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Amplifier Question


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Got a question for any of you math minded folks. Start with a Peavey CS-1200: 1200 watt 2 channel amp (600 watts/ch @ 4 ohms). If you plug a 4 ohm speaker into each side then each channel is running nominally at 4 ohms and producing 600 watts. Now bridge the amp into mono. This should in theory give you 1200 watts @ 8 ohms according to the manual. If you were to plug in one of the 4 ohm speakers mentioned above, would that give you 2400 watts into 4 ohms? (Yes, I know that most speakers can't handle 2400 watts - this is a theoretical question). Would plugging both speakers in parallel give you a 2 ohm load w/4800 watts? Yes I realise that the amp would probably explode if driven at 2 ohms in bridge mode for any length of time, but I'm curious. I don't need to do this, I don't want to do this, I'm just looking for someone to confirm or deny if my thinking is correct or whether I've had too many Irish Coffees tonight :-) -- Rob P.S.; Yes, I know that a 2 ohm load at 4800 watts would draw something like 49 amperes and is thus almost completely outside the realm of usability. But this is only a theoretical situation after all :wave:
I have the mind of a criminal genius.....I keep it in the freezer next to mother.
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im guessing but it means that 1200 is the max and 8ohms is the minimal resistance so puttin a ohm on would blow it,thats why you cant use 2 ohms because as the power increases you have to have enough resistance im totally lost here,i hope someone can explain it ive been trying to figure out ohms for years
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Lee, You're a guitarist. Babe Richards. Why? -- Rob Drummer, vocalist, sound tech, burgeoning studio tech, and one hell of a paperweight
I have the mind of a criminal genius.....I keep it in the freezer next to mother.
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The short answer is that you blow your amp by dropping below the minimum resistance. IANA Scientist, but think of it this way: yes, a load from 8 ohms dropped to 4 ohms would almost double the output potential of the circuit, but if the electronics aren't rated to handle that much power it's like driving your car into redline. You can get away with it at lower revs (underloading the ohms at low power) but if you open it up, it's toast.
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[quote]Originally posted by Dwarf: [b] Why? Rob Drummer, vocalist, sound tech, burgeoning studio tech, and one hell of a paperweight[/b][/quote]ROFL!!!! Well, it's just that I really prefer to hear drummers play naked... I mean, errrhhh.... unamplified. Yeah, that's it. :D
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[quote]Originally posted by Franknputer: [b]The short answer is that you blow your amp by dropping below the minimum resistance. IANA Scientist, but think of it this way: yes, a load from 8 ohms dropped to 4 ohms would almost double the output potential of the circuit, but if the electronics aren't rated to handle that much power it's like driving your car into redline. You can get away with it at lower revs (underloading the ohms at low power) but if you open it up, it's toast.[/b][/quote]good post,
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voltage squared devided by impedence equals power in watts. V^2/Z = Power So, assuming an ideal amplifier, yes you are correct. If Z is halved, P is doubled. But in the real world, due to loading effects, V would also start to drop is Z gets to low, and like everyone else was to quick to mention, you would blow fuses, or ruin components in the amp. Does this help?
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[quote]Originally posted by velvetoceansound: [b]voltage squared devided by impedence equals power in watts. V^2/Z = Power So, assuming an ideal amplifier, yes you are correct. If Z is halved, P is doubled. But in the real world, due to loading effects, V would also start to drop is Z gets to low, and like everyone else was to quick to mention, you would blow fuses, or ruin components in the amp. Does this help?[/b][/quote]Does it help? I don't know, lemme get my decoder ring so I can figure out what the hell you said! :) You seem to be saying that yes, my thinking is correct as far as wattage goes. And no worries about blowing anything since there's no way I'd run the CS-1200 at 2 ohms in bridged mode. If I [b]needed[/b] that kind of power I'd probably invest in Crown Macrotech MA-5002VZ amps - 1775W/ch @4ohms in stereo mode. 3670W @8 ohms bridged. -- Rob
I have the mind of a criminal genius.....I keep it in the freezer next to mother.
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If you study the spec sheets of amplifiers you will find that the power doesn't double as the load impedance is halved. Due to inefficiencies in the amp (lower impedance means higher current and more power dissipation in the amp) the relationship isn't a straight line. There are better ways of cooking your burgers than using your amp as a griddle.
It's not a successful climb unless you enjoy the journey
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Running an amp in bridged mode essentially connects both amps in series, so that the negative output of channel one feeds the positive of channel 2, which is why you use the two hot leads for the outputs. Since each amp can at best handle a 2 ohm load, you cannot effectively drop below 4 ohms, as the amps combined circuit impedance would perceive this as effectively a short. There would be little to no advantage in power running this particualr amp in bridged mode. When you bridge the amp, the amps divide the load, which can be no more than 4 ohms. Each amp produces 600w@4ohms, so the combined power of bridged would be 1200w@4ohms. Hope this is helpful.

Hope this is helpful.

 

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[quote]Originally posted by where02190: [b]Each amp produces 600w@4ohms, so the combined power of bridged would be 1200w@4ohms. [/b][/quote]Actually, according to the manual it's 1200W at 8 ohms. But I don't think I'd trust a Peavey in that kind of power range. Luckily this was only an intellectual excersize so no one needs get hurt :) -- Rob
I have the mind of a criminal genius.....I keep it in the freezer next to mother.
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