amp?

[mcbuggin 1]

What are the current draws and rateings? Also welcome.

[Incriminator 16]

40.1's draw around 450 amps at 12 volts, 20.1's draw around 210 amps at 12 volts.

[mcbuggin 1]

40.1's draw around 450 amps at 12 volts, 20.1's draw around 210 amps at 12 volts.

man your quick thanks. does the draw go down with a 16 volt system I think that it would and what kind of power would they produce @16 volts

[lee 2]

If voltage goes up, current has to go up if resistance stays the same

[mcbuggin 1]

If voltage goes up, current has to go up if resistance stays the same

well makes sense don't know what I was thinking. thanks

[jmenace 0]

just lookin arroun for some amps for my future projects and i wanted to know where would i be able to buy your products

[amatt85 0]

i was wondering how much the 40.1s are? (shoot me a pm if you dont/cant post it in here plz)

also, with the proper electrical, can the 40.1s be burped at a .35load?

thanks

[jmenace 0]

no reply PM me too if possible i will be interested in the 20.1s if i can find where to buy them

[johnnylemoine 0]

If voltage goes up, current has to go up if resistance stays the same

so its different than A/C voltage?? cause i know that amp draw drops as voltage increses with A/C voltage

[TheOtherSide 0]

Remember the old formula: P= VI? This means that disregarding load, which is constant, and assuming that you are using the same wattage (lets say 1800W @ 1 ohm) and the voltage is 12V, the current draw is 200A (assuming 75% efficiency, I know it's 86% at 4 ohm, but lets use 75% at 1 ohm).

Now multiply that voltage to 16V, the same wattage, you get 150A load (assuming 75% efficiency).

So what this means is that if the output is regulated to be 1800W at 1 ohm, then yes, your current draw would go down. But if you are assuming the output wattage will increase proportionally to the voltage, then the new output (at 16V) would be about 77% higher, or roughly 3200W. Using this with the increased voltage, the current draw would increase to about 267A.

Hope this helps.

Edited September 23, 2007 by TheOtherSide

[lee 2]

Remember the old formula: P= VI? This means that disregarding load, which is constant, and assuming that you are using the same wattage (lets say 1800W @ 1 ohm) and the voltage is 12V, the current draw is 200A (assuming 75% efficiency, I know it's 86% at 4 ohm, but lets use 75% at 1 ohm).

Now multiply that voltage to 16V, the same wattage, you get 150A load (assuming 75% efficiency).

So what this means is that if the output is regulated to be 1800W at 1 ohm, then yes, your current draw would go down. But if you are assuming the output wattage will increase proportionally to the voltage, then the new output (at 16V) would be about 77% higher, or roughly 3200W. Using this with the increased voltage, the current draw would increase to about 267A.

Hope this helps.

Your formula is incorrect. Ohms law is, Voltage [ Electromotive force ] divided by Intensity [ amps or current all the same ] gives you resistance [ aka ohms ]. V divided by resistance gives you Intensity. Resistance x Intensity or Intensity x Resistance gives you Voltage.

Watt Law is Amps x Volts or Volts x Amps = Watts. Watts divided by Amps gives you Volts or Watts divided by Volts gives you amps.

So going off what you have at the top, If you know you have 200 amps of current draw and 12 volts, you're producing 2400 watts. Hows that you say, well amps [ 200 ] x volts [ 12 ] = watts. And your resistance is .06 Ohm, not very amp stable. How did I get .06 ohm, well 12 volts divided by 200 amps.

Now at 16 volts divided by your 150 amps is .107 ohms of resistance, again ohms law. How many watts is that, 16 volts x 150 amps = 2400 watts, again watts law. Why is it the same watts with 16 volts x 150 amps, and 12volts x 200 amps, that's because resistance increased. If you read my previous post carefully, I said if voltage goes up current has to go up if resistance stays the same. Well in this case, resistance did change, it went up from .06 ohm with 12 volts and 200 amps of current to .107 ohm with 16 volts and 150 amps of current. To answer your question about having less current with higher voltage is, voltage is the push, current is the flow of electrons. So when you have a higher opposition to flow, resistance [ ohms ] current goes down. If resistance goes down current flow goes up.

The third problem is 16 volts x 267 amp draw = 4,272 watts @ .06 ohm load

Hope this didn't confuse anybody, but those are the correct #s for the examples given.

Edited September 24, 2007 by Earthman

[TheOtherSide 0]

I wasn't trying to be too confusing but this is all "theoretical" Your basing on 100% efficiency which is not possible, ever. Also, you are basing your math on the WRONG numbers to get 0.06 ohm. The output V, I are not the same as the input. I am not trying to be rude, bashing, or confusing, but read further, I'll explain. Also, if you look at your calculations, 0.06 ohm is pretty good for the input stage of a device. Almost no resistance...that is a good thing.

Your interpretation of resistance is a little askew though. Your output voltage is not 12V or 16V as you think. It is based on the load that is presented at the output of the amplifier stage. My #'s are based on input only and are not output dependent. If you are producing a theoretical 1800W at 1 ohm and you know the input voltage, you can assume the current draw. That is, you know the output power, you know the load (resistance), and you want to know the current draw or as you put it "intensity."

