signal across a load resistor.
.
Figure 13a: Schematic Diagram with diode device in a circuit.
Figure 13b: Waveform of the applied signal
using a diode across the load resistor..
Problem:
5.1. Line voltage is 120 volt rms but may be as low as 105 Vrms or as high as 125
Vrms. Calculate the peak value for each of these extremes.
𝑉𝑝𝑘 = √2 𝑉𝑟𝑚𝑠
a. 𝑉𝑝𝑘 = √2 (105) = 148.5 𝑉
b. 𝑉𝑝𝑘 = √2 (125) = 176.75 𝑉
5.2 The transformer of the previous example has a turns ratio of N1:N2 = 4:1.
What is the peak voltage across the load resistance? The average voltage?
𝑁1
𝑉𝑝𝑟𝑖
=
𝑁2
𝑉𝑠𝑒𝑐
𝑉𝑝𝑘 = √2 𝑉𝑟𝑚𝑠
𝑉𝑃𝑟𝑖 = √2 (120) = 169.7 𝑉
𝑉𝑠𝑒𝑐 = 𝑉𝑝𝑟𝑖
𝑁2
1
= 169.7 = 42.43 𝑉𝑜𝑙𝑡𝑠 𝑝𝑘
𝑁1
4
𝑉𝑟𝑚𝑠 = (0.707)𝑉𝑠𝑒𝑐 = 0.707 (42.43) = 29.99 𝑉𝑜𝑙𝑡𝑠
Peak Voltage across the Load:
𝑉𝑝𝑘 = √2 (29.99)𝑉 = 42.43 𝑉𝑜𝑙𝑡𝑠
Average Voltage after the diode:
𝑉𝑑𝑐 =
𝑉𝑠𝑒𝑐𝑝𝑘
42.43
=
= 13.50 𝑉𝑜𝑙𝑡𝑠
𝜋
𝜋
5.3. If the transformer from above is 2: 1, what is the peak load current? The DC
load current?
𝑉𝑠𝑒𝑐 = 𝑉𝑝𝑟𝑖
𝑁2
1
= 169.7 = 84.85 𝑉𝑜𝑙𝑡𝑠 𝑝𝑘
𝑁1
2
𝑉𝑟𝑚𝑠 = (0.707)84.85 = 0.707 (42.43) = 59.99 𝑉𝑜𝑙𝑡𝑠
Peak Voltage across the Load
𝑉𝑝𝑘 = √2 (59.99)𝑉 = 84.85 𝑉𝑜𝑙𝑡𝑠
Peak load current
𝐼𝑝𝑘𝑙𝑜𝑎𝑑 =
𝑉𝑝𝑘𝑙𝑜𝑎𝑑
84.85
=
= 0.16968 𝐴 = 169.68 𝑚𝐴
𝑅𝑙𝑜𝑎𝑑
500
Average load voltage at the load
𝑉𝑑𝑐 𝑙𝑜𝑎𝑑 =
𝑉𝑠𝑒𝑐𝑝𝑘
84.85
=
= 27 𝑉𝑜𝑙𝑡𝑠
𝜋
𝜋
𝐼𝑑𝑐 𝑙𝑜𝑎𝑑 =
𝑉𝑑𝑐𝑙𝑜𝑎𝑑
27
=
= .054 𝐴 = 54 𝑚𝐴
𝑅𝑙𝑜𝑎𝑑
500
Center Tapped Rectifier
Same circuit from the above circuit
Average Value and Output Frequency
The average value or dc value of a full – wave signal is
𝑉𝑑𝑐 =
2 𝑉𝑠𝑒𝑐𝑝𝑘
𝜋
Example:
The average value of full – wave signal is 170 V, the average value is
𝑉𝑑𝑐 =
2 (170)
= 108.22 𝑉𝑜𝑙𝑡𝑠
𝜋
The period of the output signal is half the frequency of the period of the input signal.
