Capacitor
05-25-2006, 02:47 PM
#41
50 Watt CAFz'r
Join Date: Nov 2004
Posts: 167
recharge rate:
time in seconds = (0.5*FARADs*TARGETVOLTS*VOLTSIN)/Power supply in J/s
effects of ESR on your overall voltage:
ir=v
where i=amperage of system
r=ESR rating of the capacitor
v=voltage drop experienced
the fact is that a capacitor cannot create current, but it can store it temporarily
although the average measured voltage over a given period of time will not change because the current draw is still the same, the dips and peaks will be closer to this average. music, and therefore current draw, is very transient in nature; a capacitor simply helps with those transient peaks and dips.
time in seconds = (0.5*FARADs*TARGETVOLTS*VOLTSIN)/Power supply in J/s
effects of ESR on your overall voltage:
ir=v
where i=amperage of system
r=ESR rating of the capacitor
v=voltage drop experienced
the fact is that a capacitor cannot create current, but it can store it temporarily
although the average measured voltage over a given period of time will not change because the current draw is still the same, the dips and peaks will be closer to this average. music, and therefore current draw, is very transient in nature; a capacitor simply helps with those transient peaks and dips.
05-25-2006, 03:02 PM
#42
Originally Posted by GPGT1
I know that a cap filters in AC but we're dealing with DC
What? This is 12 volts DC like i said there is no filtering going on you should read up on your electrical theory, and thicker wire would have much less loss at our power levels smaller wire could never carry that load.
well heres some crude math, now not taking into effect what the altenator is putting out or anything. lets say you were running a 60hz tone at and the amp is drawing 1000watts. for half a wave it takes 0.0083 seconds. at that power it will take a 1 farad charged to 14.4volts cap 0.2seconds to discharge on its own with no power supply(alt or batt.). After your half wave was done playing you would be at 13.81volts, not too bad. Now if the batt and alt were connected, that cap would have a chance to recharge between peak cycles. Now with caps when they are empty they take longer to discharge then to charge and when they are full more time to charge then to discharge. giving that 1 farad cap the same time to charge as we did to discharge it would be at 14.37 volts from 13.81. so this is where having an good charging system comes in. To give that cap slightly more time to charge, the alt would need to be able to handle a sufficiant load.
I hope you dont work with electricity
What? This is 12 volts DC like i said there is no filtering going on you should read up on your electrical theory, and thicker wire would have much less loss at our power levels smaller wire could never carry that load.
well heres some crude math, now not taking into effect what the altenator is putting out or anything. lets say you were running a 60hz tone at and the amp is drawing 1000watts. for half a wave it takes 0.0083 seconds. at that power it will take a 1 farad charged to 14.4volts cap 0.2seconds to discharge on its own with no power supply(alt or batt.). After your half wave was done playing you would be at 13.81volts, not too bad. Now if the batt and alt were connected, that cap would have a chance to recharge between peak cycles. Now with caps when they are empty they take longer to discharge then to charge and when they are full more time to charge then to discharge. giving that 1 farad cap the same time to charge as we did to discharge it would be at 14.37 volts from 13.81. so this is where having an good charging system comes in. To give that cap slightly more time to charge, the alt would need to be able to handle a sufficiant load.
I hope you dont work with electricity
I would like to understand the quick math that you did and I have some questions:
1: what amplifier efficiency did you use to calculate the current draw, and what was that current draw that you used in the calculation?
2: what ESR did you assume for the cap
3: what current was the alternator suppying, or what was the internal resistance of the alternator, and what was the resistance of the wire from the alternator to the cap/amp?
And by the way, it takes cap the same time to charge and to discharge from one voltage to another, given the same current. It is not just the voltage that determines the charge rate, but the RC time constant: (Rsource +ESRcap)xC.
