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Lesson 5 (continued)

We know that if 100 amps of current flows out of our cap, those 1.7 volts will drop across the ESR of .017 ohm. This will cause the output to drop to 12.5 volts just like it did with the unlimited cap.


This means that the load (100 ohms resistance) will be consuming 1250 watts from our cap. 12.5 volts x 100amps = 1250 watts. The total wattage output produced by the cap is 1420 watts. 14.2 volts x 100 amps = 1420 watts. Unfortunately 170 watts of power will be lost in heat in the ESR of the cap. This represents a loss of 13% of our total usable joules (960) at this point. Now tonight’s question is if we increase the current draw to 300 amps (300amps x 14.2volts = 4260 watts), how many watts will be dissipated in the ESR of the cap and what percentage of the total 4260 watts does it represent? Of our total usable 960 joules, what percentage will be left for the stereo?


Lesson 6

Ok before the next lesson let’s review lesson five. When I checked the posts no one had the correct answer of 56% but some were close. The important part is that everyone seems to understand the loss mechanism. From lesson five we see that the energy we can get out of a cap is inversely proportional to the rate that we try to take it out. This is because the ESR that is in series with the output stays constant regardless of the load. At very high power levels, this ESR can amount to a sizeable amount.

In an earlier lesson we learned that the ESR causes a voltage drop proportional to current flow. When voltage is dropped across a resistance heat is created. Lesson five taught us that with 100 amps (flowing from a cap with .017 ohm ESR) we lose 13% of our joules as heat when we try to remove them. If a cap has an ESR of .017 ohms, and 300 amps flows we will lose 56% of the stored energy when we try to remove it. In our giant cap example with 300 amps of current, we will lose this as 1530 watts of heat. This is the same loss mechanism that causes a battery or amp or power supply to get hot when they are delivering high power levels. Virtually all voltage sources have at least some ESR. At this point we should have a good understanding of how ESR affects a component. The next logical thing to cover is ESL.

ESL stands for equivalent series inductance. Just like the ESR it can be modeled as an inductor in series with the output of our capacitor. Now everyone in Car audio knows what inductors do. They resist a change in current flow. Their most common use is in Speaker crossovers. When used in series with a woofer they let the slowly changing low frequencies pass, but stop the fast changing high frequencies. The reason an inductor does this is because it behaves like a resistor that changes value with frequency. Unlike a capacitor that decreases in value with increasing frequency an inductor decreases in value with decreasing frequency.

Lesson 6 (continued)

Now I have been told that the ESL value of the giant cap is 0.2 mh. Somebody check my math but I think this would put the reactance of the cap near .063 ohms at 50 Hz. This means that if we wanted to refresh our amps at a rate of 50 Hz (seems reasonable if we were playing bass real loud) our ESL of .07 ohm would be in series with our .017 ohm ESR for a total value of .08 ohms.


Now we know from ohms law that if we try to get 100 amps through .08 ohms we will have a voltage drop of 8 volts and at 300 amps the drop would be about….well it’s pretty clear that we will be left with less than a fraction of a volt if we start out with only 14.2. Is everybody still with me? I know it’s not good news but I’m not making this stuff up.

Now for tonight’s lab lesson to prepare us for lesson 7. Tomorrow, I will post the results of the following test. If you want to check me, go to Radio Shack and buy the following: Bulb # 272-1127, Socket # 272-360, and a nine volt alkaline battery. For the battery a Radio Shack is ok but a Duracell is better. Make sure it is fresh!!!!!

Wire the socket and connect it to the nine volt battery and record how long the bulb stays lit. Be prepared to wait for a couple hours. Charge a giant cap to 14.2 volts and do the same with it. Be prepared to wait about an hour. Charge a 1 or 1.5 Farad cap to 14.2 volts and do the same. This will take only a few minutes. Record the times and we will discuss the importance of this in our next lesson.


Lesson 7

Ok in last lesson I left everyone with instructions to duplicate the results of the test I am going to post tonight. The purpose of this test was to put the capacity of even a giant cap in perspective. As I have pointed out in earlier lessons storing electrons in the form of a charge on a plate is not really very efficient. Some folks think we should stand in awe of a value like 2000 Joules. Well our test tonight puts some reality in this value. If we perform a test like described in the end of lesson 6 we come up with the following results.

1.5 Farad cap lights the bulb for about …………5 minutes and 28 seconds

a giant cap lights the bulb for about……………. 54 minutes

a nine volt alkaline does so about …………………. 2 hours and 14 minutes

did anybody get results similar to these…….are we in agreement on these numbers ?


Lesson 7 (continued)


As for the relationship of these numbers, each of these units has a higher ESR than the previous one. The highest ESR in the group was the nine volt battery. It actually has enough energy to light the bulb far longer but since its ESR is fairly high it loses a lot of its energy as heat internally. But even still it should be apparent that it holds more energy than the giant cap and a whole lot more than a 1.5 farad unit

For now I do not care to concern ourselves with the meaning of this ---we will cover it in the closing. Before going on let’s review a few facts. In lesson 3 we learned that a giant cap can hold 1960 joules at 14 volts. In lesson 4 we learned that only 960 of them sit at a potential above 10 volts. In lesson 5 we learned that if we want to use them at a rate of 100 amps we will lose 13% of the 960 that are left.

If we use them at a rate of 300 amps we will lose 56% of the 960 which will leave us with only about 500 usable joules. And these losses are only for the ESR mechanisms—they do not include the ESL mechanisms that could actually be higher if the demands are quick enough.

It has been suggested that the purpose of these giant caps is to provide quick energy. It has also been suggested that they are for slow energy.

I am not sure what is being claimed so I guess I need to cover both situations. As for slow energy I think the previous test could put that thought out to pasture. For long term energy one of these units is less useful than a nine volt battery and to compare it to a car battery is really useless. After all what good is 500 useable joules when we have over 2 million in the car battery? It should be obvious if one of these devices can be of any use at all it will have to be able to provide energy faster than a car battery. But before we get to that issue lets cover the behavior of alternators and batteries under dynamic load conditions.

Tomorrow is Saturday and I will have time to measure the response time of a few alternators. This will enable me to model my closing explanations more exactly. I will post the results of these tests tomorrow night.