ELECTRONICS --> BASIC THEORY ( UPDATED NOV 27 )

Post by **PrecisionBoost** » Sun Sep 05, 2004 4:18 am

Electronics General Theory on DC circuits

PART 1 Batteries

I thought I should start out by explaining the physics of electricity and how batteries work... this might be a little confusing at first but it's not really important for you to understand what I'm talking about.

If it gets too confusing simply jump to PART 2

In a battery there are two sides one is positive and one is negative.

On the negative side of the battery you have a whole pile of electrons and they want to get back to their natural home on the positive side of the battery.

The postive side of the battery consists of a whole bunch of positive IONs waiting patiently for the electrons to come back to make them neutral again.

These postive IONs are commonly called "holes"

When you put a battery charger onto your battery it does "work" and strippes the electrons out of their natural "neutral" habitat and then deposits it on the other side of the battery where they are trapped.

So for every electron on the negative side there is one "hole" waiting to accept it on the postive side.

The only way for the electrons to get back to the positive ions and create a neutral state is to do some work such as turning a motor or lighting an LED.

Current is actually a measurement of how many electrons move from one side to the other in a given time period.

Electrons are measured in Coulombs and an Ampere is simply one Coulomb per second.

Now you might say how much is a Coulomb of electrons?

1 Coulomb = 6.3 X 10^18 electrons

So if someone says something uses 1 amp of electricity you know that there are 6,300,000,000,000,000,000 electrons going through the wire every single second

A single electron can't do much by itself but when you get a few hundred trillion of them pushing things start to happen.

The confusing part is the direction in which everyone things power moves.... power actually runs from the negative side to the positive side of the battery.

In physics all diagrams are written showing current flowing from the negative terminal to the positive terminal but all electronics theory was written showing the current flowing from the positive to the negative.

So even though we know that power doesn't flow from positive to negative we still use this in electronics diagrams because it's allways been taught this way.

Now that we have this simple battery theory out of the way we can get to something more important.

Post by **PrecisionBoost** » Sun Sep 05, 2004 4:58 am

PART 2 Units of Electricity and Simple forumulas

Voltage is typically measure in "Volts" and is represented by the letter V when being used in formulas

Current is typically measure in "Amps" and is represented by the letter A when being used in formulas

Resistance is typically measured in "Ohms" and is represented by the letter R when being used in formulas.

"Ohm" is typically replaced by the greek letter

Since this isn't part of my regular keyboard text I will simply continue to type them out as ohms instead of the regular greek symbol

Ok.... here is the very first and most important formula

Voltage equals Current multiplied by resistance V = I * R

This can be rewritten as I = V / R and R = V / I

Here are some examples....

You have a 4 ohm resitor and you measure the current running through it to be 10A..... What is the Voltage of the battery?

V = I * R

V = 10 A * 4 ohms

V = 40 Volts

You have a 15V battery and you measure the current running through a resistor to be 5A..... what is the resistance?

R = V / I

R = 15V / 5A

R = 3 ohms

You have a 12V battery and a 4 ohm resistor.... what is the current level going through the resistor?

I = V / R

I = 12V / 4 ohms

I = 3A

Hopefully that is simple enough.... it's time to get to basic resistance calculations.

The completely useless info for the day is........

A typical battery with a "reserve capacity" rating of 105 minutes will be able to produce a current level of 25 amps for just under two hours without going completely dead.

During this 105 minutes the amount of electrons that pass from the negative terminal to the positive terminal is..... 9.9225 X 10^23

That's 992,250,000,000,000,000,000,000 electrons

Post by **PrecisionBoost** » Sun Sep 05, 2004 4:58 am

PART 3 Simple resistance calculations

Ok... there are two types of basic resistive circuits... one is called parallel and the other Series.

When you take two resistors and put them in parallel the total resistance decreases

When you take two resistors and put them in series the total resistance increases.

Example of a parallel circuit

Example of a series circuit

The formula for calulating the total resistance for a Parallel circuit is:

Rtotal = 1 / ( 1/R1 + 1/R2 )

So if R1 = 8 ohms and R2 = 8 ohms you get:

Rtotal = 1 / ( 1/8 + 1/8 )
Rtotal = 1/ ( 1/4 )
Rtotal = 4 ohms

So Rtotal will allways be less than the smallest resistor.

Example #2

R1 = 10 ohms R2= 100 ohms

When put into parallel the total resistance is:

Rtotal = 1 / ( 1/10 + 1/100 )
Rtotal = 1 / ( 1/110 )
Rtotal = 9.09 ohms

When calculating total resistance in Series you simply add the resistance.

