# BASIC ELECTRICAL CIRCUITS AND ANALYSIS

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```					  BASIC ELECTRICAL CIRCUITS AND ANALYSIS
Ref:   Horowitz, P, & W. Hill, “The Art of Electronics,” 2nd. ed., Cambridge (1989).

CHARGE - symbol C, unit is coulomb; 1 coulomb = 6.2415×1018 electrons

CURRENT - symbol I or i, unit is amp(ere) (A); flow of electrons per unit time

1 A = 1 coul/sec

VOLTAGE - symbol E or V, unit is volt; potential difference, 1 joule of work is
needed to move 1 coulomb of charge through a potential difference of 1 volt (V)

By convention (thanks to Ben Franklin), current flows from more positive (less
negative) to more negative (less positive). Electrons are the real carriers; they go in
the opposite direction.

RESISTANCE - symbol R, unit is ohm

OHM’S LAW: a potential difference of 1 volt will cause a current flow of 1 ampere
through a resistance of 1 ohm.

E=IR

RESISTORS IN SERIES - RT = R1 + R2 + R3 + .. + Rn

1  1  1  1                       1
RESISTORS IN PARALLEL -              = +   + +                  +
RT R1 R2 R3                      Rn

R1R2
TWO RESISTORS IN PARALLEL:              RT =
R1 + R2
E2
POWER:        P = EI = I R =
2

R
VOLTAGE DIVIDER:
R1
R2
Vo = VB                                 VB
R1 + R2
R2       Vo
KIRCHHOFF RULES
The law of conservation of energy and conservation of charge lead to two
fundamental rules for circuit analysis. These rules are called Kirchhoff's rules, and
they are used to determine the value of the electric current in each branch of a
multi-loop circuit.
The first rule, the node equation, states that the sum of the currents into a specific
node (junction) in the circuit equals the sum of the currents out of the same node.
[A junction is a place in a circuit where more than two wires join.] Electric charge is
conserved: it does not suddenly appear or disappear; it does not pile up at one point
and thin out at another.

I1                     I3
I2

I1 + I2 - I3 = 0                       ∑I   n   =0

The second rule, the loop equation, states that around each loop in an electric
circuit the sum of the emf 's (electromotive forces, or voltages, of energy sources
such as batteries and generators) is equal to the sum of the potential drops, or
voltages across each of the resistances, in the same loop. All the energy imparted
by the energy sources to the electrons carrying the current is equal to that lost by
the electrons in useful work and/or heat dissipation around each loop of the circuit.

R1                                 R2

V2
V1
R3

V1 - IR1 - IR2 + V2 - IR3 = 0                          ∑V   n   =0
SUPERPOSITION: The total current in any part of a circuit is equal to the algebraic
sum of the currents produced by each source separately. To calculate, short all
other voltage sources and open all other current sources.

VOLTAGE SOURCES: (for ideal voltage source, Rs = 0)
I
R
s

IDEAL
V

V

Thevenin's theorem: Any network of linear sources and resistors with two
terminals can be replaced by an equivalent circuit comprising an ideal voltage
source and a series resistance.

Rth

BLACK                     Vth
BOX

To determine the Thevenin equivalents, you can either calculate them or make two
measurements.
1.    Open-circuit voltage Voc. This is the Thevenin equivalent voltage.
2.    Short-circuit current Isc. The Thevenin resistance is then Voc/Isc.

CURRENT SOURCES: (for ideal current source, Rsh=∞
I         IDEAL

I                   Rsh

V
Norton's theorem: Any network of linear sources and resistors with two
terminals can be replaced by an equivalent circuit comprising an ideal current
source and a shunt resistance. This is analogous to Thevenin’s theorem.

BLACK                                 In           Rn
BOX

The Norton resistance is the same as the Thevenin resistance, and the Norton
current is the ratio of the Thevenin voltage to the Thevenin resistance.
Norton and Thevenin are equivalent. Thevenin is preferred for high-impedance or
constant-voltage circuits and Norton is preferred for low-impedance or constant-
current circuits.

Maximum power transfer theorem: For a given voltage source VB and series (source)
resistance RS, the value of load resistance RL that maximizes the power delivered to the load is
equal to the source resistance.
PL = I 2 RL
RS
VB
I=
RS + RL
I                                                            VB2 RL
VB                                                          PL =
( RS + RL )
2
RL
Set dPL/dRL = 0 to find

RL = RS
CAPACITANCE

Capacitance of two parallel plates:
C         0.224KA
C=
d
K = dielectric constant
A = area
d = spacing

Capacitors in series:
1  1  1   1              1
= +   +   +        +
CT C1 C2 C3               Cn

The total charge is distributed among
the series capacitors, and the total
capacitance is decreased

For two series capacitors:
C1C2
CT =
C1 + C2
Capacitors in parallel

CT = C1 + C2 + C3 + ….. + Cn

All circuits have "stray" capacitance.
EX: MOST Coaxial cables have capacitance ~100 pF/meter.
CAPACITIVE REACTANCE

1
XC =
jω C
j merely indicates that the phase difference between the
current and the voltage φ is 90°. The voltage and current are
out of phase, and the power dissipation is zero.

