The following is a series of excepts from CarSound written by Richard Clark which examines capacitors in every detail. Read first...then ask questions.
Also this is a conversation between Mr. Clark and another individual (a very cocky individual and a firm believer that caps will make you louder)
Lesson 1
Ok “powertrip” how about we have a discussion in basic electrical theory? At the end of this thread you should be the one that can explain to the world that according to ohms law it is impossible for these things to do any good. That is of course if you can admit that they do obey ohms law. We will do this a little at a time so how about you humor me and stick to my questions. We will do them a couple at a time so everyone can follow along. Let’s do a little calculation. Suppose we have a resistor that is .017 ohms (seventeen milliohms). I think that is what you say the ESR of the giant caps is.
The ones I have seen have measured higher but I will give you the benefit of the doubt. According to ohms law how many volts are dropped across .017 ohms if 100 amps of current are flowing? How about if we up the current to 300 amps? Let’s establish the answers to these questions before we go any farther. If we can't agree on the answer to this there is no hope we will ever get to the truth.
Lesson 2
Thanks David you are exactly right. If anyone wants this explained please ask David to clarify it. If everyone is going to follow this and understand fully the final conclusion it is important that no one miss any steps. There will be about ten lessons. Since power trip has left the building we will continue with the rest of the class. ESR stands for equivalent series resistance. This means exactly what it sounds like. It means that if we have a source of voltage it will behave exactly as if it has a resistor of the same value in series with its output. An amplifier has ESR, a power supply has ESR, a battery has ESR, and yes, a cap has ESR. Components have ESR’s because we do not have perfect conductors to make things from.
And now for the homework. Last night we learned that if 100 amps flows through .017 ohms there will be a voltage drop of 1.7 volts. And if the amp flow increases to 300 amps the voltage drop will increase to 5.1 volts.
For the sake of theory only let’s say we have built the largest cap in the universe and it has billions and billions of Farads. Its plates are made of a newly discovered material we'll call unobtanium. This new material has no resistance therefore our super cap has an ESR of ZERO ohms. We then charge the capacitor to 14.2 volts. We then place a resistor with a value of .017 ohms in series with one of the terminals of this cap. The question is: If we place a load that draws 100 amps from this cap what will the resulting voltage be on the load side of the resistor? What will the voltage be on the cap side of the resistor? What about if we increase the load to 300 amps? What will the voltages be on each side of the resistor?
Lesson 3
Ok now that we have studied ESR and understand what it is and it’s effect on the working of a circuit we will move on to another subject. But don’t forget about ESR as it is one of the important final building blocks in our search for truth about caps and we will come back to it. Today we will review the important concepts about total energy storage in a device like a cap. This has been covered in earlier posts (and I will say quite correctly) but I am going to expand on it as well as reiterate it for those who did not get to read it. Besides, I think I can simplify it a little.
In electronics, we measure power in watts. Wattage tells us how much work a device can do. But a wattage rating does not tell us anything about how long we can sustain that work. When we add the element of time to our wattage, we use a value we call Joules. A joule is a watt second. This means that one Joule of energy can provide a watt for a second. Ten joules can provide a watt for ten seconds or ten watts for one second or five watts for two seconds one hundred watts for a tenth of a second, and so on.
The formula for determining the total joules stored in a capacitor is very simple. We take one half the capacitor’s value in farads and multiply it times the squared charge voltage. For example a one farad cap charged to 14 volts would be .5 X (14x14) = 98 or .5 X 196 = 98 Joules. A 20 farad cap charged to 14 volts would be 10 X (14x14) = 1960 Joules.
There is a very important concept to understand about energy storage. A capacitor actually stores electricity.
Batteries don’t. Batteries have the potential to produce electricity by means of a chemical reaction but caps actually store electrons on their plates in the form of an electrostatic charge. In our next two lessons we will learn why this is important to know.
But first, the homework. This is a “think about it question”. We have learned that a Joule is a watt second. A Yellow top battery is rated at 65 amp hours. This means it can provide 65 amps for an hour. The question is how many Joules does this represent? Since this is a thought question, it would really help if whoever answers would show us your math.
Lesson 4
In the actual real world the voltage of the battery would drop a little from its open circuit voltage of 12.8 volts with a 65 amp load. In the case of the yellow top its actual voltage at 65 amps is about 12.2v when fully charged. By the end of the hour it would be down to about 10v. If we use 11 as an average our answer would be........ 2,574,000. Now that's still a lot of joules! Now actually this is not enough to totally kill the battery but at this point there isn't much left in it. This brings us to a very important fact. The energy in a battery will be depleted almost completely by the time it is down to 10 volts.
