Electronics Lessons: The Transistor Simple Circuits 2 (Lab 3)

Now that we've looked at resistors, transistors, and capacitors. Lets see if we can put them to some use to build another really simple circuit, this will be a bit more complicated than using the transistor as a switch.

Lets start by build an amplifier;

I don't expect that you know how to design the amplifier yourself (that's why you're here?), so I'll hold your hand all the way through the design process, we'll be using at least one of the components in a way that I haven't described that it could be used like that yet, so there will be a little deviation to describe that, but hey, this is only lab3 and you're already building something pretty cool... you know the components from the past 5 lessons, and you're aware of one of the building blocks that we'll be using from a past lab lesson (lab 1).

Perhaps the holly grail of all analogue circuits is the amplifier, I don't know why if I'm honest, but just about everyone seems to want to know how to make amplifiers, and those that do know how to make amplifiers, well, in spite of the maths involved seem to get a kick out of making them, and even those that shy away from the maths have a really good go following reference circuits.

This is the first amplifier circuit I was ever introduced to:

Common Emitter Amplifier
The last amplifier that we looked at was an emitter follower or common collector amplifier, that's a device that amplifies current, whilst leaving voltage unchanged, we used that to boost the amount of current available from a low power source so that we could turn on a light.

The amplifier that we'll be looking at now doesn't amplify current, it amplifies voltage.
The first thing to do is introduce you to the circuit.

Firstly, lets break the circuit down,
The little round thing with the waveform shape in it is a signal source,

For this demonstration we'll set the output of the generator to a sine wave, with a frequency of 1KHz and we'll set the amplitude (size of the signal) to 1v.

This means that the waveform outputted by this generator is:
A nice curvy wave form, that is modelled by the math function sin(x).
It goes between +1v and -1v.
And it does this one thousand times a second (a complete cycle take 1millisecond -1ms.)

The next component we see is a capacitor, in the lesson on capacitors I talked about the capacitor acting like a storage tank for electricity, saying that you could charge a capacitor and discharge a capacitor, here we are using a different property of the capacitor, it's ability to block DC (direct current) signals, and allow AC (alternating current) to pass.

The reason that we want to use this is because we need to set a bias on the transistor so that it's working in a linear region, -don't worry about that for now, I'll explain more later, right now we'll look at the DC blocking capabilities of the capacitor.

When we use a capacitor like this it's called a coupling capacitor.

Coupling capacitors
The problem with the signal source that we have is that it's waveform is centred around 0volts.
fair enough the transistor will turn on then the voltage rises above zero volts, but when it's in it's negative half of the cycle (or anywhere in the cut-off region), the transistor won't be conducting at all. So we need to move the centreline of the wave form up so that it's centred around a positive voltage, that'll let our transistor work in a region where it's on a little right up to on a lot, (remember how we talked about the transistor as a variable switch? now we're going to use it as a variable switch.)

Before we introduce the transistor into the mix we'll look at how we get that wave form to climb up to a more positive voltage using coupling capacitors.

The diagram below shows some things that should be familiar to you from lab 1.

Ignore the capacitors and signal generator for now.
There is a positive voltage source (+9v) and a zero voltage source (0v) (just like the battery terminals.) And there are two resistors set up as a potential divider, followed by two more resistors set up as potential dividers.

Using the maths learned in Lab 1 we can see that the voltage at the blue probe is 6v, and the voltage at the pink probe will be 3v. -remember Vout = (Z2/(Z1+Z2)) * Vin)?

So we know that there is a DC voltage of 6v at the blue probe, and a DC voltage of 3v at the pink probe.

(now we'll include the capacitors and signal source again).

The resistors used are significantly high enough to not interfere with the source when the AC signal is introduced.

Reading from left to right.
The first red probe, measures voltage a +/-1v centred around 0v.

The capacitor then blocks any DC voltage going to, or from that point in the circuit moving onto the next point, (blue probe).

