Problemo

Maybe someone can help, been wondering this for a while.

If an amp can put out 70w RMS @ 4 ohms, and 120w RMS @ 2ohms, would it be possible to run 2x 4 Ohm speakers in parallel (I believe its parallel which halves impedance, no doubt someone will correct me if I'm wrong !) thus causing the amp to see a 2ohm load.

Would this damage the amp in any way ? I say this because I'm thinking of running 2x sets of components up front (I say *thinking*) from one amp. Whats your thoughts, other than it might not be necessary as one set would be loud enough, but I'm helping someone build a show car, and door builds look pretty trick.

Any suggestions would be appreciated folks.

- Prob

zanthrax

almost right.

both imdepance and electric resistance are measured in Ohm.
but to calculate the total amount of Ohms in a circuit the formulas are reversed.
in imdepance it is:

parallel: just add them up:
so, 2 x 4 ohms in series = 8 ohms.

series: 1/Lt = 1/L1 + 1/L2
where
Lt = total imdepance
L1 = speaker 1's imdepance
L2 = speaker 2's imdepance
since L1 = L2 you can asume:
Lt = L1 (or L2)/2 (the number of speakers)

so 2x4ohms parallel = 4/2 = 2 ohms.

I have no clue doing that could damage your amp.
But I would recommend getting a 2 channel amp.

vteg

The amp would have to be 1 ohm stereo stable to drive a 2 ohm bridged load.

You will damage the amp.

1.8monster

it depends on what kind of amp you have really. there are some amps that say they are stable at 2 ohm but you can run it lower. most amps have an automatic shut off anyway. i would get a mono amp also to run it safely.

rjr162

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by zanthrax &raquo;</TD></TR><TR><TD CLASS="quote">almost right.

both imdepance and electric resistance are measured in Ohm.
but to calculate the total amount of Ohms in a circuit the formulas are reversed.
in imdepance it is:

parallel: just add them up:
so, 2 x 4 ohms in series = 8 ohms.

series: 1/Lt = 1/L1 + 1/L2
where
Lt = total imdepance
L1 = speaker 1's imdepance
L2 = speaker 2's imdepance
since L1 = L2 you can asume:
Lt = L1 (or L2)/2 (the number of speakers)

so 2x4ohms parallel = 4/2 = 2 ohms.

I have no clue doing that could damage your amp.
But I would recommend getting a 2 channel amp.
</TD></TR></TABLE>

actually the correct formula for parallel is

1
-------------
1 1 1
- + - + - ETC
A B C

or written another way:

1/((1/A) + (1/B) + (1/C) + etc)


Plus you mixed up parallel and series.. series you just add, and you can use the same formula for Impedence as you can resistance... the only issue with using it for resistors is the tolerance of resistors will make your final value off compaired to the actual value of the resistors because of the error factor.


EDIT&gt;&gt; wow this thing really chopped up my spacing job on the first "equation"

nsxxtreme

rjr is correct

hehe I got one for you. What is the resistance from A to B

vigor5spd

If you are running two speakers in parallel then the amp will only see a 2ohm load and you will be fine.

However you stated that you wanted to 2 sets of components which is 4 mids and 4 tweets. Now if you do this setup in parallel, then the amp will see a 1 ohm load and is not stable which will run hot and eventually get damage.

Clearify how many speakers you want to hook up.

nsxxtreme

Depending on the amp you may be ok. If the amp is rated for 4 ohmsX1 then the manufacturer does not recommend connecting it this way. There are many reasons why they may not recommend doing it this way. An Autotek mean machine I happen to know can power way more than what they are rated for. I have ran 1/4 ohm loads bridged with no problems other than heat. Which I used temperature controlled fans to remove.

So if you hook it up this way watch the heat. And use common sense if it keeps shutting down then stop running it that way.


Modified by nsxxtreme at 1:42 AM 11/23/2003

Problemo

In total I will be running 2 sets of components, 2x one side, 2x the other side using two of the available channels. As far as I understand it you can run either of the channels in 4 or 2 ohm mode. So I would have F1 and F2 channels running 2x component sets each at 2ohm. Does that make a bit more sense ?

Thus, the amp see's 2 ohms on each channel, which would be correct, it's just like running 1 2ohm speaker I believe.

Thanks in advance.

Prob

rjr162

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by nsxxtreme &raquo;</TD></TR><TR><TD CLASS="quote">rjr is correct

hehe I got one for you. What is the resistance from A to B

</TD></TR></TABLE>


Only took a quick glance, have to get to work, but all paths add up to 1500 ohms (or so it looked, maybe I'm off)... so the path of least resistance from A to B would depend on which resistors in your diagram had the highest error factor in a negitive fashion if you get what I mean...

zanthrax

here's a nice, complete explanation:
http://www.eatel.net/~amptech/...i.htm

I thought it was the other way around for imdepance.... my bad.

rcurley55

Do not run two comp sets up front! 95% of the time, multiple drivers playing the same frequencies is a BAD thing!

nsxxtreme

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by rjr162 &raquo;</TD></TR><TR><TD CLASS="quote">


Only took a quick glance, have to get to work, but all paths add up to 1500 ohms (or so it looked, maybe I'm off)... so the path of least resistance from A to B would depend on which resistors in your diagram had the highest error factor in a negitive fashion if you get what I mean...</TD></TR></TABLE>

The answer is R = 1/4 (3R) + 1/4 (1/R+1/R+1/R)

rjr162

heh BAD drawling! Where are the solid black dots to indicate connections between wires, and the §Ù (ok not quite what I was looking for but the closest thing I could find in character map) U shaped symbols to indicate non-connected wires that cross paths on the drawling. No wonder you tricked me! lol

vigor5spd

Prob,

Question, a component consist of a seperate mid and seperate tweet wired to a passive crossover. What is the ohm rating when the mid and tweet is hooked up the passive crossover? 2 ohms?

If it is 2 ohms and you want to run the left/right side in parallel the amp will see a 1 ohm load.

rjr162

no, it's 4 ohms. The way the cross-over network works out with the tweeter and mid, it ends up being a 4 ohm impedence.

vigor5spd

Yes, but he wants two sets on the left and two sets on the right.
Thats 4 speakers @ 4ohms wired in parallel would achieave a 1 ohm load.

Problemo

It doesnt work how you think. If an amp is 2 ohms stable, that means each channel (if working in a normal 4 channel/2 channel mode) would have 2 to be 2 ohms. In my case the amp is either 2ohm or 4ohm stable. So I could have 2x 4ohm speakers in parallel on one channel and the same on the other. This would provide the amp with a 2ohm load.

As far as I am aware (the more I read) this wouldn't cause the amp any problems as long as it did not run lower than specified on the amp (I.e. produce a 1 Ohm load and frig the power supply.

I *think* thats right ....

Let me know what ya all think.

- Prob

vigor5spd

You bridging the amp?

Problemo

Nah, it is purely being used to power full range speakers if you like. The sub is powered by another seperate sub.

- Prob