The charging and discharging of the cap, the voltage pulsing form 13.8 to 14.4 is changing current and voltage. Last time I chequed, when a voltage and current changes, it is an AC waveform on a DC. The cap is a filter that filters out the pulses caused by the amp drawing current. We can look a this problem from a frequency domain point of view and say that the cap is FILTERING out the changing ALTERNATING waveform from the DC part, or in a time domain point of view we can say that the cap is evening out the pulses. It is the same regardless.
And since we seem to be in a discussion about credentials: Yes, as a matter of fact I do work in the electonics field, have an electrical enginering degree, and have spent the last 20 years in design, simulation and measurements of electrical systems.
05-25-2006, 03:10 PM
#43
Originally Posted by Haunz
I agree with dukk.....
voltage drop in a vehicle is primarilly caused when we exeed the current draw that the alternator can provide which is related to its RPMs and the IR losses in the windings... from there we should expect that our batt can provide gobs of current, however batteries also suffer from IR loss as well as voltage drop due to concentration polarization (and a few other factors)....
voltage drop in a vehicle is primarilly caused when we exeed the current draw that the alternator can provide which is related to its RPMs and the IR losses in the windings... from there we should expect that our batt can provide gobs of current, however batteries also suffer from IR loss as well as voltage drop due to concentration polarization (and a few other factors)....
Originally Posted by Haunz
If you believe in RF, they can show you a test report showing thier 100F carbon caps can significantly improve amplifier performance.... of course we all know that the test was probably flawed and that RF's marketing department loves to spew fluff... 
I liked your post. Lots of researched information.
05-25-2006, 03:18 PM
#44
50 Watt CAFz'r
Join Date: Nov 2004
Posts: 167
Originally Posted by zoomer
a 100F cap would probably do the trick, but this may be only worth it for extreem SPL applications. If it for SQ, it would be cheaper to buy a better amp!
I liked your post. Lots of researched information.
I liked your post. Lots of researched information.
05-26-2006, 04:22 AM
#48
points I dont think anyone is considering is the dynamics of the situation and how the alt/batt/cap will interact...
we can say a 1f cap holds around 75 joules of usefull energy before there would be an audible drop in amp output level (14v-11v = 1db drop)... that is enough to sustain a bass note at 1000 watt levels for .075 of a second... (transient portions of a musical passage may last less then 1/10 that time while some may last 10x or even longer)
If we say we have a true 75 amp @ 14v alternator we may say we can sustain > 2kw for that same period or probably even longer considering we may get over 120amps at 11v from the same system.... If we throw the battery into the picture things get more complicated...
But the key idea behind using a cap is that it is supposed to dump current into the amplifier faster then the battery can; and sustain current draw untill the battery can catch up....
(this is of course assuming we are already exceeding what the alternator can provide)
It is thus important to consider that the discharge rate of the battery or cap must be pictured as an RLC circuit where discharge rate is related to voltage differential, resistance, and inductance...
In lamans terms the key idea being that we could have a battery that sits at 14.4V all day long but if it's series inductance or resistance is too high we may never get power to the amps from the battery in time to reproduce short lived transients...... in the same sense if the inductance of our cap is too high it will suffer the same fate.....
for some real world examples I suggest looking into RC's archives over at carsound.com
we can say a 1f cap holds around 75 joules of usefull energy before there would be an audible drop in amp output level (14v-11v = 1db drop)... that is enough to sustain a bass note at 1000 watt levels for .075 of a second... (transient portions of a musical passage may last less then 1/10 that time while some may last 10x or even longer)
If we say we have a true 75 amp @ 14v alternator we may say we can sustain > 2kw for that same period or probably even longer considering we may get over 120amps at 11v from the same system.... If we throw the battery into the picture things get more complicated...
But the key idea behind using a cap is that it is supposed to dump current into the amplifier faster then the battery can; and sustain current draw untill the battery can catch up....
(this is of course assuming we are already exceeding what the alternator can provide)
It is thus important to consider that the discharge rate of the battery or cap must be pictured as an RLC circuit where discharge rate is related to voltage differential, resistance, and inductance...