Rtotal = R1 + R2

So with R1=10 ohms and R2 = 100 ohms the total resistance is:

Rtotal = 10 + 100
Rtotal = 110 ohms

So when resistors are in series the total resistance will allways be higher than the largest resistor.

Now you might say.... who cares about total resistance.... well these two simple formulas will allow you to take a whole pile of resistors and figure out the exact voltage and current through each and every resistor as well as the total amount of current and power the battery has to supply.

Next I will give you a more complex example that uses both series and parallel resistors and we will break it down into a simple single total resistance number.

Lets say that R1 = 100 ohms R2 = 200 ohms R3 = 500 ohms

To solve the problem you would first combine the R1 & R2 because they are in series and do not have anything else connected between them.

So R1 + R2 = 100 ohms + 200 ohms = 300 ohms

Now we know that the resistance down this pathway is 300 ohms

Then we look at the parallel combination of R1 + R2 parallel R3

So we need to solve for R1 + R2 = 300 ohms in parallel with R3 = 500ohms

Rtotal = 1 / (( 1/ R1+R2) + ( 1/ R3 ))

Rtotal = 1 / (( 1 / 300) + ( 1 / 500))

Rtotal = 1 / ((0.003333) + ( 0.002 ))

Rtotal = 1 / 0.53333

Rtotal = 187.5 ohms

Post by **PrecisionBoost** » Sat Nov 27, 2004 5:47 am

Ok... I know it's been a while since I updated this but I'm going to get into how current flows next.

With the next example we will use our resistor calculations to figure out how much current, voltage and power goes through each resistor.

In this example R1= 4 ohms R2 = 8 ohms R3 = 2 ohms

First we have to find out Rtotal.

R1 and R2 are in series so we calculate their value by simply adding them.

R1+ R2 = 4ohms + 8 ohms = 12 ohms

Lets call the resultant resistance R12

R12 is in parallel with R3 so we use the formula from above to figure out the total resistance.

Rtotal = 1 / ( ( 1/R12) + ( 1/R3 ) )
Rtotal = 1 / (( 1/12 ) + (1/2) )
Rtotal = 1 / (0.0833) + (0.500)
Rtotal = 1 / 0.5833
Rtotal =1.714 ohms

So we know that the total resistance the battery sees is 1.714 ohms

We can now calculate the current the battery is putting out by using the formula Current = Voltage / Resistance.

Battery Current = 12V / 1.714 ohms

Battery Current = 7.00 amps

Now we will figure out how the total current gets distributed.

The voltage accross the terminals of R3 is allways 12V so we can figure out how much current goes accross it by simply using the same forumla we just used to find the battery current.

Current= voltage/resistance

I = V / R

I = 12V / 2 ohms

I = 6 amps

Now we know that the battery is putting out 7 amps and R3 is using 6 amps so that leaves us with just 1 amp going through resistors R1 & R2

Lets check it out and make sure....

R1+ R2 = 12 ohms

So if I = V / R and the voltage accross R12 is 12volts we can calculate it as...

I = 12V / 12 ohms
I = 1 amp

Now we know that 1 amp goes through R1 and R2 and 6 amps goes through R3.

Now you might be thinking..... what is the voltage accross each resistor?

The answer is 12V accross R3 and 12V accross R1 & R2

Given that the current through R1 & R2 is 1 amp we can figure out the individual voltage by using the same forumula from above.

I = V / R

V = I * R

Voltage accros R1 = 1 amp * R1 = 1 amp * 4 ohms = 4 volts

Now if the total voltage across R1 & R2 is 12V and the voltage across R1 is 4 Volts then we can assume that there is 8V accross R2

Just to be sure we can plug it in our formula.

V = I * R
V= 1 amp * 8 ohms
Voltage accross R2 = 8 Volts.

Next we will calculate the power which is measured in Watts...

Power = Voltage X Current

So we can take the above example and figure out how much power goes through each resistor.

R1 = 4 ohms and we've allready calculated the current passing through it to be 1 amp and the voltage accross it to be 4 volts.

So the power through R1 = 4V X 1A = 4 Watts

Next we can look at R2....

Power R2 = 1 A X 8 Volts = 8 Watts

Next we can look at R3.....

Power R3 = 6 A X 12V = 72 Watts.

Now the total power should allways equal the sum off all the individual powers...... so 4W + 8W + 72W = 84 Watts.

