Electronics, contents
Lecture 1: DC, circuit analysis and theorems, passive components
Lecture 2: AC, complex representation, phasors
Lecture 3: Passive filters, active components,diode,transistors
Lecture 4: Operational amplifiers
Lecture 5: Digital electronics I. binary, hex, logic gates,
Lecture 6: Digital electronics II, ADC, DAC, buses,
Lecture 7: Microprocessor function and how to use
Lecture 8: ASICs, PAL,FPGA
FYSN15,electronics
Formulas from last
time
Resistors:
U= R · I
P= U · I = R · I2
Time dependent
u(t) = R i(t)
Inductors:
L = Ф/I = μ ·N2 ·A /l
Capacitors:
C= Q / U
W=(1/2) · C · U2
C=ε · A/l
Time dependent
u(t)=C-1 ∫i(t)dt
i(t) = Cdv(t)/dt
Time dependent
u(t) = Ldi(t)/dt
FYSN15,electronics
complex numbers I
The graphical way of handling sinus currents and voltages works well for small circuits and phasors
illustrate nicely what its all about. For more quantitative analysis we need something more powerful.
We shall learn how to use complex number represenation instead. The maths of complex numbers is
quite analogous to adding vectors. Actually, simple operations on complex numbers give the same results
as the derivation and integration of sinus functions as we have done previously
(a,b)
Recall complex numbers: complex number (A) can be written as
A=a + jb. It contains 2 real numbers, a and b and the symbol j. We
know that j2=-1. j is not a real number. We can regard j as a factor
which is multiplied by b. b is the imaginary part of the complex number
and a is the real part. A can be viewed as a point in the complex plane
where the horisontal coordinate is the real part and the vertical is the
imaginary (the unit on the vertical axis is j)
You can also use polar koordinates (|A|, α) where |A| is the length of
the vector ending in (a,b). We obtain
A= |A|(cos α + j sin α) and |A| = (a2 + b2)½ och tan α = b/a
Eulers formula: ejα = (cos α + j sin α). Exponent form is often the
most convenient way of expressing complex numbers in electronics
FYSN15,electronics
complex numbers II
But whats the point? If we add the two complex numbers
(a,b) och (c,d) we get (a+c,b+d) whihc is the same as if we
had done vctor adding on the two vectors ending in (a,b) och (c,d)
To multiply a complex number with jis the same as a
phase shift, counter-clockwise by 90 degrees
j(cos α + j sin α)=- sin α + j cos α
(a,b)
To multiply by j does the same as derivation of
sinus
To divide a complex number with j is the same as a
phase shift, clockwise by 90 degrees
j-1(cos α + j sin α)=j j-2(cos α + j sin α)= -j (cos α +j sin α) = sin α - j cos α
To divide by j does the same as integration of sinus
FYSN15,electronics
Summed up
The current in complex form
i(t) = I0 e jωt
and voltage
v(t) = v0 e jωt
v(t) = R i(t)
Determine impedanse for resistor,
capacitor
inductor
These impedance are in complex
form. The voltage amplitudes
become correct (I0R, I0/ωC and
I0ωL). Also the phase is correct
since for the capacitor we divide
by j (phaseshift -90 degr) and for
the inductor we multiply with j
(phaseshift + 90º).
v(t) = (1/j ωC) i(t)
v(t) = jωL i(t)
So we have voltage, current and impedance in complex form . The relation v(t)=Z i(t) holds for
resistive, capacitive and inductive. Ohms law in complex form. Relations that we have derived for
resistors for DC take the same form for sinusoidal time dependencies if we work with the complex
Counterparts. Examples could be KVL and KCL, node – mesh analysis, resultant or serial and
parallel connection.
FYSN15,electronics
An example. RL circuit in
complex form
i(t)
U
u(t)
0=
|Z|
I
0
U0L=jωLI0
U0LR= RI0
I0
Resultant impedance when serial:
Z= ZR + ZL= R + jωL
u(t) = Z i(t) = (R + jωL) i(t).
The absolute value of Z: |Z| = [R2 + (ωL)2]½ so
U0= [R2 + (ωL)2]½ I0
and
Phase angle is given by
tan ϕ = ωL/R
FYSN15,electronics
Passive filters
vin(t)
vut(t)
Transfer function H(t) = vut(t)/vin(t)
lowpass
highpass
bandpass
bandstop
IRL
You loose some amplitude
Cutoff is not sharp
Define cutoff as H=1/√2
Power 50%.
FYSN15,electronics
Low pass RC filter
Cutoff at ω=1/RC
|H(ω)| = 1/(1+ (ωRC)2)½
Θ=-arctan(ωRC)
If v(t) step voltage
V0
0
Vut(t)=V0[1-exp(-t/RC)] = V0exp(t/RC)
Vut(t)=V0exp(-t/RC)
FYSN15,electronics
General problem
-All traces on a circuit board have:
-Resistance, inductance and capacitance.
-Values are small but results in a finite,
exponential rise and fall of digital pulses and
clocks.
-This limits the frequency of digital circuits
due to the loss of timing accuracy compared
to infinitely sharp pulse edges.
FYSN15,electronics
RLC,bandpass filter
ω0= √LC
|ωc1-ωc2|=R/L
FYSN15,electronics