Therefore,
𝑓𝑜𝑢𝑡 = 2𝑓𝑖𝑛
So if fin = 60 hz
𝑓𝑜𝑢𝑡 = 2𝑓𝑖𝑛 = 2(60) = 120 ℎ𝑧
The reverse voltage of the non-conducting diode is twice the secondary voltage
𝑃𝐼𝑉 = 2𝑉𝑚 = 2 𝑉𝑠𝑒𝑐
Example:
In the figure below, the maximum voltage at the secondary winding is 28.3 Volts.
Ignore the diode drop and calculate the load voltage. Also what are the output
frequency and peak inverse voltage?
Solution:
𝑉𝑑𝑐 =
2 (𝑉𝑠𝑒𝑐_𝑝𝑘) 2 (28.3)
=
= 18.01 𝑉𝑜𝑙𝑡𝑠
𝜋
𝜋
𝑓𝑜𝑢𝑡 = 2(60) = 120
𝑃𝐼𝑉 = 2𝑉𝑚 = 2 𝑉𝑠𝑒𝑐 = 2(28.3) = 56.6 𝑣𝑜𝑙𝑡𝑠
The Bridge Rectifier
During the positive half cycle of the secondary voltage, diodes D2and D3 are forward
biased and during the negative half cycle, D1 and D4 are forward biased. Th is why the
load voltage is a full – wave signal.
The average value or dc value of a full – wave signal is
𝑉𝑑𝑐 =
2 𝑉𝑠𝑒𝑐𝑝𝑘
𝜋
Its output frequency is
𝑓𝑜𝑢𝑡 = 2𝑓𝑖𝑛
And the Peak Inverse Voltage is
𝑃𝐼𝑉 = 𝑉𝑚 = 𝑉𝑠𝑒𝑐
Ideal Average Rectifiers
Number of Diodes
Transformer Necessary
Peak Rectified Output
DC output (unfiltered)
Peak Inverse Voltage
Output Freq

Halfwave
1
No
Vm
Vm/Pi
Vm*
fin
Center - Tapped
2
Yes
Vm
2Vm/Pi
2 Vm
2fin
Bridge
4
No
Vm
2Vm/Pi
Vm
2fin
With a capacitor input filter, PIV = 2 Vm
By making the accuracy of the calculations,
𝑉𝑝𝑘𝑙𝑜𝑎𝑑 = 𝑉𝑚 − 𝑑𝑖𝑜𝑑𝑒 𝑏𝑎𝑟𝑟𝑖𝑒𝑟 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 (half wave or center tapped)
𝑉𝑝𝑘𝑙𝑜𝑎𝑑 = 𝑉𝑚 − 0.7
𝑉𝑝𝑘𝑙𝑜𝑎𝑑 = 𝑉𝑚 − 2𝑥𝑑𝑖𝑜𝑑𝑒 𝑏𝑎𝑟𝑟𝑖𝑒𝑟 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 (bridge)
𝑉𝑝𝑘𝑙𝑜𝑎𝑑 = 𝑉𝑚 − 1.4
The Choke – Input Filter
The uses for a pulsating dc voltage are limited to charging batteries, running dc motors
and other limited applications. To convert half - wave and full – wave signals into
constant dc voltage we must filter or smooth – out the ac variations using the choke
input filter.
𝑉𝑑𝑐 =
𝑅𝑙
𝑅+𝑅𝑙
𝑉’𝑑𝑐
Where
𝑉𝑑𝑐 = 𝑑𝑐 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
R = dc resistance of choke
Rl = load resistance
𝑉’𝑑𝑐 = dc voltage from full wave rectifier
Output Ripple
𝑉𝑟 =
𝑋𝑐
𝑋𝑙
𝑉′𝑟
Where
𝑉𝑟 = 𝑟𝑚𝑠 𝑜𝑢𝑡𝑝𝑢𝑡 𝑟𝑖𝑝𝑝𝑙𝑒
𝑉 ′ 𝑟 = 𝑟𝑚𝑠 𝑖𝑛𝑝𝑢𝑡 𝑟𝑖𝑝𝑝𝑙𝑒
For a ripple frequency of 120 hertz
𝑉𝑟 =
5.28 𝑥 10𝐸−7 𝑉𝑝
𝐿𝐶
(Full Wave)
Ripple Factor
Ripple factor is defined as the ratio of RMS value of an alternating current component in the
rectified output to the average value of rectified output.
