Outlaw Audio home shop products hideout news support about
Page 3 of 3 < 1 2 3
Topic Options
#10411 - 09/10/02 08:28 PM Re: I'd like to bi-wire
charlie Offline
Desperado

Registered: 01/14/02
Posts: 1176
A good way to look at it is considering what the effective impedence will be. Driving a full range signal into, say, a properly crossed over tweeter will result in 'normal' current flow only in the treble range, so taken as a whole (voltage v. current) the apparent impedence of that load would be very high. So your treble amp might see an equivelent (depending on signal) of 20+ ohm load while the mid-bass amp might see a 11 ohm load (just pulling numbers out of air for illustration.)

The amp may run cooler, suffer from less rail sag, etc, but you won't actually get rated output except in limited conditions, such as an input signal that falls completely within it's frequency range. In this case the effective impedence would be back to the expected value.
_________________________
Charlie

Top
#10412 - 09/11/02 08:51 AM Re: I'd like to bi-wire
MrSandman Offline
Gunslinger

Registered: 04/20/01
Posts: 128
Loc: Charlotte, NC, USA
Steves & Gonk, I think there may be a semantic difference in the way you are describing the same thing. I agree that each side of the X-over would see a maximum of 100W (using the 7100 with 2 channels bi-amping). But there are both a high and a low pass X-over (assuming typical 2 way bi-wireable speaker with passive X-overs) therefore the speaker cabinet has a total potential of delivering 200W.

If the same speaker had jumpers between the bi-wire terminals, then the voltage would be the same, but the current would be divided between the two crossovers. Assuming (the impossible!) that the characteristics (impedance, phase shift etc) of the low and high X-over and their related drivers were identical in every way, then half the current would go each way and therefore the wattage would be 50W per X-over for a total of 100W per speaker cabinet.

The above is assuming that every item acts as the ideal and losses/shifts etc are not accounted for. I am not considering Charlie's suggestion of an active X-over, as I have absolutely 0 experience. What he said makes sense, though. If there is an obvious omission, please correct, as we are all learning every day.

Last thought: Since you have to double power for a + 3db gain, is it worth bi-amping for an SPL approach only? Gut feeling is no. . .

S.

Top
#10413 - 09/11/02 10:24 AM Re: I'd like to bi-wire
BenjaminRigby Offline
Gunslinger

Registered: 03/15/02
Posts: 120
Loc: McHenry, IL, USA
Gonk seems to have the right idea with what I was asking. I was kind of nervous about going with a larger amp as the center and surrounds are rated to 175 and 125 Watts of input power(both are 8 ohms nominal, 4 ohms minimum). I could go with the more powerful surrounds that can handle 300 watts at 6 ohms, but this would cost more. I'm all for cost effectiveness as I'm sure many of you are too.

The 4 ohms min is what gets me going about using the 770 or 755. 300 Watts into 125 or 175 recommended seems to be kind of far over. I would not be playing at extremely loud levels, but I would like to eliminate the chance of overdriving them as much as possible.

This leads me to another question, with the resistance ratings of these speakers, will the amp always try to output at 4 ohms? Or will it be a little lower? I realize that technically the resistance is not entirely constant, but on average, what will it do?

Mr. Sandman, I know doubling the power will only give 3db more, but at the same volume level the sound will come through better.

Top
#10414 - 09/11/02 11:05 AM Re: I'd like to bi-wire
BenjaminRigby Offline
Gunslinger

Registered: 03/15/02
Posts: 120
Loc: McHenry, IL, USA
Here is a message straight from Scott. Less than two hours for a reply too!

Hello Ben,

Thanks for contacting the Outlaws!

Amplifiers increase or decrease amperage based upon the impedance or resistance that a speaker pushes back on the signal. This means that the lower the impedance, the higher the amperage and in turn the higher the output wattage. So in answer to your inquiry of "if the speakers are capable of handling a 4 ohm load, will the amp always try to output at 4 ohms", the amp will increase or decrease the amperage based on whatever impedance it sees. As your speakers increase and decrease their impedance over the frequency spectrum, the amp will adjust automatically. Since they have a nominal impedance of four ohms, the amplifier will operate those channels at 4 ohms for the majority of the time.

As for your bi-amping inquiry, the mids and highs would receive 150 watts and the low frequency drivers would also see 150 watts, bringing the total to 300 watts. As a side note, you may want to check with Mirage to ensure that you do not need an external cross-over when bi-amping their speakers.

Please do not hesitate to let me know if you have any other questions.

Best Regards,

Scott

Top
#10415 - 09/11/02 11:34 AM Re: I'd like to bi-wire
charlie Offline
Desperado

Registered: 01/14/02
Posts: 1176
I'm not sure where you stand as far as electrical theory in your background. If this seems overly simple please forgive me - I'm just trying to offer a complete explaination.

Scott is correct in that you would have N x 2 watts AVAILABLE, however that would not really be delivered.

Watts are computed by multiplying amps by volts. So an 8v RMS sine wave into an 8 ohm impedence would drive one ampere, resulting in an 8 watt power disipation across the load.

Amplifiers are a voltage source - the output voltage is designed to be a magnified version of the input voltage instant by instant. They are also capable of supplying a relatively large amount of current, but the actual current draw is not controlled tightly by the amp. The load is allowed to draw whatever it needs at the currently applied voltage, instant by instant.

A speakers impedence will not be altered by your amplifier.

If a uniformly distributed (noise) signal is used as an example, and it is driven into a speaker with a crossover that splits the current (and power) in half, and the overall speaker impedence with this signal is 8 ohms, then under these conditions each half of the speaker is exhibiting a 16 ohm impedence. If the speaker inputs are split between two amps each amp (all else not changed) would see a 16 ohm effective load. Total power available to this system would be whatever the amps can deliver into 16 ohms (NOT 8) x 2.

So you may get a bit more power, but I doubt it would be much, as most amps are optimized to drive loads 8 ohms and below.

[This message has been edited by charlie (edited September 11, 2002).]
_________________________
Charlie

Top
#10416 - 09/11/02 01:44 PM Re: I'd like to bi-wire
steves Offline
Desperado

Registered: 06/18/01
Posts: 356
Loc: Oregon
Quote:
Steves & Gonk, I think there may be a semantic difference in the way you are describing the same thing.
I think you are right. While trying to simplify things, I was ignoring the SUM of the watts available to both sides as I figured it might be confusing and not really give Benjamin the answer to what I thought he was looking for. That will teach me for thinking! All the follow up replies have been excellent-even I understand now.

Top
Page 3 of 3 < 1 2 3

Who's Online
0 registered (), 68 Guests and 0 Spiders online.
Key: Admin, Global Mod, Mod
Newest Members
modajoico, Andre, UfFrank, BFC, Rick Abbott
8655 Registered Users
Top Posters (30 Days)
Forum Stats
8,655 Registered Members
88 Forums
11,312 Topics
98,654 Posts

Most users ever online: 386 @ 01/14/20 03:43 PM