Ignoring the phase term,

1
XC =
ωC
XC → ∞ at ω = 0 (DC)
XC → 0 when ω → ∞
CHARGING A CAPACITOR

Immediately upon closing the switch,

dQ
I=
dt
Applying Kirchhoff's loop law

Q    dQ Q
VB = VR + VC = IR + = R   +
C    dt C
This differential equation has as a solution

Q = CV (1 − e − t / RC )
VB
I = (1 − e − t / RC )
R
RC is time constant τ
CAPACITOR CHARGE vs. TIME

at t = 0,                  as t → ∞,
Q=0                       Q → CVo
VC = 0                     VC → Vb
VR = Vb                     I→0
Vo
I=
R

at t = RC, capacitor is 63% charged
CAPACITOR DISCHARGE

Place resistance R across capacitor C. Close switch at t=0.
LOW-PASS FILTER I

Every system has/is one. It is a voltage divider with a
frequency-dependent component.

R

C

Simple one characterized by a time constant τ = RC

TIME DOMAIN RESPONSE (to a step function)

RISE                               DECAY

−t                               −t
V (t ) = V0 (1 − e        τ
)       V (t ) = V0e         τ

1

0.8
RESPONSE

0.6

0.4

0.2

0              1                  2                     3
TIME / TIME CONSTANT
LOW-PASS FILTER II

FREQUENCY DOMAIN TRANSFERT FUNCTION T(ω)

1                            τ = RC
T (ω ) =
ω c = 1τ
1+ω τ     2 2

Cut-off frequency ωc where ω2τ2 = 1. T(ω) is down 3dB = 0.707.
Falloff at high frequencies is –6dB/octave = –20dB per decade.
1

0.8
RESPONSE

0.6

0.4

0.2

0        1        2       3           4        5
RELATIVE FREQUENCY

1
RESPONSE

0.1

0.01
0.1               1            10             100
RELATIVE FREQUENCY
HIGH-PASS FILTER

Many systems have a high-pass filter to block response at DC
and low frequencies.

Simple HPF also characterized by τ = RC,                                  ω c = 1τ

Cut-on frequency ωc where ω2τ2 = 1. T(ω) is down 3dB = 0.707.
Rise at low frequencies is 6dB/octave = 20dB per decade.

C

R

FREQUENCY DOMAIN RESPONSE

ωτ                              ωτ
T =                                =
1+ω τ      2 2
ω2
1+ 2
ωc
1                                                  1

0.8
RESPONSE

RESPONSE

0.6
0.1
0.4

0.2

0                                               0.01
0   1             2            3                  0.01        0.1         1      10   100
RELATIVE FREQUENCY                                          RELATIVE FREQUENCY
BANDPASS FILTER I

Combine LPF and HPF

R
C
C
+                       R

If they do not interact,

1                      ωτ
T=                                      ×
1+ω τ  2 2
LP          1 + ω 2τ HP
2

1

0.8
RESPONSE

0.6

0.4

0.2

0       5            10       15    20
RELATIVE FREQUENCY

1
RESPONSE

0.1

0.01
0.01     0.1         1       10      100
RELATIVE FREQUENCY
BANDPASS FILTER II

If CLP = CHP AND RLP = RHP OR τHP = τLP = τ = 1/ω,

ωτ       ωτ
T=         =
1+ω τ
2 2
ω2
1+ 2
ωc
0.6

0.5
RESPONSE

0.4

0.3

0.2

0.1

0                       1                 2          3
RELATIVE FREQUENCY

1
RESPONSE

0.1

0.01
0.01      0.1         1       10   100
RELATIVE FREQUENCY
SIGNAL BANDWIDTH

Signal bandwidth of a low-pass filter
• Frequency where filter transmission (voltage or
current) falls to -3dB (1√2 = 0.707) of transmission at
DC (typically 1)
• Also referred to as half-power point, where power
transmission has dropped to 0.5 (-6dB)
• Power transmission = [voltage transmission]2

Signal bandwidth of a band-pass filter
• Frequency where filter transmission (voltage or
current) falls to -3dB (1√2 = 0.707) of transmission at
the maximum transmission

Without a high-frequency cutoff, the signal bandwidth of a high-
pass filter is undefined.
TRANSFORMER

Some inevitable losses (heating)
Impedance ratio goes as square of turns ratio NS/NP.

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