We can check this by using our power formula and out Battery Current.

The battery was putting out a total of 12V and 7 amps.

Power Battery = 12V X 7 A = 84 Watts.

You may have noticed that no matter how you calculate it everything will add up to the total accross the battery.... this is called "conservation of energy" and it's the main prinicpal behind electronics.

Now instead of resistors you can imagine those as speakers.... and you can see just how the power is split up depending on how you arrange the speakers.

Post by **PrecisionBoost** » Sat Nov 27, 2004 6:25 am

Ok now we will look at another example using speakers....

Let's say you just bought a nice amplifier that is rated for 100W at 4 ohms

Speaker #1 = 4 ohms = R1 Speaker #2 = 4 ohms = R2

Now we use our forumla to find the resistance that the amplifier is seeing.

R amp = 1 / ( (1/4) + (1/4) ) = 1 / 0.500 = 2 ohms

So the amplifier is going to see 2 ohms.

Now given that the amplfier is rated at 100W at 4 ohms we can figure out the voltage output of the amplifier.

Power = Voltage X Current

and Current = Voltage / Resistance

So we can mix things around and get....

Power = Voltage X (Voltage/Resistance)

Power = (Voltage^2) / Resistance

Voltage ^2 = power X resistance

Voltage = square root of ( power X resistance )

Voltage = root ( 100W X 4 ohms )

Voltage = square root of 400

Voltage = 20 Volts

Just to prove this is correct lets use a few formulas to make sure....

I = V / R

I = 20V / 4 ohms

I = 5 Amps

So the amplifier is putting out 20V and 5 amps into a load of 4 ohms.

Power = Voltage X Current = 20V X 5 amps = 100 Watts.

Now this is all and fine if the amplifier is seeing 4 ohms accross it's terminals but in this example is seeing 2 ohms.

Amplifiers allways put out the same voltage no matter what the load is (in theory) so the only thing that can change is the current.

As a result the power output must also increase.

Amplifier power = V^2 / R total

P = ( (20V)^2 ) / 2 ohms

Amp power = 400 / 2ohms = 200 Watts

So with 20V accross R1 & R2 we can calculate the current and power into each as...

Current through R1 = V / R = 20 Volts / 4 ohm = 5 amps
Power = 20V X 5 amps = 100W

Current through R2 = 20 Volts / 4 ohms = 5 amps
Power = 20V X 5 amps = 100W

And now you understand why an amplifier will put out twice as much power if you put two 4 ohm speakers in parallel.

Now I did say that in theory the voltage will stay at the same level but in reality it never works perfectly due to "internal resistance" inside the amplifier.

The amplifier itself has some resistance so when you start loading it by adding more speakers it tends to waist a little more power than it would have if it were at it's "design load"

So for example you might find that an amplifier will be rated at 35W at 4 ohms and only 50W at 2 ohms.

In theory the amp should put out 70W at 2 ohms but due to several subject that are way too advanced for this simple example you never really double the power.

As well most amplifiers can only handle so much "load" before they start to become unstable.

If you took a plain old amplifier that is rated at 100W and tried to run 8 subwoofers off it the amplifier is going to cut out due to overload or it will load it down so much that the voltage drops.

Some amplifier companies design their amps to handle almost any load... so they will continue to bang out power no matter how many speakers you throw at them.

So in theory if you put 8 speakers that have 4 ohm resistance in parallel the amplifier would see.....

Ramp = 1 / (1/4)+(1/4)+(1/4)+(1/4)+(1/4)+(1/4)+(1/4)+(1/4)
Ramp = 1 / 2.000
Ramp = 0.5 ohms

Now most amps will die with this kind of load.... typically anything under 2ohms will only work on a nice quality amp.... so you get what you pay for.

Now lets take that same example of a 100W output at 4 ohms.

This is what I would guess the output voltage of a decent amp to be....

4 ohms --> 20 Volts
2 ohms --> 19 Volts
1.5 ohms --> 18 Volts (this is just a guess to illustrate my point)
1.0 ohms --> 17 Volts
0.5 ohms ---> 16 Volts

So with a really decent 0.5 ohm stable amp you might see the rating of 100W go up by a factor of about 6.5

Power = V^2 / R
Amp power = 16 ^2 / 0.5 ohms

Amp power output = 256 / 0.5 = 512 Watts

So because the speakers are in parallel and all equal to 4 ohms that power gets divided equally between all speakers.

As a result each and every speaker gets 512W / 8 speakers = 64 Watts.