Lesson 4 (continued)
By the time we have removed those 2.5 million joules from the battery it probably doesn't have more than a hundred thousand joules left. We can almost totally deplete the battery's energy and never drop below 10 volts. This is because the battery doesn't store electricity. It stores chemicals. A chemical reaction produces the electricity. Storing actual electrical charges is very inefficient.
Look at our poor capacitor. Even if we made one as big as a battery it would still only be good for perhaps fifty to one hundred thousand joules---less than that left in a nearly dead battery. But if that were not enough there's more bad news. This exercise will be tonight’s homework.
A capacitor is like a gas tank in a car. The pump can only remove gas down to the pickup point. Any gas below this point can never be removed by the pump. If we charge a 20 farad cap to 14 volts we know from previous lessons that it will contain 1,960 joules. If we use that cap in a system and load it till it drops to 10 volts along with our battery how many joules will we have removed from the cap? How many joules will remain in the cap that we can never benefit from if our system never drops below 10 volts?
Lesson 5
In our last lesson we learned that caps actually store charges on their plates. And of the 1960 joules stored in a 20 Farad cap, 1000 of them sit at a potential below 10 volts. This means there is no way they can ever be used by an operational audio system. Today we will look at another loss factor. It has to do with the loss factor due to the ESR of the cap.
We have already studied voltage drop due to ESR but now let’s view it from an energy/watts standpoint. Let’s clarify things. The power delivered to the stereo by the battery and alternator bypass the cap. They merely flow by its terminals. If the cap charge is lower than the battery/alternator potential current will flow INTO the cap until it reaches equilibrium with the Battery/Alternator. If the B/A potential is lower than the charge potential of the cap current will flow OUT of the cap to the battery and or the amp.
Always remember that voltage always flows from the highest potential to the lowest potential, just like water. Current does not however flow into the alternator even if it is lower than the battery and cap because it has diodes on its output that only let current flow FROM its output. Now whenever any current flows into or out of the cap it must pass thru the ESR of the cap. The resistance is really distributed throughout the cap but it behaves just like it was right on the output terminal as in a series circuit location in the circuit loop does not matter. Now suppose our 20 farad cap is charged to 14.2 volts and we place a load on its output. This load is the same one that we used in lesson 2 to cause 100 amps of current to flow from our unlimited capacity cap. Only now we have our smaller 20 farad cap.
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.
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 ?
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.
Lesson 8
For this lesson I have done some actual measurements. Here are the test conditions: To measure voltage we used an Audio Precision with a DCX module. It is accurate to three decimal places. For sample time we chose 40 samples per second. For the non audio system test I used a KAL carbon pile load tester. It can do power tests on 12 volt charging systems up to 1200amps continuous. The audio system consisted of a couple of Rockford 1100 amps bridged into four ohm nominal speakers. The alternator was a stock Delco 80 amp CS type unit.
Its case temperature was monitored by a Raytek ST2L IR sensor. Engine speed was regulated with a Thexton #398 IACV tester. The music material was the SPL track # 30 from the IASCA competition disc. The battery was a Stinger spb-1000. All voltage measurements were done directly at the terminals of one of the amps.
Chart 1 Alternator/cap/battery test with 200 amp dummy load
For this test we monitored the voltage of the car with the stereo turned off. With the car running the voltage can be seen to be stable at about 13.7 volts. After 22 seconds (The horizontal scale is 100 seconds-10 sec per major division) we applied a 200 amp load. The voltage can be seen to drop to 11.6 on both traces. This test obviously exceeds the ability of the alternator to keep its regulation set point so its voltage falls. The drop can be seen to be nearly instant (steep curve) until about 12.5 volts where the battery starts to supply a significant amount of the power.
Ultimately the voltage drops to 11.6 and at 26 seconds we turn off the load. The voltage then starts to rise to the regulator set point as the battery is recharged (yellow curve) and as the battery and cap (green curve) are recharged. At a time of 50 seconds I turn the motor off so the alternator stops. The voltage then droops down to the float voltage of the battery—about 12.7. The only reason for the small difference at 50 seconds is because I couldn’t get the timing of the engine shut-off exactly the same every time. I did it several times and these two are within one second. That was as close as I could get it.
I am able to see no difference from these measurements. There are microscopic differences but I believe they are due to the alternator temperature. Alternator regulators are usually temperature sensitive. As they get hotter they tend to fold back. For this reason we let the unit cool off between each test and closely monitored the case temp throughout the tests. For this reason I believe that none of these measurements are meaningful to more than a couple tenths of a volt.
Chart 2 Music tests with an audio system
Note: Between each test the alternator was allowed to cool and the battery was charged until an automatic charger said it was topped off.
