dead amp?
09-20-09, 12:09 AM
#1
dead amp?
Can a Capacitor (Cap) fry an amp? I have a JL 500/1 that I just got back from being warranty by JL. The amp was working fine when I first installed it, but as I was putting everything in my trunk back together in and was finishing up, I went to start up my car to see how the sub was hitting. But the amp wouldn't turn on. So did the cap fry my amp?
09-20-09, 08:50 AM
#2
Zombie Slayer
Can a Capacitor (Cap) fry an amp? I have a JL 500/1 that I just got back from being warranty by JL. The amp was working fine when I first installed it, but as I was putting everything in my trunk back together in and was finishing up, I went to start up my car to see how the sub was hitting. But the amp wouldn't turn on. So did the cap fry my amp?
09-20-09, 10:13 PM
#4
Zombie Slayer
If the cap is bad it wont hold a charge. This excerpt may help a bit.
The following is borrowed from mobileelectronics.com.au
AND is an excerpt from CarSound written by Richard Clark. (yes, that richard clark.....for those of you that are unfamiliar google his name and witness how he pioneered and advanced car audio!!!)
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.
Now, is it possible to have a steady 14 V because we added caps? I don’t think so.
The following is borrowed from mobileelectronics.com.au
AND is an excerpt from CarSound written by Richard Clark. (yes, that richard clark.....for those of you that are unfamiliar google his name and witness how he pioneered and advanced car audio!!!)
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.
Now, is it possible to have a steady 14 V because we added caps? I don’t think so.
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