You can use the formulas above to calculate it but I assure you this is correct.

Now if that amp had managed to put out the full 20V at 0.5 ohms each speaker would get 100W for a total output of 800 Watts..... but this is virtually impossible due to that thing I mentioned called "internal resistance".

Perhaps some really expensive amps might manage something closer to 800W but all amplifiers have some "internal resistance" so it's unfortunatly a fact of life that you can't get double the power with double the speakers.

Now just to give you an example.... this is what a cheap amp might put out.....

4 ohms --> 20 Volts
2 ohms --> 15 Volts
1.5 ohms --> 10 Volts
1.0 ohms --> 3 Volts
0.5 ohms ---> 1 Volts

Typically a cheap amp will protect itself if you try to put too much load onto it.... and because it has a high internal resistance it will get really hot when you try to make more power.

Post by **PrecisionBoost** » Sat Nov 27, 2004 7:43 am

I might as well try to explain internal ampifier resistance as simple as I possibly can....

Here is what a diagram would look like...

So the Rint is the internal resistance of the amplifier.

This comes from things like such as....

1) Quality of circuit board (gold or silver plated helps)
2) Thickness of power supply wires inside your car
3) Length of power wires inside your car
4) Size of internal amplifier power supply
5) Quality of Mosfet amplifier parts inside amplifier
6) Power disipation ability

There are many other factors but many are well beyond what most of you will care to know about an amplifier.

So lets jump into some calculations....

Expensive amps have a very low internal resistance and they use expensive components and methods such as gold or silver plating to stop the circuit board from oxidizing. (copper is a great conductor but copper oxide is a great resistor....so oxidation of copper will decrease power handling ability)

Cheap amps will use inexpensive components and are designed to simply work at one given resistance level.

Ok lets use two examples....

Amplifier #1 has an internal resistance of 0.1 ohms

Amplifier #2 has an internal resistance of 0.75 ohms

Now you might think that there isn't much difference..... just 0.65 ohms.... but when you do your calculations you will see just how much that difference can mean.

Both Amp#1 & Amp #2 are rated for 100W at 4 ohms

So with a 4 ohm speaker they both put out 20V of signal

If we have 20V accross the speaker we will get a current of 5A

The votage drop accross Rint can now be calculated for each amp...

Amplifier #1 --> Rint = 0.1 ohms
V = I X R = 5 amps X 0.1 ohms = 0.5 V

So in reality amplifier #1 is actually putting out 20.50 V but only 20V is getting to the speaker due to losses associated with internal amplifier resistance.

Amplifier #2 --> Rint =0.75 ohm

V = I X R = 5 amps X 0.75 ohms = 3.75 V

So in reality amplifier #2 is actually putting out 23.75 V but only 20V is getting to the speaker due to losses associated with internal amplifier resistance.

Here is where we see the huge difference..... lets now put two 4 ohm speakers in parallel accross the amplifier terminals.

The amplifier is still going to try and put out 20V but it's seeing 2 ohms so it's going to increase the level of current going out the terminals.

*************************************************

So amplifer #1 is still putting out 20.5V

We now need to calculate the total resistance....

Rtotal = Rint + R speaker = 0.1 ohms + 2.0 ohms
Rtotal = 2.1 ohms

I = V / R = 20.5V / 2.1 ohms
I total =9.76 amps

Now that we are putting out more current the voltage drop accross Rint will be higher.

Vdrop = I total X Rint

Vdrop = 9.76 amps X 0.1 ohms = 0.976 Volts.

So now the amp will be putting out 20.5V and 0.976V will be waisted on internal resistance leaving 19.524 Volts for the speaker.

We can calculate the power into the speakers two ways...

Method 1 -->
Power = V^2 / R =( ( 19.524 )^2) / 2 ohms
Power = 381.19 / 2 ohms
Power = 190.5 Watts

So about 95 Watts will go through each speaker for a total of 190.5 Watts

Method 2 -->
Power = V X I = 19.524 X 9.76 Amps
Power =190.5 Watts

Now if we want to see how much power is waisted internally we can calculate it as follows....

P = I X V = 9.76amps X 0.976 V =9.5 Watts

*******************************************

Here is where amplifier #2 is going to show it's true colors.....

So amplifer #2 is still putting out 23.75 Volts

We now need to calculate the total resistance....

Rtotal = Rint + R speaker = 0.75 ohms + 2.0 ohms
Rtotal = 2.75 ohms

I = V / R = 23.75/ 2.75 ohms
I total =8.636 amps

Now that we are putting out more current the voltage drop accross Rint will be higher.