Purple curve
For our first test we played the system with the engine off and no cap. The result was the purple trace at the bottom. We played the system as loud as we could get it that seemed to produce no audible distortion. This was track 30 of the IASCA disc. It starts off with fairly low level sounds for the first 34 seconds. In order to insure the electrical system was stable we did not start the measurement until we were 20 seconds into the song. This means that our 0 starting point is: 20 on the CD counter. The battery was able to maintain its voltage just below 12.5 until the loud bass hits at 34 seconds (14 seconds into our chart) At this time it dropped to about 11.5 and had a few large variations due to the music. According to the computer calculations (third chart) the average voltage for this test was 11.7volts. This test was done as a baseline for the following tests.
Yellow curve—no cap
For this test the volume was left as it was for the baseline test. The engine was started. Notice that at low volume the alternator was able to maintain about 14 volts. When the loud music hit the voltage dropped to about 12.5 where it remained except for a few short moments where it actually climbed back to over 13.5 volts. The computer averaged calculations for the average voltage during the 100 seconds of this test was 12.973 volts.
Red curve—cap added
This test was identical to the previous test except the cap (15 farad type) was added 6 inches from the amp with 4 gauge wire—no relays or fuses. The red curve seems to overlay the yellow except that the actual peaks don’t rise as fast or as high during the brief quiet moments. I feel this would be due to the alternator having to recharge the cap. The voltage on loud passages hovered around 12.5 volts. The computer averaged calculations for this test show the average voltage to be 12.878 volts. I see no meaningful differences with or without the cap. I certainly don’t see the voltage sitting solid at 14 volts.
One note I might add is that this was a two thousand watt system driven right to clipping and the average voltage stayed above 12.8 with a stock 80 amp alternator. Under these conditions the battery would never discharge!
The green and blue curves were done just for kicks while we had the system set up. In both these tests we turned the volume up until the system was very distorted. This placed a severe load on the alternator and caused the voltage to dip as low as 12 volts. The curves seem to follow each other so closely that unless you have a good monitor it is doubtful you can tell there are two curves. The average voltage for these two curves was both 12.277 and 12.295 volts. If this volume were sustained for very long periods of time this battery would discharge.
Lesson 9
Now that we have had time to study theory in each of the 8 lessons and the results of the actual tests on a real system it is finally time to bring this discussion to a close. Unfortunately, when this thread started I was unable to explain the concept, as it was obvious that many of the people posting responses just didn’t have a good grasp of the way things really work. Those of you who have taken the time to follow the lessons should know by now why I was so frustrated at the arguments that were so illogical. It is important to keep in mind that this is a technical forum, not a marketing forum. I do not care or want to know about companies or brand names.
Nothing I have said was ever meant to disparage a particular product or company and I would appreciate it if in the future we could always keep that in mind. We should be able to discuss the merits of radial vs. bias ply tires without caring if they are made by Michelin or Goodyear.
In car audio we have little choice of how we are going to power our systems. Presently we have only four things that are practical. Each of them has its own characteristics that incorporate good points and bad points.
Let’s review them
The battery--this device has the ability to provide a very large amount of current. But due to its nature the current is provided at a voltage that is less than optimum –at least for a high powered stereo. Since its float point is 12.8 volts if fully charged, it can provide current only at voltages that are proportionally lower than 12.8 Volts.
The alternator—this device is electronically regulated at a point that allows it to recharge the battery. The alternator is usually designed to output voltage in the 13.8 to 14.5 volt range. Because its output is actively regulated it attempts to maintain this voltage with varying load conditions up to the point where it’s output cannot keep up with the load at which time it’s output drops off very rapidly. While relatively tight regulation is the strong point of the alternator it’s weak point is that it simply is not practical to obtain one that can provide large amounts of current like a battery is capable of.
The capacitor. The advantages of a cap are that it can charge up to whatever the highest voltage source in the system is, (in a car this would be the alternator) and provide current at this elevated voltage. The down side of a cap is that it cannot store very much total energy and only a portion of this energy is available at a usable voltage potential. The fourth type of device is an electronic voltage regulator. These devices have not been part of this discussion so I will pass over them for now.
Now modern car audio amplifiers are capable of consuming enormous amounts of power. Even with efficiencies in the range of 60% to 90% an audio system is capable of drawing hundreds or thousands of amps from the cars electrical system. Typically, the audio system is larger than any other electrical device in the car including the engine starter. Fortunately for the car, the demands of an audio system are rarely continuous in nature. The very nature of music rarely demands more than a duty cycle of 10% to 20% from a power standpoint. This means that the audio system is demanding short term, but repetitive peaks of current from the electrical system.
The primary source of this power is the alternator. It should be considered primary for two reasons. The alternator is the only first generation source of power. It ultimately provides all the power for the system either directly or indirectly by restoring power to the battery or cap. It is also primary as it is the power source with the highest voltage potential. In an electrical system current always flows from the source of highest voltage to circuits of lower of lower potential.
