ESR = Equivalent Series Resistance
The following have ESR
Power supply
Amp
Battery
Cap
So if you had 100 amps flowing through a resistor of .017 ohm the voltage drop would be
1.7v
300 amp flowing through would be a drop of 5.1volts
Watt= How much work can be performed but does not tell you for how long it can sustain that work.
Joule= can tell you, it’s a watt second, so 10 Joule can provide 1 watt per 10 sec. or 10 watts for one sec. Or how ever you can divide it up.
Formula for determining Joules stored in a capacitor
Take on half of the capacitors value (farads) and multiply it to its voltage squared.
1 farad cap at 14 volts
.5 x (14x14) = 98 Joules
A cap just stores energy
Batt. Don’t they have the potential to produce electricity by a chemical reaction
Cap just stores electrons on their plates in the form of electrostatic charge.
Ei. What is the joules of a 65 amp hour battery
So a fully charged batt. With a 65 amp draw would have about 12.2volts and by the end of the hour it will be down to 10volts about dead.
Using the voltage at about 11volts joules would be 2,574,000 joules
So after removing the 2.5 million joules from the battery it probably doesn’t have but more then hundred joules left.
You can almost drain the batt. And never drop below 10 or so volts. If a cell or something goes bad yes but its pretty much drained around there.
So if you charge a 20 farad Capacitor to 14 volts we know with our formula its 1960 joules. Using that cap in a audio system and load it till it drops like with the battery and drops to 10 volts how many joules will we have removed from the cap? What is left in the cap will be useless cause our system will never drop below the 10 volts unless battery is bad.
The cap will have 1000 joules that we can not use because the voltage is not dropping any more than the 10 but most amps etc. will not work at this voltage.
Power that is delivered to the stereo by the battery and alternator bypass the cap. They merely flow by its terminals if the cap drops below the battery or alternators voltage the current will flow into the cap until it reaches the same voltage as battery and cap.
But if the cap has more voltage than the battery or alternator the current will flow out of the cap to the battery or amp.
Now we know that the cap has an ESR so if this 20 farad cap has this of .017 and we put a load on it of 100amps. Its charged up to 14.2volts now what will our voltage be at when we draw the 100 amps out of it?
Will be 12.5 volts
This is 100 ohms of resistance. The wattage is 1250 watts from you cap . 12.5x 100amps = 1250 watts
The cap is actually putting out 14.2volts x 100amps = 1460 watts but 170 watts of power is actually lost in heat in the ESR of the cap. That’s a loss of 13%
So if we change the 100 amp draw to 300amp draw it equals to 56% of the stored energy is lost. So there are 1530 watts lost in heat.
ESL = Equivalent Series Inductance
The following have ESR
Power supply
Amp
Battery
Cap
So if you had 100 amps flowing through a resistor of .017 ohm the voltage drop would be
1.7v
300 amp flowing through would be a drop of 5.1volts
Watt= How much work can be performed but does not tell you for how long it can sustain that work.
Joule= can tell you, it’s a watt second, so 10 Joule can provide 1 watt per 10 sec. or 10 watts for one sec. Or how ever you can divide it up.
Formula for determining Joules stored in a capacitor
Take on half of the capacitors value (farads) and multiply it to its voltage squared.
1 farad cap at 14 volts
.5 x (14x14) = 98 Joules
A cap just stores energy
Batt. Don’t they have the potential to produce electricity by a chemical reaction
Cap just stores electrons on their plates in the form of electrostatic charge.
Ei. What is the joules of a 65 amp hour battery
So a fully charged batt. With a 65 amp draw would have about 12.2volts and by the end of the hour it will be down to 10volts about dead.
Using the voltage at about 11volts joules would be 2,574,000 joules
So after removing the 2.5 million joules from the battery it probably doesn’t have but more then hundred joules left.
You can almost drain the batt. And never drop below 10 or so volts. If a cell or something goes bad yes but its pretty much drained around there.
So if you charge a 20 farad Capacitor to 14 volts we know with our formula its 1960 joules. Using that cap in a audio system and load it till it drops like with the battery and drops to 10 volts how many joules will we have removed from the cap? What is left in the cap will be useless cause our system will never drop below the 10 volts unless battery is bad.
The cap will have 1000 joules that we can not use because the voltage is not dropping any more than the 10 but most amps etc. will not work at this voltage.
Power that is delivered to the stereo by the battery and alternator bypass the cap. They merely flow by its terminals if the cap drops below the battery or alternators voltage the current will flow into the cap until it reaches the same voltage as battery and cap.
But if the cap has more voltage than the battery or alternator the current will flow out of the cap to the battery or amp.
Now we know that the cap has an ESR so if this 20 farad cap has this of .017 and we put a load on it of 100amps. Its charged up to 14.2volts now what will our voltage be at when we draw the 100 amps out of it?
Will be 12.5 volts
This is 100 ohms of resistance. The wattage is 1250 watts from you cap . 12.5x 100amps = 1250 watts
The cap is actually putting out 14.2volts x 100amps = 1460 watts but 170 watts of power is actually lost in heat in the ESR of the cap. That’s a loss of 13%
So if we change the 100 amp draw to 300amp draw it equals to 56% of the stored energy is lost. So there are 1530 watts lost in heat.
ESL = Equivalent Series Inductance