If there were no DC component here we'd measure +/- 1v centred around 0v, however, the potential divider here means that there is a 6v DC voltage at this point, so that +/-1v is superimposed on this 6v source. meaning that the wave form is now alternating between +7v and +5v (+/-1v centred around 6v).

Moving a little further on there is another capacitor that stops this +6v DC component from moving to or from any other part of the circuit. Here we reach the pink probe.

Now the wave form has had all it's DC component removed and is once again 0v +/-1v.
but our potential divider has set a DC voltage of 3v onto this part of the circuit. So once again the AC voltage is superimposed on the DC voltage giving a wave form that's +/- 1v centred around 3v (+4v to +2v).

Here is what the wave forms are looking like at each part of the circuit.

Biasing
I suppose that the first question that needs to be answered is Why do we bias amplifiers. I gave a pretty poor description above about the fact that it doesn't turn on on the negative half cycle of the input wave, (and that is true).
but to answer the question properly again we're going to need to look at the data sheets for the component again.

Lets take a really simple (and now really rather old transistor, the BC108).

http://www.datasheetcatalog.org/datasheet/MicroElectronics/mXuvuyv.pdf

As before there are a couple of important things to look at in the data sheet.

The first is Base - Emitter voltage (Vbe) you can see that this is 0.6v

Next look at the gain value of the transistor, (this is called hfe) this is (for group A components typically 170).
this large hfe value means that we know that the collector current is much much larger than the base current.

so Vbe = 0.6
Gain = large
Ic >> Ib
Ie = Ic + Ib

You'll find a graph in that data sheet called "output characteristics",
The output characteristics these define the voltages that you need across the device to make it work in a region called the linear region, this is the region where the device is going to amplify small and large signals equally without distorting either.
There are several lines on the device, all are in a shape that goes up at the start, and has a sharp knee, then settle into a straight line, the different line correspond to different base currents.

On the graph below I've marked two areas in red, you want to stay out of these regions because either the transistor is off or behaving in a non linear way (very on), (the one closest to the X axis is the cut off region) and the one closest to the Y axis is the saturation region.

The first thing that you're going to need to determine is the Resistor RL, this is the load resistor for the amplifier,

First, you're going to need to extend the graph so that we can see it working inside the voltages that you're working with, (the x axis only goes up to 5v, and we're going to use a 12volt supply).
(I'll assume by now that you've murdered an ATX power supply and therefore have a 12v supply (and are no longer relying on 9v batteries).

Now you need to draw line on the graph between the point the you want to set as ICmax, (on the Y axis, and your supply voltage (Vcc) on the X axis.

Set your ICmax somewhere below the highest marking on the chart, I've chosen 1.75mA

This new red line is called the DC load line, and in the middle of this line is a point that's at the middle of the linear operating region of the amplifier, this is the quiescent point, or Q point.

You can see the green dot in the middle, set at around 5.8v (you might also be realising that this is not an exact science!)

Right... so what we know now is:

Ic(max) = 1.75mA
and Vcc = 12
Vce(q) = 5.8
Ic(q) = 0.9mA

We need to make a decision now as to what we want the voltage of our emitter to sit at, some guides suggest splitting the voltage equally between the Rc, Rb and the transistor itself, however, I think that setting Vre as 1v doesn't seem like a terribly bad thing to do! -you could chose a different value.

To find a value for Rc
We use the forumla

Ic(max) = (Vcc - Vre)/Rl

0.00175 = (12 - 1)/Rl

Which works through to give us the equation
Rl = 11/0.00175 = 6285
The closest standard value is 5.6K or 6.8K

It'll be perfectly fine to use a 5.6K this will increase your IC(max) value, but you're still far below the absolute maximum rating given on the first page of the data sheet. the higher the voltage here the more restrictive you're being on the input signal voltages that could be used before the amplifier clips the signal.

