Summary Electronics
by Ruben Tol
HC1: Measurement systems & signals
1.7 Ohm’s law
The relationship between voltage, current and resistance can be represented in a
number of ways, including:
V = IR,
V
I= ,
R
V
R= .
I
1.8 Kirchhoff ’s laws
Current law
At any instant, the algebraic sum of all the currents flowing into any junction in a
circuit is zero:
X
I = 0.
Voltage law
At any instant, the algebraic sum of all the voltages around any loop in a circuit is
zero:
X
V = 0.
1.9 Power dissipation in resistors
The instantaneous power dissipation P of a resistor is given by the product of the
voltage across the resistor and the current passing through it. Combining this result
with Ohm’s law gives a range of expressions for P . These are
P = V I,
P = I 2 R,
V2
P = .
R
1
,1.10 Resistors in series
The effective resistance of a number of resistors in series is equal to the sum of their
individual resistances:
R = R1 + R2 + R3 + · · · + Rn .
1.11 Resistors in parallel
The effective resistance of a number of resistors in parallel is equal to the reciprocal
of the sum of the reciprocals of their individual resistances:
1 1 1 1 1
= + + + ··· + .
R R1 R2 R3 Rn
1.12 Resistive potential dividers
To calculate the voltage at a point in a chain of resistors, one must
determine the voltage across the complete chain, calculate the voltage
across those resistors between that point and one end of the chain,
and add this to the voltage at that end of the chain. For example, in
this figure,
R2
V = V2 + (V1 − V2 ) .
R1 + R2
1.13 Sinusoidal quantities
The frequency of a waveform is related to its period by the expression
1
f= .
T
2.2 Sine waves
2.2.1 Instantaneous value
For a voltage waveform, the peak value of the waveform is Vp , so this waveform
could be represented by the expression
v = Vp sin (θ).
2.2.2 Angular frequency
The number of radians per second is termed the angular frequency of the waveform
and is given the symbol ω:
ω = 2πf,
ω
f= .
2π
2
,2.2.3 Equation of a sine wave
By replacing the phase angle θ by ωt, the equation of a sinusoidal voltage waveform
becomes
v = Vp sin (ωt) = Vp sin (2πf t).
A sinusoidal current waveform might be described by the equation
i = Ip sin (ωt) = Ip sin (2πf t).
2.2.4 Phase angles
Sinusoidal waveforms not starting at the origin on the time measurement (t = 0)
can be described by adding the phase angle ψ at t = 0:
v = Vp sin (ωt + ψ) = Vp sin (2πf t + φ).
2.2.6 Average value of a sine wave
We can calculate the average of a sinusoidal waveform by integrating over half a cycle
(180° / π rad) and dividing by half the period. For a sinusoidal voltage waveform,
this implies
1 π
Z
Vp 2Vp
Vav = Vp sin (θ) dθ = [− cos (θ)]π0 = .
π 0 π π
Therefore,
2
Vav = Vp .
π
Similarly, for a sinusoidal current waveform,
2
Iav = Ip .
π
2.2.7 The r.m.s. value of a sine wave
For instantaneous expressions, we write a bar above an expression to note the mean
of that expression. The term v 2 is referred to as the mean-square voltage and i2 as
the mean-square current. While these are useful quantities, we more often use the
square root of each quantity. These are termed the root-mean-square voltage Vrms
and the root-mean-square current Irms , where
Vrms = v 2 ,
Irms = i2 ,
Now, considering a sinusoidal voltage v = Vp sin (ωt), and integrating this over a
complete cycle and dividing by the period, we can see that
T 12 2 Z T 12
Vp
Z
1 1 Vp
Vrms = Vp2 2
sin (ωt) dt = (1 − cos (2ωt) dt =√ .
T 0 T 0 2 2
3
,Therefore,
1
Vrms = √ Vp .
2
Similarly,
1
Irms = √ Ip .
2
For the average power dissipation with alternating quantities, we get
Pav = Vrms Irms ,
Vrms
Pav = ,
R
2
Pav = Irms R.
2.2.8 Form factor and peak factor
The form factor of any waveform is defined as
r.m.s. value
form factor = .
average value
The peak (or crest) factor for a waveform is defined as
peak value
peak factor = .
r.m.s. value
HC2: Network basics, models, components
3.2 Current and charge
An electric current represents a flow of electric charge. The magnitude of a current
is clearly related to the rate at which charge is flowing, and therefore
dQ
I= .
dt
Rearranging this expression for charge, we get
Z
Q = I dt.
3.5 Resistance and Ohm’s law
The resistance of a material is proportional to the resistivity ρ, the length l, and the
cross-sectional area A:
ρl
R= .
