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Posted: Mon Dec 04, 2006 11:59 am
by black00
Jlongo wrote:Very good stuff:) I thouroughly enjoyed reading it, and yes this should be a sticky on this forum and many other's.

good idea.

Posted: Mon Dec 04, 2006 12:20 pm
by Thoraudio
almond wrote:i don't think you can really tell an amps potential by looking at the fuses though. when you do amps X voltage = wattage, what is the impedance. someone explain this for me?

Impedence doesn't matter.

Current times Voltage times efficiency gives the RMS output of the amp.

Low impedence will drop the efficiency however.

Posted: Mon Dec 04, 2006 1:31 pm
by almond
so if the amp pushing say 500 watts at 4 ohms and 1000 watts at 2 ohms the amp will pull say 50 amps at 4 ohms and 100 amps at 2 ohms?? i'm thinking because the 2 ohms is allowing more current??

Posted: Mon Dec 04, 2006 2:15 pm
To say it correctly, it is because it is making more power, not just more amps.
Just in general, think of it like this:

The amp's outputs have a maximum voltage that can be present before clipping.

Voltage = current * resistance, so with a set voltage, if the impedance drops, the current increases.

Power = voltage * current, so more current = more power.

Now regardless of output current, voltage, or whatever impedance is on the output, you just need to know how much power is on the output. The power in = power out divided by the efficency.

Power = current * voltage, the input is 14.4 volts, so you can determine the current.

If the amp is 1000 watts, and it is 80% efficent, then it has to have 1250 (1000/.8) watts at the input. At 14.4 volts, it needs 87 amps to make 1000 watts on the output. Noticed I never had to know the load or output current to determine that; It does not matter if that is 1000 watts at 4 ohms, or 1000 watts at 1 ohm.

As already stated, the big factor is that the amp's efficency is going to change as the load impedance drops. From the formulas above, if the efficency changes with different loads, the input current will not be a linear relationship. And just as a side note, a lot of amps to not actually double their output when the load impedance is cut in half.

In your example, the amp may need 50 amps with the 4 ohm load, but it may need 110 amps with the 2 ohm load, and it may not actually make double the power on the output.


Posted: Mon Dec 04, 2006 4:19 pm
by SQBubble
it really has 6 40amp fuses or were you sarcastic? lol

Posted: Mon Dec 04, 2006 4:25 pm
It has 6 - 3 of them are internal, but they are all in parallel.

Look at this picture:

power/gnd connections - three fuses - speaker connections - three more fuses


edit - picture pops up really small. Go to the original page you posted and look at the two middle pictures on the top row.

Posted: Mon Dec 04, 2006 4:51 pm
Thoraudio wrote:for 2000 watts (RMS), a class D amp would need almost 185 amps of current (at 14.4v and 75% efficiency).

Well, as an expert on US Amps I can say that I guess it's a good thing that we have 6 40 amp fuses ( 3 external and 3 internal ) on the MD3D for a total of 240 :cwm18:

Posted: Mon Dec 04, 2006 5:55 pm
by SQBubble
ohhh now I see it!!


now i am more confident and have a good eye on that amp:)

I have been told that sometime some compagnies put on some fake fuse to make it look like it has alot and make alot of power... how can you know if an amplifier has fake fuses?

Posted: Mon Dec 04, 2006 6:02 pm
Man I like them. . . .

Posted: Tue Dec 05, 2006 1:29 am
by almond
i also forgot that when i clamped my amp it was doing 900 watts rms at 3 ohms. and i also realize that the amp won't usually see a 1 ohm load after impedance rise.