The current gain of the transistor we know from the data sheet is 170

the gain is the value of the collector current, over the base current

hfe = Ic /Ib

hfe = 170 and Ic = 1.75mA
Therefore Ib = 1.75ma / 170 = 10uA


Given that we now know the values of Vre, Vbe and Ib we can start to set the bias of the base leg of the transistor.

To find the value of R2 we use the following formula.

R2 = (Vre + Vbe)/(10 * Ib)

R2 = (1 + 0.6)/10 * 0.00001)
R2 = 1.6/100 * 100^-6 = 16000 -15k is the closest standard value

and we can also figure out the value for R1

R1 = (Vcc -(Vre + Vbe))/(11*Ib)
R1 = (12 - 1.7)/(11*0.00001)
R1 = 10.3/0.00011 = 93636 - 100k is the closest value


Now we need to go about setting that emitter resistor Re.
We calculate this using Ohms law
We know that the emitter current is the sum of the base and collector current
Ie = Ic + Ib
Ie = 1.75ma + 10uA
Ie = 1.76mA

We know that Ohms law says V/I = R

Vre / Ie = Re
1 / 1.76mA
1/0.00176 = 568 Ohms (and the closest standard value is 560Ohms)

Earlier we talked about coupling capacitors,
on that last amp circuit you'll see red and blue probes, this is what the signal looks like at those points.

Input vs. Output
Finally, let's move our probes to measure the input and output signals.

Look closely at that chart.

The red line is the input signal, and the blue line is the output. We see that the blue line is much bigger than the red line, but also notice that as the red line is going up, the blue line is going down.
This amplifier has switched this signal upside down.

This type of amplifier is called an inverting amplifier.

Gain
The gain of am amplifier is its output divided by its input.

remember the current gain of the transistor was the collector current, over the base current.
(actually it's emitter current over the base current, but since the gain factor is so large we assume that Ie is just about equal to Ic)

The voltage gain works the same.
The input voltage is +/-1v, (e.g 2v).
and the output voltage swing is +/- 4v (eg 8v)

So the gain of this amplifier is 8/2 = 4

The signal coming in is made 4 times larger.

Electronics Lessons: The Transistor Simple Circuits (Lab 2)

So in the last lesson I talked about the transistor, I talked about how you could turn the transistor on to a fully conducting state.

I also mentioned that transistors could be used as current amplifiers.

You'll use a current amplifier when you need a little more power to switch on an output than that output can supply.

We discussed before how the transistor was off, and how it could be turned on.

So, lets make a device that puts a light on with the presence of a voltage, even if that voltage source doesn't have the power (current sourcing capability) to light up the light itself.

In this tutorial we're interested in two areas of the transistors output, because we're going to use the transistor as a switch we'll either have the transistor in the cut-off region (off), or the saturation region (on) of it's output characteristics. (There are the areas marked in red in the chart below).

The Circuit
What we do is connect the input to the base of the transistor, we use a resistor to ensure that not too much current is pulled from the voltage source that we're detecting.

We'll also put a resistor between the circuit voltage source, and the transistor to limit the current being drawn from the source.

Our light will be an LED, we're expecting that the voltage going to the LED will be 5V at most, but as little as 3v so we look at some data sheets for LEDs and we find that the there are a couple that we can't use, (some have Vmax as 4v), and others will tolerate a higher voltage, but won't light with only 3v.

Eventually we come across the L-53GD-5V made by Kingbright

It has the following characteristics.

So we can see that it'll be on, and bright at 5v, and it will turn on, (though only be half as bright) at 3volts, (we're interested in sensing 5v Logic levels and 3.3v Logic levels right?)

Elsewhere in the data sheet it tells us that the maximum current is 17mA
We know that our greatest voltage going through the LED will be 5v
So we use ohms law to determine the resistor needed.

5/0.017 = 294

So we really want to make the total resistance between the supply voltage and the indicator LED around 300Ohms
In case you're interested

3.3/300 = 0.011 or 11mA available for the 3.3v logic level to light the LED.