Vdrop = I total X Rint

Vdrop = 8.636 amps X 0.75 ohms = 6.48 Volts !!!!!!!!!!!!!!

So now the amp will be putting out 23.75 and 6.48V will be waisted on internal resistance leaving just 17.27 Volts for the speaker.

We can calculate the power into the speakers two ways...

Power = V X I = 17.27 X 8.636 Amps
Power = 149 Watts

So about 75 Watts will go through each speaker

Now if we want to see how much power is waisted internally we can calculate it as follows....

P = I X V = 8.636 amps X 6.48 V =56 Watts !!!!!!

*****************************************

So now you can see why the more expensive amplifier will kick the ass of the cheap one when you start adding speakers.

Amp #1 put out 190.5W at 2 ohms with just 9.5 Watts waisted internally.

Amp #2 put out 149W at 2 ohms with a huge 56 Watts of power waisted.

Of course this extra 46Watts gets turned into heat so amplifer #2 will run way hotter than amplifier #1.

*******************************************

Lets take this one step further and calculate at 0.5 ohms ( eight 4 ohm speakers in parallel configuration )

I will jump right to the numbers and skip the calculations....

*************************************************
Amp #1 with 0.1 ohms internal resistance

Rtotal = 0.1 ohms + 0.5 ohms = 0.6 ohms total resistance
Current output = 34.17 amps
Internal voltage drop = 3.417 Volts
Voltage across speaker = 17.08 Volts
Power output to speakers = 583.68 Watts
Power to individual speakers = 73 Watts
Power waisted in Rint = 117 Watts

*************************************************
Amp #2 with 0.75 ohms internal resistance

Rtotal = 0.75 ohms + 0.5 ohms = 1.25 ohms total resistance
Current output = 19.0 amps
Internal voltage drop = 14.25 Volts
Voltage across speaker = 9.50 Volts
Power output to speakers = 180.5 Watts
Power to individual speakers = 22.6 Watts
Power waisted in Rint = 271 Watts

**************************************************

So at the end of the day the more expensive amplifier with less internal resistance was able to pump out 584 Watts into eight speakers while the cheap amplifier with a high internal resistance was only able to push out 181 Watts into eight speakers.

In reallity the second amplifier would overheat very quickly and then shut down until the heat levels dropped.

So hopefully everyone understands the importance of decent amplifiers that are stable to low impedance loads ( such as 0.5 ohms )

The better amplifier managed more than three times the power of the crappy amplifier.

So don't be fooled by these junk amps that claim similar power levels to those of more expensive brands.... they might put out similar power with very little load but when it come to crunch time the cheap amp will die every time.

Post by **PrecisionBoost** » Sat Nov 27, 2004 8:04 am

I'm briefly going to touch on the subject of impedance.

Most of you will notice that speakers are typically rated in impedance levels and not in resistance.

This is because the speakers use coils to move the cone.... and coils are measured as "inductance" which is similar to resistance but not exactly the same.

Inductance naturally changes with frequency.... so at low frequency a 4 ohm speaker might actually be reading something closer to 1 ohm.... and at higher frequencies it might actually read something closer to 400 ohms.

This is why a subwoofer won't play 7000 hz tweeter sounds very well.

This is also why crossovers are important..... so you only send sounds of a particular frequency range to the speaker it's intended for.

Speakers also have simple resistance..... ideally the manufactureres will match the coil impedance to the internal resistance of the speaker as this is the most efficent setup.

So for example... I'm looking at the specifications for one of my JL Audio 8W1-4 subwoofers...

Nominal impedance = 4 ohms
DC resistance = 3.34 ohms

The reason I'm touching on this is to explain also that subwoofers will "load" an amplifier way more than a tweeter.

Typically they will measure the inductance at a specific frequency but the actual impedance will change depending on what sound frequency is being played.

To make things more complex speaker enclosures can actually effect the impedance of the speaker..... so "true impedance" calculations are really really complex with very heavy mathmatical calculations.

To get back to my point.... since subwoofers load the amplifiers more due to dropping impedance levels at low frequency the difference between a good amplifier with low internal resistance and a cheap ass knock off is very noticeable.

So keep in mind that low ohm stability is one of the most important features when looking at a subwoofer amplifier.

Post by **PrecisionBoost** » Mon Mar 06, 2006 1:29 am

Is there anything other electronics theory anyone would like me to explain???

PM me with requests

Chris