Looking at it from the output stage, if you want 1800W at 1 ohm, you must set the voltage accordingly...thus the Gain Setting of the amplifier. You can vary the voltage of the amplifiers output, not the current. What does that tell you? The input is steady, if you draw 200A at 12V at full load, you will be outputting say, 47.4A at 47.4V [remember, you want 1800W output at 1 ohm thus P=V^2/R and P=I^2*R] (test the output of an amplifier and tell me if it ever reads 12V...typically, it isn't even close). That does not factor into Ohms law. You are thinking of both the output and input voltages as being identical...they aren't. Plus, the output current is not equal to the input current. There is loss, heat dissipation (efficiency), and other parts of the amp that use power that is not directly correlated to output power. In a situation like this, you must assume the load and power as being constant (non-factors) and the only variables are the input voltage and current. Increase either and the other will also increase as the amplifier produces more power.

This is about the easiest way to explain it. If you are even more confused or think I am wrong, message me and I will try to explain it better.

Edited September 24, 2007 by TheOtherSide

[mcbuggin 1]

I wasn't trying to be too confusing but this is all "theoretical" Your basing on 100% efficiency which is not possible, ever. Also, you are basing your math on the WRONG numbers to get 0.06 ohm. The output V, I are not the same as the input. I am not trying to be rude, bashing, or confusing, but read further, I'll explain. Also, if you look at your calculations, 0.06 ohm is pretty good for the input stage of a device. Almost no resistance...that is a good thing.

Your interpretation of resistance is a little askew though. Your output voltage is not 12V or 16V as you think. It is based on the load that is presented at the output of the amplifier stage. My #'s are based on input only and are not output dependent. If you are producing a theoretical 1800W at 1 ohm and you know the input voltage, you can assume the current draw. That is, you know the output power, you know the load (resistance), and you want to know the current draw or as you put it "intensity."

Looking at it from the output stage, if you want 1800W at 1 ohm, you must set the voltage accordingly...thus the Gain Setting of the amplifier. You can vary the voltage of the amplifiers output, not the current. What does that tell you? The input is steady, if you draw 200A at 12V at full load, you will be outputting say, 47.4A at 47.4V [remember, you want 1800W output at 1 ohm thus P=V^2/R and P=I^2*R] (test the output of an amplifier and tell me if it ever reads 12V...typically, it isn't even close). That does not factor into Ohms law. You are thinking of both the output and input voltages as being identical...they aren't. Plus, the output current is not equal to the input current. There is loss, heat dissipation (efficiency), and other parts of the amp that use power that is not directly correlated to output power. In a situation like this, you must assume the load and power as being constant (non-factors) and the only variables are the input voltage and current. Increase either and the other will also increase as the amplifier produces more power.

This is about the easiest way to explain it. If you are even more confused or think I am wrong, message me and I will try to explain it better.

well earthman and otherside even though this is a lil bit over my head it was a great read. thanks and I guess that i'll have to practice some math now

[TheOtherSide 0]

well earthman and otherside even though this is a lil bit over my head it was a great read. thanks and I guess that i'll have to practice some math now

To be quite honest, I may have confused myself. As long as the amp is unregulated, and it can handle the increased voltage, the current draw shouldn't change that much. For example, at 14.4V, an amp is making 3600W into a 2 ohm load will draw roughly 250A (lets assume it's 100% efficient to cut out some BS). At 17V, it makes 4000W into the same 2 ohm load. In this case, the current draw is about 236A. Because we are neglecting the changes in efficiency based on the load and voltage, the current is different but you get the idea hopefully. (BTW these are published numbers for an amp...maybe a little incorrect but published nonetheless...http://www.overstock.com/Electronics/Clif-Designs-4000W-Digital-Clas-X-Amplifier/1406009/product.html)

I can see slight changes in the current occuring but nothing major...maybe 5 amps. Yes, I know the above numbers don't exactly work out but it gives you an idea. There is a lot more to determining these variables than just pluging and chugging the numbers into an equation. You have to understand heat loss, efficiency, work, electrical bleed off, etc. It's a lot to take into consideration. If you want to really think about it, why then do we use the same fuse for the amplifier at steady voltage (12V from the battery) and when the car is running (typically 14.4 V)? You have to assume the current isn't going to change thus resulting in the use of the one fuse rating.

So to correct myself, the current shouldn't change between 12, 14.4, 16 volts as long as the amp is unregulated. The power, however, will.

[RXZILLA 0]

I can't wait till the new amps come out....I know they will be awsome.

[jmenace 0]

where are the IA 40.1 amp availible to purchase

[ia2 21]

email us for purchase info [email protected]

[damonhargroves07 0]

lol wheres devildriver to put his input on all this arguing over wattage and voltage. lol!

[BKOLFO4 29]

40.1's must be pretty darn strong. . . . .I only have two of mine in right now running them at 2 ohms each. That should be a total of 3600 watts to my (4) 15's. From the output and the amount of cone excursion I am getting, I am going to guess they may be putting a little more than 900 watts per driver.

Going to be interesting to get all 4 installed running at 1 ohm each!

Brian

[johnnylemoine 0]

well that was a good read about the voltage and amp draw relations. didnt mean to start an argument or anything. i was just curious about the first statement because i know at work where i deal with A/C voltage electric motors for air conditioners and whatnot, i know i have less amp draw the higher the voltage. didnt know if it was different with D/C voltage or the fact that resistance changes so much in car audio and whatever else may affect the amp draw verses the voltage.