I say the 3.3v, remember this is a current amplifier, not a voltage amplifier. The voltage is going to remain the same as what's going into the transistor base.
Anyway, at 3.3v the LED only draws some 6mA, so there is plenty of current available. (and that 11mA is below the devices max draw of 17mA

Schematic
Here is the schematic of the circuit.

And here's what happens when a logic level of 1 (+5v) is applied to the base resistor

When the circuit is on it pulls around 6ma from the logic source, and the current going through the LED is about 14mA.

Stay tuned to see this idea scaled up!

Electronics Lessons: The LED

I briefly touched on LEDs previously in the output devices post a while back.

That post was designed to be a little bit of a primer into all kinds of output devices, to get you thinking about the kind of output that you want to put on your projects.

Now I'm going to start looking at those output devices in a little more detail.
I'll skip over light bulbs for the moment, other than small signal lamps, (torch bulbs) there aren't a lot of light bulbs that are particularly practical to drive from beginner projects, they don't solder directly to boards (requiring bulb holders), and lots require either mains voltage, or the type of current that's really going to burn you if it goes wrong. -I'll come back to light bulbs as output devices later in the series mostly because it'll be necessary to explain a project that I've written up and am waiting to publish.)

LEDs
LEDs are light emitting diodes.

When you buy an LED you'll normally find that appears in a round shape with a dome top.
the round shape has a small ledge of skirt at the bottom of it and one of the sides of this next to a leg is flat.
You'll also find that when you buy the component that one leg is slightly shorter than the other.

The reason that a diode has these characteristics is:
The round body: this is a nice easy package type to make, however, LEDs to come in all different shapes and sizes.
A look here: http://www.rapidonline.com/Electronic-Components/Optoelectronics will help to show just how many different shapes and sizes!

The reason that LEDs generally have a dome shaped top is so that they can be seen from a variety of different angles. if you look at surface mount LEDs these will generally have flat tops, this is so that light guides, that will carry the light from your board to your front panel can fit on top. (a light guide is like a fibre optic cable, except that it's generally made of clear plastic (not glass) and is quite thick (measured in millimetres not microns), and is rigid.

The reason that there is one leg shorter than the other, and the reason for the flat spot on the case is that these show which leg is negative.

The component is a diode and will only conduct one way (when used within specification). and when the diode is conducting it will let you know by giving out light.

Circuit symbol
The LED is a diode, therefore is shares the same basic symbol as the diodes, (arrow with a flat bar on it) however, as this device emits light there are also two arrows the come out of the diode.

I've only drawn the basic, and now standard symbol for the LED.
however, you may find that there is a circle around the diode symbol, you may find that the arrows have a zig-zag in the middle, the arrow heads may or may not be filled.

Some experiments to try
There are some experiments that you may wish to try.

The first experiment that you should do is either using a power supply with a variable voltage. or using batteries, the examples that I'll use will involve batteries.

For this experiment you're going to need 3AA batteries, (1.5 Volts).
and a 1K resistor (note this is not the same colour bands as shown in the pictures, that's just a picture).
and a 100Ohm resistor

First connect all the batteries end to end inside a battery holder.
Connect the 1k resistor to the battery packs' negative terminal.
Now connect the other leg of the resistor to the negative leg of the LED
Connect the positive leg of the LED to the first battery in the battery pack.

You'll see that the LED doesn't come on at all.

That's because in order to make the LED light up around 2volts is used up. since you don't even have 2 volts this thing won't light up.

so now connect the positive leg of the LED to the second batteries positive terminal.

Now you've got 3v as a supply voltage, so you've got enough to make the LED light up.
but it'll be very dim.

That's there isn't much current flowing through the device, and the amount of current going through the device is proportional to how bright it looks.

To figure out how much current goes through the device,
start with your supply voltage (3v)
take away the forward voltage of the LED (2v) you're left with 1v

Now use Ohms law to determine how much power is flowing through the circuit
1/1000 = 1miliamp (barely enough to even make it work!)

