Fundamentals. This is a wide audience so I will try to cater for all. Feel free to ask questions. We will break down a reference circuit diagram into manageable sections Later we will look at how to put them back together. The reductionist approach. BITX20 bidirectional SSB transceiver.
The reductionist approach
Voltage “V” is the potential difference between two points. (Imagine as height)Current “I” is the rate of flow of charged particles. (Imagine as water flow)
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 downincreasing resistance to the current.
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
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 numbersIn a superconductor the electrons don’t ever collide with the atoms so the resistivity is zero.
Commercial resistors are often carbon or metal film.6 inches of HB pencil (Carbon) is about 16 OhmsIf we connect 6V across it we get 1 volt per inchIf we connect 12V across it we get 2V per inch (and it catches fire)
For an ideal resistor “R”
The current “I” increases with applied voltage “V” (Electromotive force)
The greater the resistance “R” to the current the less current I flows.
I = V / R
Amps = Volts / Ohms
Our resistors both resist the current. Its like one longer resistor (or pencil). We add the resistances R1 and R2 so:
R = R1 + R2
I = V / (R1+R2)
Just like in our pencil the voltage will distribute itself proportional to the resistance.
E.G if R1 is twice R2 then 1/3 of the voltage will be across R2.
So V will be 4 volts.
V2 = V1 * R2 / (R1+R2)
(We can prove this from Ohms law)
I = V1/(R1+R2)
I = V2/R2
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
So: 1 / R = 1 / R1 + 1 / R2
Or: R = (R1 * R2) / (R1 + R2)
First we combine the parallel resistors. Using R = (R1 * R2) / (R1 + R2)
So R = 2000*2000/(2000+2000) = 1K
We now have a potential divider with 1K at the top and our combined 1K at the bottom.
Using V2 = V1 * R2 / (R1+R2)
So V = 12Volts * 1000 / (1000+1000)
So V = 6 Volts.
Sometimes you can (and should) calculate what you want just using the principles of:
Resistors in series.
Resistors in parallel.
However for network analysis you need Kirchoff’s laws.
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)
For each node we can write an equation. In this case its:
0 = I1 + I2 - I3 - I4
using positive for arrows pointing to the node and negative if they point away.
Alternatively I1 + I2 = I3 + I4
For each loop we can write an equation. Since voltages add we can work round the loop:
0 = (V2-V1) + (V3-V2) +
Mathematically this is true. (But that doesn’t prove Kirchoff's law)
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
We write an equation for each node. Assuming V0 and V2 are the supply:
(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