𝑟=
.
Where
𝑉𝑟 = 𝑟𝑚𝑠 𝑜𝑢𝑡𝑝𝑢𝑡 𝑟𝑖𝑝𝑝𝑙𝑒
𝑉𝑟
𝑥 100%
𝑉𝑑𝑐
𝑉𝑑𝑐 = 𝑑𝑐 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑐𝑟𝑜𝑠𝑠 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
The lower the r the better.
Critical Inductance
Is defined as the minimum amount of inductance that gives good filtering. The
critical inductance for a full wave rectifier at a line frequency of 60 hz is given by
𝐿𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 ≅
𝑅𝑙
1000
Example
In the figure , Rl = 750 ohms, choke is 10 Henries, C = 500 microfarad, the full – wave
signal at the input to the choke has a peak value of 25.7 v and an average value of
16.4 volts. If the choke has a dc resistance of 25 ohms, what is the dc output voltage?
The output ripple? The ripple factor?
𝑉𝑑𝑐 =
𝑉𝑟 =
𝑅𝑙
𝑅+𝑅𝑙
750
𝑉’𝑑𝑐 = (25+750 ) (16.4) = 15.9 𝑣𝑜𝑙𝑡𝑠
5.28 𝑥 10−7 𝑉𝑝
5.28 𝑥 10−7 (25.7)
=
= 2.71 𝑚𝑉
(10)(500𝑥10−6 )
𝐿𝐶
𝑟=
𝑉𝑟
2.71 10−3
𝑥 100% =
𝑥 100% = 0.017%
𝑉𝑑𝑐
15.9
The ripple is so small and that it is similar to that from a battery.
The Capacitor – Input Filter
Using a capacitor instead of a choke changes the operation from an average
detection to peak detection. During the first quarter cycle of input voltage, the diode is
forward biased. Ideally it looks like a closed switch. Since the diode connects the
source directly across the capacitor, the capacitor charges to the peak voltage Vp.
Just passed the positive peak value, the diode stops conducting, which means
the switch opens. The capacitor has much + Vp volts across it. With the source
voltage slightly less than +Vp the capacitor will try to force the current back through the
diode. This reverse biases the diode. With the diode off, the starts to discharge through
the load resistance RL. The idea of a peak rectifier (also a peak detector): the RLC time
constant is much greater than the period of the input signal. Because of this , the
capacitor will lose only a small part of its charge. Near the next positive input peak, the
diode turns on briefly and recharges the capacitor.
At the output, the smaller the ripple the better.
Long Time Constant
In full wave circuits, the line frequency is 60 hz, fout is 120 hz.
T = 1/120 = 8.33 ms. Long time constant, RLC must be at least 10 times 8.33ms.
RLC >> 83.3 ms.
When this condition is satisfied, the following approximations can be used for FW
rectifiers.
Vdc = (1 −
Vr =
0.00417
.0024𝑉𝑝
Where Vr is the rms ripple voltage.
𝑅𝑙𝐶
𝑅𝑙𝐶
) 𝑉𝑝
@ any frequency
Given the ripple factor in percent and the load resistance, the formula to calculate the
minimum capacitance needed for filtering.
𝐶𝑚𝑖𝑛𝑖𝑚𝑢𝑚 =
0.24
(𝑟)𝑅𝑙
Example:
Given a secondary voltage of 30 V in figure 5 -10, what are the DC output voltage and
ripple? Ignore diode drop.
Figure:
Schematic Diagram of a Bridge Type FW Rectifier.
Given: Vsec = Vp = 30 Volts
Solution:
For long time constant:
RLC = 470 *220 x10 -6 = 103 ms
RLC is greater 83.3ms, therefore using
Vdc = (1 −
Vr =
0.00417
0.00417
𝑅𝑙𝐶
103 𝑚𝑠
.0024𝑉𝑝
𝑅𝑙𝐶
=
) 𝑉𝑝 = (1 −
.0024∗30𝑉
103 𝑚𝑠
= 0.6999
) 30 = 28.8 𝑉𝑜𝑙𝑡𝑠
Example:
A Full Wave peak rectifier has to meet these specifications: r =2 and RL = 10kohms.