Now swap your 1k resistor for the 100Ohm, resistor.
the LED should appear to be brighter. the reason is that more current is flowing
1v / 100 Ohms = 10milliamps of current flowing.

Now repeat the same steps on the positive voltage of the last battery.

This time the LED should be reasonably bright with the 1k resistor and much brighter with the 100Ohm resistor

this is due to the fact that there is now 4.5v supply. (subtract 2 for the LED) leaves 2.5v being dissipated in the resistor.
2.5/1000 = 2.5milliamps.

Before you substitute the 1k resistor for the 100Ohm resistor be sure to take a look at the resistor from the top (birds eye view looking down on the dome,) and from the side.

Notice how it looks brighter at the top. and as you come down to look at the side view, the light almost disappears completely?

Now put that 100 Ohm resistor in the circuit.
When you put the 100ohm resistor into the circuit
2.5/100 = 25milliamps.

You may now find that either the LED is lit very brightly, or that it sort of lit brightly for a bit, but now doesn't work.
It's not going to have exploded! but it might have failed inside. (if you pump enough power through an LED it will explode though!)

Looking at data sheets
To find out why your LED doesn't work any more we'll have a look at the data sheet.

The data sheet that I'm using for reference is:
http://datasheet.octopart.com/L-53GD-5V-Kingbright-datasheet-578943.pdf

Data sheets give you all kinds of useful information, some are better than others, this one just happens to give a lot of information.

So... lets look at the data sheet and find out why the things that were happening above happened.
why varying the resistor made a difference to the brightness, why varying the voltage made a difference, and why the viewing angle made a difference.

We'll start with the viewing angle:
At the end of the data sheet is a diagram called spatial distribution. this diagram is a little confusing at first, but here is now to read it.

firstly there is a circle at the bottom, you need to imagine that the dome of the LED is inside this circle, now you notice that there is a straight line going up to the marking 0 degrees.

This means that you're looking straight down on the LED.
you can see on the blob shape in the middle touches this 0degree line at the top.
moving along this reaches 1 on the brightness scale.

Now if you change your viewing angle to 30degrees, you find that the blob touches a different brightness line, this line is 0.5
This means that at 30degrees that LED is half as bright as when looking at it from the top.

Now if you move anywhere in the 60 - 90degree region, you'll see that the blob doesn't even touch the line, basically, you'd be lucky to see any light at all.

Clearly this is important if you're using the light to indicate anything.

For a start, lets say that your device sits flat at one end of a room, sitting down at the other end of the room your viewing angle will be so shallow that you won't see it at all, OK so this might mean that you just don't know if your cat is inside of outside, no big deal.

Now consider that you're using these LED as brake lights? you're braking in one lane, people behind you can see, but what about people who are only slightly behind and alongside. they can't see that you're braking. what if you try to use these as direction indicators/turn signals? now it's really dangerous as people who need to know your intentions are now given no warning of your intentions.

You might think those LED bulbs are super bright and look cool (and they do) But what if they aren't bright enough? what of the viewing angle? what if they are too bright? this is why (in Europe at least) all car parts have to be E marked, to say that they are tested and approved.

Next let's look at the voltage characteristics:
What these charts are telling us is that as more voltage is applied, the device will get brighter.

And that as more voltage is applied, more current is passed through the device.
This is exactly what we saw in the experiment, as voltage increased, the device got brighter.
As the current limiting resistor was decreased, more current flowed through the device, which was proportional to the voltage "used" in the device, and so the device got brighter.

Next lets look at the maximum ratings:
The maximum ratings are there to tell us what conditions we can place on the device, this this case the maximum rating tell us a few things.

Firstly: forward current (max) the maximum current permitted to flow through this device is 17.5mA (when the voltage is 5v)
This means, select your current limiting resistor such that only this much current can flow.

