Transcript Document

Fundamentals
This is a wide audience so I will try to cater
for all.
Feel free to ask questions
The reductionist approach
We will break down a reference circuit
diagram into manageable sections
Later we will look at how to put them back
together
BITX20 bidirectional SSB transceiver
A BITX20 single stage
A simplified single stage
+12 V
R1
R2
0V
A potential divider
+12 V
+12 V
R1
R1
V
R2
0V
R2
0V
A Single resistor
+12 V
R1
V
+V
I
R2
0V
R
0V
Voltage and Current
Voltage “V” is the potential
difference between two points.
(Imagine as height)
+V
I
R
Current “I” is the rate of flow of
charged particles. (Imagine as
water flow)
0V
What are resistors and conductors?
An applied voltage across an object causes an electric
field across the material.
The electric field accelerates any “free” electrons in
the material. This motion is the electric current.
Electrons collide with atoms which slows them down
increasing resistance to the current.
What factors determine resistance?
The greater the length of an object the more
resistance it will have.
The greater the cross sectional area the less resistance
it will have.
Resistance = Resistivity * Length / Area
What properties affect resistivity?
Some materials (e.g. copper) have lots of free
electrons and have low resistivity, (good conductors).
Others (e.g. glass) have almost no free electrons.
Semiconductors (e.g. silicon) have modest numbers
In a superconductor the electrons don’t ever collide
with the atoms so the resistivity is zero.
Some real resistors
Commercial resistors are often carbon or metal film.
6 inches of HB pencil (Carbon) is about 16 Ohms
If we connect 6V across it we get 1 volt per inch
If we connect 12V across it we get 2V per inch (and
it catches fire)
Ohms Law
For an ideal resistor “R”
The current “I” increases with applied
voltage “V” (Electromotive force)
+V
I
The greater the resistance “R” to the
current the less current I flows.
R
I=V/R
Amps = Volts / Ohms
0V
Resistors in series
V
R1
I
I
Our resistors both resist the
current. Its like one longer
resistor (or pencil). We add the
resistances R1 and R2 so:
R = R1 + R2
R2
I = V / (R1+R2)
0V
The potential divider
+12V
+12V
R1
I
Pencil
V
I
R2
0V
0V
The potential divider
+12V
Just like in our pencil the voltage
will distribute itself proportional
to the resistance.
R1
I
V
I
R2
0V
E.G if R1 is twice R2 then 1/3 of
the voltage will be across R2.
So V will be 4 volts.
The potential divider
V1
V2 = V1 * R2 / (R1+R2)
R1
I
V2
I
(We can prove this from Ohms law)
I = V1/(R1+R2)
I = V2/R2
R2
0V
Resistors in parallel
V
I1
I2
Our resistors both carry current so
its like one thicker resistor. We add
the currents so I = V / R1 + V / R2
From Ohms law we have I = V / R
R1
0V
R2
So:
1 / R = 1 / R1 + 1 / R2
Or:
R = (R1 * R2) / (R1 + R2)
A simple network
First we combine the parallel resistors. Using
R = (R1 * R2) / (R1 + R2)
+12V
1K
I1
So R = 2000*2000/(2000+2000) = 1K
I3
I2
2K
2K
V
We now have a potential divider with 1K at
the top and our combined 1K at the bottom.
Using V2 = V1 * R2 / (R1+R2)
0V
So V = 12Volts * 1000 / (1000+1000)
So V = 6 Volts.
Kirchoff’s laws
Sometimes you can (and should) calculate what you want
just using the principles of:
Resistors in series.
Resistors in parallel.
Potential dividers.
However for network analysis you need Kirchoff’s laws.
Kirchoff's 2 Laws
We have already started using these.
1) Current is conserved. So if you add up all the current
into a connection (a node) it will add to zero.
2) Voltage measurements are consistent. So if in a loop
you add up the voltage differences between
successive nodes the result will be zero.
There are some pitfalls so we need to look in detail.
Current is conserved
I3
I1
I2
I4
To apply this rule we mark an
arrow on every link in a circuit
and label the links I1, I2 etc.
You regard current in the direction
of your arrow as positive,
otherwise its negative.
(It doesn’t matter which way you
mark the arrows)
Current is conserved
For each node we can write an
equation. In this case its:
I3
I1
I2
I4
0 = I1 + I2 - I3 - I4
using positive for arrows pointing
to the node and negative if they
point away.
Alternatively I1 + I2 = I3 + I4
Voltage measurements are consistent
V1
I1
V2
I2
Loop
For each loop we can write an
equation. Since voltages add
we can work round the loop:
0 = (V2-V1) + (V3-V2) +
(V4-V3)+ (V1-V4)
V4
V3
Mathematically this is true. (But
that doesn’t prove Kirchoff's
law)
Direction of voltages
V1
V2
I
Loop
V4
R
V3
If we mark a current arrow I on a
component (e.g. a resistor)
then for V= I * R we must
regard the tail of the arrow as
positive.
So in this example
V2-V3 = I * R
Previous example using Kirchoff’s laws
+12V
I2+I3=I1
1K
(V-0)/2K+(V-0)/2K= (12-V)/1K
I1
I3
I2
2K
V
V/2+V/2= 12-V
V=12-V
2K
2*V=12
V=6 volts
0V
Using Kirchoff’s current law again
+12V
I2+I3=I1
1K
(V-3)/2K+(V-0)/2K= (12-V)/1K
I1
I3
I2
0V
(V-3)/2+V/2= 12-V
V-3+V=24-2*V
2K
2K
3V
V
+
4*V=27
V=6.75 volts
The current law for multiple nodes
We write an equation for each node.
Assuming V0 and V2 are the supply:
V2
R2
R3
I1
I4
I3
V1
V3
R5
I2
R1
V0
I5
R4
(At V1)
I1+I2+I3=0
(At V3)
I4+I5 -I3=0
Ohms law gives equations for I e.g.
I2 = (V0 – V1)/R1
Note not V1-V0 !!
Solve using simultaneous equations.
7 unknowns: V1, V3, I1, I2, I3, I4, I5
7 equations: 2 above plus 5 Ohms Law
Questions?
Feedback?