What is the minimum capacitance needed?
Solution:
𝐶𝑚𝑖𝑛𝑖𝑚𝑢𝑚 =
0.24
0.24
=
= 12 𝑚𝑖𝑐𝑟𝑜 𝑓𝑎𝑟𝑎𝑑
(𝑟)𝑅𝑙
(2)(10,000)
The Zener Diode
The Zener diode is made for operation in the breakdown region. By varying the doping
level, the manufacturer can produce a breakdown voltages from 2 volts to 200 volts by
applying reverse voltages that exceed the Zener breakdown voltage, and it will act like
a constant voltage source.
Zener Diode Breakdown Voltage and Power Rating
Where
Vz is the breakdown voltage
Iz is the reverse current
Therefor the product of Vz and Iz is the power dissipation of the Zener diode.
Pz = Vz x Iz
For instance, Vz = 12v , Iz = 10mA, then
Pz = 12 x .01 = 0.12 Watts
Zener Impedance
In the breakdown region of a Zener diode, a small increase in voltage produces a large
increase in current. This concludes that a Zener diode has a small impedance which is
Zz =
∆𝑣
∆𝑖
Example:
If a curve tracer shows changes of 80 mv and 20 mA, the Zener impedance is
Zz
.08
.02
= 4 𝑜ℎ𝑚𝑠
Exercises:
Read each of the following and provide the missing words.
1. The symbol of a diode points from the p side to the n side. Thus, _____current
flows the p side to the end side.
2. The silicon diode, its knee voltage is equal to its barrier potential which is
approximately ______ volts. Likewise for a germanium diode its knee voltage is
equal to _____ voltage.
3. Any metallic conductor has some resistance, including the p and n region of
semiconductors. The sum of this resistances is called _____ resistance.
4. In a circuit, the diode acts like an automatic switch. When current flows to the
direction of the arrow of the diode, the diode state is ______. And when the flow of
current is going to the opposite direction of the arrow symbol of the diode, the diode
is _______.
5. Using the 2nd approximation of a diode, the voltage drop across the diode is ____v.
6.
7. When a diode is reverse biased, it has a small reverse ____. One way to estimate
this diode parameter is with the reverse resistance, defined as the ratio between the
reverse voltage and _______ ______.
8. If the a diode was suddenly reverse biased, the ______charges can flow in the
reverse direction for a while. The greater the ______ , the longer these charges can
contribute to reverse current.
9. The time it takes to turn off a forward biased diode is the reverse recovery time. As a
guide, trr is approximately the time it takes a diode to drop to 10% of forward current
when a diode is switched from forward to reverse bias.
10. The LED is a light emitter, and the photodiode is a light detector. A varactor has a
variable ______. The Schottky diode has no charge storage, which allows it to switch
on and off much faster than a bipolar diode. The _______ diode is the second most
widely diode.
Bipolar Junction Transistors
History of Transistors
The transistor was invented by a team of three men at Bell Laboratories in 1947. The
three individuals credited with the invention of the transistor were William Shockley, John
Bardeen and Walter Brattain. The introduction of the transistor is often considered one of the
most important inventions in history. They were awarded the Nobel Prize in Physics in 1956 for
the transistor.
Definition and Function of Transistors
•
A transistor is a semiconductor device with at least three terminals for
connection to an electric circuit. It is also called a solid-state device. Transistors
are used for amplification and switching.
Transistors regulates current or voltage flow and acts as a switch or gate for signals.
•
The term transistor is a combination of the words “transfer” and “resistor”
because it is a resistor or semiconductor which can amplify electrical signals as
they are transferred through it from input to output terminals.
As an example, FM receiver tunes in the signals from the range of 88 MHZ to 108
Mhz. The received signal can be weak due to the disturbances such as EMI noise,
weak signal due to the distance where it is coming from. Now if this signal is read as it
is, you cannot get a fair output. amplification is needed to boost the signal. It
also regulates the incoming current and voltage of the signals.
Amplification means increasing the signal strength. Amplification is needed
wherever the signal strength has to be increased.
•
A transistor also acts as a switch to choose between available options in digital
applications.