Then we have absolute maximum ratings:
maximum power: 85mW this means select your input voltage, and current limiting resistor such that less power than this is able to flow, (for lower voltages you can use lower value resistors.)

The absolute maximum voltage rating is 6v, putting more voltage than this through the device will make it stop work, (possibly in a spectacular way.)

Last, but not least we have temperature ratings.
now these are important if you're planning on making a project that'll go in a particular place.
With temperatures of -40 - 60 degrees Celsius, these devices aren't going to be suitable for making stuff that you plan to use in the Antarctic, nor will they be suitable for devices that are left in the baking hot equatorial sunshine.

But lastly, they tell you how long you can hold a soldering iron on the leads before the heat will damage the component. this goes back to the Electronics workbench shopping list of equipment.

If you buy a low power soldering iron, you will have to wait longer for things to heat up. That's because it take the heating element longer to heat up, in the same way that a 2000w kettle boils faster than a 1000w, a 40W soldering iron heats up the components quicker than a 15 or 20Watt one, meaning that you don't have to hold it on the work as long. meaning that you're less likely to damage the components.

Electronics Lessons: The resistor and Ohms law

I started to write a post on what is electricity, but soon realised that subject was far too broad, far to detailed, far to simple and far to complicated all at once...

Suffice to say, what electricity is doesn't matter so much as what you can do with it.

I'm going to create a series of blog posts as a kind of electronics 101, the idea being that anyone can start at the beginning, learn about components, and learn how to build up circuits.

Throughout these lessons I'll be making a series of good analogies, and bad analogies.

Resistors

Theory
In Physics things have potential energy when they are sat still, a weight sitting on top of a book shelf has potential energy, if you nudge the weight to the edge it'll fall off, that potential energy has been converted into kinetic energy.

For the purpose of a resistor, think of electricity as like a tank of water, sat on top of a hill, the electricity, like anything is itching to get to the ground, and will take the path of least resistance.

A resistor adds resistance to the flow of electricity.

Look at the two pictures, the first shows a tank of water with a small opening, when you think about it you'll see that in this case the water will run down slowly, the narrow channel adds resistance and stops the water just gushing out.

The second picture has a bigger opening, in this case the water just gushes out, you can't stop it because the channel is too wide, the water gushes out with such force that you couldn't just put your hand over it.

There are calculations that you could do, knowing the size of the header tank and the size of the outlet to know with what force the water is pouring through the hole.

In electricity, the size of the header tank is measured in Volts, the size of the pipe or opening is measured in Ohms, and the force that the water pushes is not measured in PSI but is measured in Amps.

Resistance, Voltage and Current
The equation is simple.

The voltage, divided by the resistance is equal to the current.
V/R = I
so say you have a 9v battery, and a 100Ohm Resistor.
9/10 - 0.09A (or 90mA)

Next you might be thinking about about how much power that actually is. (if your resistor can't handle the power it's get hot and fail, sometimes quite spectacularly.)

Voltage, Current and Power
Another electronics law tells us that Power = Voltage multiplied by Current
In the example above the voltage was 9V, and the current dissipated was 0.09A

9 x 0.09 = 0.81W

Now less than 1 Watt doesn't sound like a lot, but consider that resistors generally come in 1/8th Watt, 1/4 Watt, 1/2 Watt, you're going to need at least a 1Watt resistor, so it's going to be reasonably large.

Resistor values
Resistor values are quite easy to remember, once you've been told what the values equate to.
black is zero and brown is one, after that the colours follow the colour spectrum, Red, Orange, Yellow, Green, Blue, Purple, then Grey and white are tagged onto the end. quite why it was done like this with black and brown first, and not just the spectrum colours first then extra ones I don't know...