A
4
,3.8 Thévenin’s theorem and Norton’s theorem
Thévenin’s theorem states that any two-terminal network of resis-
tors and energy sources can be replaced by a series combination
of an ideal voltage source V and a resistor R, where V is the
open-circuit voltage of the network and R is the resistance that
would be measured between the output terminals if the energy
sources were removed (see the top figure).
Norton’s theorem states that any two-terminal network of resis-
tors and energy sources can be replaced by a parallel combination
of an ideal current source I and a resistor R, where I is the short-
circuit current of the network and R is the resistance that would
be measured between the output terminals if the energy sources
were removed (see the bottom figure).
If we connect nothing to the terminals of a circuit, their outputs
should be the same; this is called the open-circuit voltage VOC .
Similarly, if we connect the output terminals together, each circuit
should produce the same current; this is called the short-circuit
current ISC . These quantities can create a Thévenin equivalent
circuit. From Ohm’s law, we get
VOC
R= .
ISC
3.9 Superposition
When a circuit contains more than one energy source, we can often simplify the
analysis by applying the principle of superposition. This implies that the voltage or
current through a circuit can be calculated by taking the sum of all voltages/currents
as if they were supplied by one individual source.
4.2 Capacitors and capacitance
For a given capacitor, the charge Q stored is directly related to the voltage across it.
The constant of proportionality is termed the capacitance C of the capacitor, thus
Q
C= .
V
5
,4.6 Capacitors in series and in parallel
4.6.1 Capacitors in parallel
The effective capacitance of a number of capacitors in parallel is equal to the sum
of their individual capacitances:
C = C1 + C2 + C3 + · · · + Cn .
4.6.2 Capacitors in series
The effective capacitance of a number of capacitors in series is equal to the reciprocal
of the sum of the reciprocals of their individual capacitances:
1 1 1 1 1
= + + + ··· + .
C C1 C2 C3 Cn
4.7 Relationship between voltage and current in a capacitor
The voltage V across a capacitor is given by
Z
Q 1
V = = I dt.
C C
The current I across a capacitor is given by
dV
I=C .
dt
4.9 Energy stored in a charged capacitor
The stored energy in a capacitor is given by
Z V
1
stored energy = CV dV = CV 2 .
0 2
5.5 Self-inductance
The voltage V across an inductor is given by
dI
V =L .
dt
5.7 Inductors in series and in parallel
The effective inductance of a number of inductors in series is equal to the sum of
their individual inductances:
L = L1 + L2 + L3 + · · · + Ln .
The effective inductance of a number of inductors in parallel is equal to the reciprocal
of the sum of the reciprocals of their individual inductances:
1 1 1 1 1
= + + + ··· + .
L L1 L2 L3 Ln
6
,5.8 Relationship between voltage and current in an inductor
The relationship between the voltage V across an inductor and the current I through
it is given by
dI
V =L .
dt
5.10 Energy storage in an inductor
The stored energy in an inductor is given by
1
stored energy = LI 2 .
2
9.2 Charging of capacitors and energising of inductors
9.2.1 Capacitor charging
When an initial voltage V is applied through a resistor with resis-
tance R and a discharged capacitor with capacitance C at a time
t = 0, the voltage v at a time t through the capacitor is given by
(see top figure)
v = V (1 − e−t/CR ).
For the initial current I through this circuit, the current i through
the capacitor is given by
i = Ie−t/CR .
9.2.2 Inductor energising
When an initial voltage V is applied through a resistor with resis-
tance R and a de-energised inductor with inductance L at a time
t = 0, the voltage v at a time t through the inductor is given by
(see bottom figure)
v = V e−Rt/L .
For the initial current I through this circuit, the current i through
the inductor is given by
i = I(1 − e−Rt/L ).
7
, 9.3 Discharging of capacitors and de-energising of inductors
9.3.1 Capacitor discharging
When a voltage V is removed from a resistor with resistance
R and a charged capacitor with capacitance C at a time t = 0,
the voltage v at a time t through the capacitor is given by
(see top figure)
v = V e−t/CR .
For the current I through this circuit, the current i through
the capacitor is given by
i = −Ie−t/CR .
9.3.2 Inductor de-energising
When a voltage V is removed from a resistor with resistance
R and an energised inductor with inductance L at a time
t = 0, the voltage v at a time t through the inductor is given
by (see bottom figure)
v = −V e−Rt/L .
For the current I through this circuit, the current i through
the inductance is given by
i = Ie−Rt/L .
HC3: Networks & impedance
6.2 Relationship between voltage and current
6.2.1 Resistors
In a resistor, the voltage is related to the current by the expression
vR = iR.
If i = IP sin (ωt), then
vR = IP R sin (ωt).
6.2.2 Inductors
In an inductor, the voltage is related to the current by the expression
di
vL = L .
dt
If i = IP sin (ωt), then
dIP sin (ωt)
vL = L = ωLIP cos (ωt).
dt
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