•
The design of a transistor allows it to function as an amplifier or a switch. This is
accomplished by using a small amount of electricity to control a gate (base) on
a much larger supply of electricity, much like turning a valve to control a supply
of water.
The bipolar junction transistor shown in Figure (a) is an NPN three-layer
semiconductor sandwich with an emitter and collector at the ends, and a base in
between.
Transistors are composed of three parts: a base, a collector, and an emitter. The base
is the gate controller device for the larger electrical supply. The collector is the larger
electrical supply, and the emitter is the outlet for that supply
It is as if a third layer were added to a two-layer diode. If this were the only
requirement, no more than a pair of back-to-back diodes. The key to the fabrication of
a bipolar junction transistor (bjt) is to make the middle layer, the base, as thin as
possible without shorting the outside layers, the emitter, and collector. Overemphasize
on the importance of the thin base region.
The layers of an NPN transistor must have the proper voltage connected across them.
The voltage of the base must be more positive than that of the emitter. The voltage of
the collector, in turn, must be more positive than that of the base. The voltages are
supplied by a battery or some other source of direct current. The emitter supplies
electrons. The base pulls these electrons from the emitter because it has a more
positive voltage than does the emitter. This movement of electrons creates a flow of
electricity through the transistor.
Constructional Details of a Transistor
This type of connection offers two types of transistors. They are PNP and NPN which
means an N-type material between two P - types and the other is a P-type material
between two N - types respectively.
Emitter
The structure shown below shows Emitter.
• This has a moderate size and is heavily doped as its main function is
to supply a number of majority carriers, i.e., either electrons or holes.
• As this emits electrons, it is called as an Emitter.
• This is simply indicated with the letter E.
Base




The middle material shown below is the Base.
This is thin and lightly doped.
Its main function is to pass the majority carriers from the emitter to the collector.
This is indicated by the letter B
BasicElectronics
How to Calculate α (alpha) of a Transistor
α (Alpha), the factor by which emitter current is multiplied to yield collector current, can be calculated in
two ways.
If the β of the transistor is known, α can be calculated very easily by the formula:
Example
If β of a transistor is 100, calculate the α of the transistor:
If the collector current, Ic, and the emitter current, Ie, are known, then α can be calculated by the
following formula:
If the collector current Ic= 7.95ma and the emitter current Ie=8ma, then alpha calculates out to be:
How to Calculate the Base Current, IB, of a Transistor
The equation to solve for Ie is:
So we must solve for Vbb and RB in order to solve for IB.
Next we compute the value of RB:
Now we can calculate the value of the base current, IB, in the circuit:
2nd Way to Calculate Base Current IB
Using Known Values
If the emitter current, Ie, and β are known for the transistor circuit, IB can be calculated by the formula:
Example
If Ie=6ma and β=99, then
3rd Way to Calculate Base Current IB
Using Known Values
If the emitter current, Ie, and the collector current, Ic, are known, IB can be calculated by the formula:
Example
If Ie=4ma and Ic=3.96ma, then IB calculates out to be:
BasicElectronics
How to Calculate VBB of a Transistor
VBB, the base voltage of a bipolar junction transistor, or in other words, the voltage that falls across the
base of the transistor, is crucial to calculations such as when calculating the base current, I B or the
quiescent emitter current, IEQ.
VBB is calculated by the formula below:
How to Calculate VCE of a Transistor
BasicElectronics
DC Analysis of a Bipolar Junction Transistor Circuit
What is shown above is an example of a bipolar junction transistor circuit, which we will do DC analysis
on. We will show you in this article how to do DC analysis of this transistor circuit so that you can solve
for the crucial unknown DC values of it.
When doing DC analysis, all AC voltage sources are taken out of the circuit because they're AC
sources. DC analysis is concerned only with DC sources. We also take out all capacitors because in
DC, capacitors function as open circuits. For this reason, everything before and after capacitors are
removed, which in this circuit includes resistor, Rs.
Below is the schematic of the circuit above with respective to DC analysis:
Find the Vbb, Rb, Ieq, and Vceq. From this then, we can find the quiescent or just simply Q-point of this
transistor circuit.