In lamans terms the key idea being that we could have a battery that sits at 14.4V all day long but if it's series inductance or resistance is too high we may never get power to the amps from the battery in time to reproduce short lived transients...... in the same sense if the inductance of our cap is too high it will suffer the same fate.....
for some real world examples I suggest looking into RC's archives over at carsound.com
05-27-2006, 07:25 PM
#49
Originally Posted by Haunz
points I dont think anyone is considering is the dynamics of the situation and how the alt/batt/cap will interact...
we can say a 1f cap holds around 75 joules of usefull energy before there would be an audible drop in amp output level (14v-11v = 1db drop)... that is enough to sustain a bass note at 1000 watt levels for .075 of a second... (transient portions of a musical passage may last less then 1/10 that time while some may last 10x or even longer)
If we say we have a true 75 amp @ 14v alternator we may say we can sustain > 2kw for that same period or probably even longer considering we may get over 120amps at 11v from the same system.... If we throw the battery into the picture things get more complicated...
But the key idea behind using a cap is that it is supposed to dump current into the amplifier faster then the battery can; and sustain current draw untill the battery can catch up....
(this is of course assuming we are already exceeding what the alternator can provide)
It is thus important to consider that the discharge rate of the battery or cap must be pictured as an RLC circuit where discharge rate is related to voltage differential, resistance, and inductance...
In lamans terms the key idea being that we could have a battery that sits at 14.4V all day long but if it's series inductance or resistance is too high we may never get power to the amps from the battery in time to reproduce short lived transients...... in the same sense if the inductance of our cap is too high it will suffer the same fate.....
for some real world examples I suggest looking into RC's archives over at carsound.com
we can say a 1f cap holds around 75 joules of usefull energy before there would be an audible drop in amp output level (14v-11v = 1db drop)... that is enough to sustain a bass note at 1000 watt levels for .075 of a second... (transient portions of a musical passage may last less then 1/10 that time while some may last 10x or even longer)
If we say we have a true 75 amp @ 14v alternator we may say we can sustain > 2kw for that same period or probably even longer considering we may get over 120amps at 11v from the same system.... If we throw the battery into the picture things get more complicated...
But the key idea behind using a cap is that it is supposed to dump current into the amplifier faster then the battery can; and sustain current draw untill the battery can catch up....
(this is of course assuming we are already exceeding what the alternator can provide)
It is thus important to consider that the discharge rate of the battery or cap must be pictured as an RLC circuit where discharge rate is related to voltage differential, resistance, and inductance...
In lamans terms the key idea being that we could have a battery that sits at 14.4V all day long but if it's series inductance or resistance is too high we may never get power to the amps from the battery in time to reproduce short lived transients...... in the same sense if the inductance of our cap is too high it will suffer the same fate.....
for some real world examples I suggest looking into RC's archives over at carsound.com
This makes no sense at all since amp power output vs DC supply voltage will be totaly dependent on the regulation of the amps switching power supply!
05-27-2006, 08:51 PM
#50
Originally Posted by zoomer
How did you calculate the 1 db drop in amplifier output power due to a drop of DC supply from 14.4 V to 11 V?
This makes no sense at all since amp power output vs DC supply voltage will be totaly dependent on the regulation of the amps switching power supply!
This makes no sense at all since amp power output vs DC supply voltage will be totaly dependent on the regulation of the amps switching power supply!
It makes perfect sense since a regulated swicthing mode power supply must make up for the loss of voltage with a increase in current. In an ideal world, the lower the input voltage, the higher the current draw, but the transistors can only handle so much current in the power supply side of the amp, not to mention the traces and other components in the power supply are only designed for so much current. Not only these issues, but the pulse width modulator only has so much head room, once you get to 50% duty cycle, that's as far as you can go with power for the amplifier DC rails. The more current you have to use to keep up with the amplifier's needs, the faster the PWM will max out... The better regulated and higher the voltage (within the amp's limits), the more available power there will be for the amplifier's rail voltages.
edit: how he calculated this 1dB drop, I have no idea, but it does make sense that the amp cannot make as much power with a lower voltage at the input...