All three devices can be used in a system to great advantage. But the dynamic conditions present in a music system determine the role each device plays and to what degree. To understand this lets consider a low current drain condition. In this scenario the alternator will be at or near its set point.
This voltage is designed to be high enough to charge the battery meaning it will be one or two volts above 12.8 volts. This means that the battery will actually be a continuous load on the alternator and provides no power to the system. The size of load it presents is determined by the state of charge of the battery. The higher its state of charge the smaller the load will be. A cap if present in a system in this state will present a load for a finite amount of time until its charge voltage reaches equilibrium with the alternator.
Unlike the battery, the cap will cease to be a load after it is charged except for a factor known as dissipation, which for all practical purposes can be ignored in this application unless it is excessive. Under these circumstances, as long as the alternator can maintain its set point, it will provide all the power for the music system and the rest of the cars accessories. The battery and cap may as well not even be in the car.
Now if we increase the current demands of the music system to an amount that taxes the alternator its output voltage will begin to drop. Even so the alternator will continue to be a source of current to the system –i.e., the car, music system, and battery. It is at this time that the cap will begin to discharge and begin to augment the alternator as a source of current. The degree to which it provides current to the system is dependent on the actual voltage at the alternators terminals. Only when the alternator begins to drop below the caps charge potential does current flow out of the cap.
This is a continuous process and the current provided by the cap tries to maintain the voltage at its charge potential. The degree to which it can do this is dependent on two things. The current provided by the cap is limited to the total capacity of the cap and any series reactance’s (resistive or inductive components) that are part of the cap. The instant the cap starts to output current its charge potential begins to drop.
Now just what can we expect the cap to provide? Suppose we happened to have a cap charged to 14 volts, with a total reactance (made up of either resistive or inductive components) of about .017 ohm. We could figure that at the first instant of discharge it could provide ten amps at 13.83 volts. Of course if we were playing the system at a level enough to load our alternator, ten amps is not likely to provide much relief. But perhaps 30 amps might help—at this modest level our cap could begin to provide current at a potential of 13.5 volts. (lesson 2).
Of course this voltage level would drop at an exponential rate commensurate with the discharge curve that is standard with caps. No doubt the cap could help out a hundred amp alternator with the addition of an extra 30 amps even though it might be for only a brief instant. But it is sort of interesting that at even this modest power level of 130 amps (100 amps alternator + 30 amps cap) the cap is unable to maintain the voltage at 14 volts.
Of course in this scenario we are sitting at 13.5 volts for a brief instant and our poor battery is unable to help at all as its potential is at a lowly 12.8 volts. In fact the battery is still a load on the system!
Now what if we get serious with our stereo and we really crank it up? Lets say we have something like a manufacturers demo van with lots of amplifiers that can draw hundreds of amps on musical peaks. Lets pick a nice round number like 500 (“Cade” said 490) amps. Lets say we have a 200 (“Cade” said 190) amp alternator. Typically such an alternator can maintain a voltage near its set point up to perhaps 80% of its rating-after which its voltage begins to drop as it provides large amounts of current. As I am not familiar with all the different alternators lets just assume these assumptions are close and our alternator is putting out 200 amps. Well our amplifiers in an instant are asking for 500 amps so what happens?
In any constant voltage system when the current capability is exceeded the voltage drops. So let’s say our alternator voltage starts dropping. What does our cap do? Since its charge potential is at 14 volts it starts to discharge and provide a source of current. Since the cap is now sharing the load with the alternator it is called on to provide what the alternator can’t—that would be 300 (see footnote) amps.
What happens to the terminal voltage of our cap when 300amps is flowing? Well for starters, the voltage tries to drop nearly 5 volts inside the cap before it can even get out. Not in a short time but instantly. There is no time constant in the formulas for ohms law. They are instantaneous calculations! But wait. The voltage doesn’t really drop to 9 volts because we have our battery sitting in reserve waiting at 12.8 volts.
Our cap lets our poor alternator down as the voltage plummets and when things hit 12.8 volts our battery jumps in and starts to take over. The battery with its enormous storehouse begins to provide vast amounts of current until things lighten up for our poor cap and alternator. Of course we could add another cap to halve our ESR loss to only 2.5 volts but that would still cause the cap terminal voltage to drop to 11.5 volts.
Let’s see how many caps of this spec we would have to add to keep the voltage at 13.5 for even a few milliseconds. We would need a cap bank with a total ESL of about .001 ohm. Gee it looks like it would take over thirty caps paralleled to maintain 13.5 volts at 300 amps for even a brief instant. And let’s hope we don’t need to do this for long, as the total power contained in thirty units is only about what is in a dozen 9v alkaline batteries! (lesson 7)
It should be clear that if the voltage doesn’t drop the caps don’t do anything. The voltage MUST drop for them to start discharging.