Anyway,

Black = 0
Brown = 1
Red = 2
Orange = 3
Yellow = 4
Green = 5
Blue = 6
Purple = 7
Grey = 8
White = 9

The last band on the resistor is always Silver, Gold, Brown, Red, Green, Blue, Purple or Grey, these colours relate to the tolerances. (how far off of the stated values the component might be).
Silver = 10%
Gold = 5%
Brown = 1%
Red = 2%
Green = 0.5%
Blue = 0.25%
Purple = 0.1%
Grey = 0.05%

So, if you have a resistor that has colour codes

Red, Red, Brown, Silver
the values are
2 2 * 10 = 220Ohm, +/- 10% -i.e the value is somewhere in the region of 220 Ohm, 198 Ohm - 242 Ohm range (quite a range.)

You might notice that black is missing from the first column, and be thinking how do you write 1Ohm, this should be Black, Brown, Black right? (01 x 1)

Wrong, the correct way to write 1 ohm, is Brown, Black, Gold (10 x 0.1)

Symbols
Resistors, like all components have a symbol user to indicate them on electronics schematics.
the symbol for a resistor looks like a box (the box does not have a line through it) however you may sometimes find the symbol written as a zig zagged line.

So that about wraps up the humble resistor. There is of course more to learn, there is indeed always more to learn. I'll cover some more advanced stuff with resistors, and the different type of resistors later.

Calculations
As with everything electronics related, sooner or later you run into something maths related.

Even with the simplest of components, maths somehow finds it's way in.
Not to worry though, the calculations concerning resistors are really simple.

Resistors in series
This is as easy as one add one.

When resistors are in series (one linked to the other like a train) you just add all the values together.

R1 + R2 + R3 + ... + Rn = Rtotal

In the circuit above there are 4 x 8Ohm resistors.

8 + 8 + 8 + 8 = 32Ohms.

Resistors in parallel
When you connect resistors in parallel, things get a little more complicated, a mathematician would tell you that it's the sum of the reciprocal of the values of the resistors that makes the reciprocal of the total resistance.

And they would be right, but they'd also likely confuse the hell out of you.

In laymans terms, it's the sum of one over the value of all the components, that makes one over the final value, 1 over this value gives you the total resistance.
(1/R1) + (1/R2) + (1/R3) + ...+ (1/Rn) = 1/Rtotal

In this circuit the resistors are all arranged parallel to each other and connected differently from the original circuit.

The sum for creating this circuit is

(1/8) + (1/8) + (1/8) + (1/8) = (1/total)
1/8 = 0.125

So writing this out a bit more long hand we have...
0.125 + 0.125 + 0.125 + 0.125 =(1/total)

If you add all of those together you get 0.5
And 1/0.5 = 2

The total resistance of 4 x 8ohm resistors in parallel is 2 Ohms.


Series and parallel connections
Sometimes you're going to want to connect things in a mixture of series and parallel.

You might do this because you want a 1.5ohm resistor, they don't make these but you could connect 2 x 1ohm resistors in parallel

1/1 + 1/1 = 2
1/2 = 0.5

Then add a third 1 ohm resistor in afterwards in series and you get 1.5Ohm.

You might want to do this for a completely different reason than getting values that you can't get...

look at this example:

To start with break it down into two halves.
there's clearly a top half (2x 8ohm resistors) and a bottom half that's the same.

So work out the values of those first.

Top half = 8 + 8 = 16 Ohms.
The bottom half also is 16Ohms.

Now they are in parallel to each other
( 1/16) + ( 1/16) = 0.125
1/0.125 = 8

Now you may be wondering why anyone would be such a sadist as to use four components, have to go through that maths when they end up with the same value as just one of those components.

And the answer is power, as in power handling.
If you have a wire that can conduct a certain amount of power (voltage and current), and if you exceed that it'll heat up and melt, then you want to get a bigger gauge wire, you're adding more strands of wire, more paths to go down.

In the example above imagine that they were 10W resistors, but you needed to be able to handle 20W of load, buy arranging two side by side you've doubled that power handling capability to the 20W that you want.

but you've also halved the resistance, so then you add a couple more resistors in series to increase the resistance again.