I have a stack of Parasound Z amps. I have some neat Parasound mini-racks. I can put a preamp, tuner, amp, z-power controller, equalizer etc.
I would like to build a rack with a preamp, tuner and amps driving one stereo set of speakers, but the amps are only rated at 30wpc (8ohms) or 45wpc (4ohms).
The speakers would be a pair of Elac Uni-fi B5.2, single post, so I can not biwire/biamp them. Nominal 6 ohms.
Since I got so many amps laying around I want to drive more power per speaker.. I can easily fit two amps in a rack. MORE POWER.. Who can complain about MORE POWER? ;-)
The amps can be bridged easily and double their power. This increases the output voltage but keeps the current going. However, doing so will affect the speaker damping as it affects the output impedance ( higher ).
Another option would be to parallel the amps... using both channels of one amp to drive a speaker. That would keep the voltage the same but double the current.
I've been reading that I should put a very low impedance (in reference to the speaker load) on the output of each amp to prevent one channel driving into the other, but since the amps are almost identical, I wonder if I need to do this. And I can't find any equation to help out. I guess I want to minimize the output load both because I want to avoid parasitic power loses and having to buy a big resistor.
Questions then:
(1) The speakers are nominally 6 ohms... so if I put a half ohm on the (+) of each amp, a half watt should be enough? Can I get away with 1/4 ohm? What is the lowest I can safely go / find?
I can't find the output impedance spec for the amps, but they call for a damping factor of 400 at 20Hz... can I calculate the output impedance from that?
EDIT... this was a mistake.
(2) How would I connect this? I assume the speaker would go through both the (+) of each channel, the (-) wired together and the L and R inputs in phase.
(3) Will this affect the output impedance greatly ( concerned about damping factor). Assume the lowest possible output impedance.
(4) I figure the power will quadruple with twice the current. P = I^2 x R. Although the bridging spec is about 60 watts only, which I think could be because the voltage amplification creates heat in the amp and that limits the power output. Huh?
(5) Would I be better bridging them? (amps have a switch for just that).
Ultimately, what will it do to the sound quality? I prefer a tight bass.
As it currently have one rack set up, I use powered Quad 12LS speakers but I want to use the rest of the stuff and 30 watts just won't rock my world.
I would like to build a rack with a preamp, tuner and amps driving one stereo set of speakers, but the amps are only rated at 30wpc (8ohms) or 45wpc (4ohms).
The speakers would be a pair of Elac Uni-fi B5.2, single post, so I can not biwire/biamp them. Nominal 6 ohms.
Since I got so many amps laying around I want to drive more power per speaker.. I can easily fit two amps in a rack. MORE POWER.. Who can complain about MORE POWER? ;-)
The amps can be bridged easily and double their power. This increases the output voltage but keeps the current going. However, doing so will affect the speaker damping as it affects the output impedance ( higher ).
Another option would be to parallel the amps... using both channels of one amp to drive a speaker. That would keep the voltage the same but double the current.
I've been reading that I should put a very low impedance (in reference to the speaker load) on the output of each amp to prevent one channel driving into the other, but since the amps are almost identical, I wonder if I need to do this. And I can't find any equation to help out. I guess I want to minimize the output load both because I want to avoid parasitic power loses and having to buy a big resistor.
Questions then:
(1) The speakers are nominally 6 ohms... so if I put a half ohm on the (+) of each amp, a half watt should be enough? Can I get away with 1/4 ohm? What is the lowest I can safely go / find?
I can't find the output impedance spec for the amps, but they call for a damping factor of 400 at 20Hz... can I calculate the output impedance from that?
EDIT... this was a mistake.
(3) Will this affect the output impedance greatly ( concerned about damping factor). Assume the lowest possible output impedance.
(4) I figure the power will quadruple with twice the current. P = I^2 x R. Although the bridging spec is about 60 watts only, which I think could be because the voltage amplification creates heat in the amp and that limits the power output. Huh?
(5) Would I be better bridging them? (amps have a switch for just that).
Ultimately, what will it do to the sound quality? I prefer a tight bass.
As it currently have one rack set up, I use powered Quad 12LS speakers but I want to use the rest of the stuff and 30 watts just won't rock my world.
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Way to destroy a pair of amps... they will 'fight' each other and possibly also oscillate. You can fix that by adding series resistors on each output. I'd start high, like at least 1ohm, then take it down as far as dare goes.
But really, you will only get potentially twice the current. The load is the same isn't it? So the current will be excactly as before. The sound will certainly be worse as the signal gets mucked up by two devices trying each their own transfer and getting messed up by the other...
Very bad idea.
The only practical option is bridging. But often that also messes the sound a bit, the two different but same amp halves trying to agree on driving the load. Do you really need more power? Get more speakers, drive a wall of sound....
But really, you will only get potentially twice the current. The load is the same isn't it? So the current will be excactly as before. The sound will certainly be worse as the signal gets mucked up by two devices trying each their own transfer and getting messed up by the other...
Very bad idea.
The only practical option is bridging. But often that also messes the sound a bit, the two different but same amp halves trying to agree on driving the load. Do you really need more power? Get more speakers, drive a wall of sound....
Hmm... did you read my post or are you knee jerking your answer based on the title of the post?
(1) I did specify series resistors on the output.... my question what is the minimum value?
(2) The channels in the amps are identical, as identical as they can be... using both channels of a single stereo amp in parallel., so they ought to be very well matched. If they both have the same output impedance, and bias, they should not drive each other, huh, So I doubt they will oscillate.
(3) The power is the square of the current.. if the current doubles, the power quadruples.
(4) Bridging adds its own problems, mostly the output impedance.
(5) I got lots of speakers, amps, preamps, receivers.... so I just want to play around with these. Heck, I might even add a Reel To Reel.
So, my question again... given TWO VIRTUALLY EXACTLY ALIKE amplifier outputs... how do I calculate the output impedance based on a damping spec and how what would be the minimal output resistor (on each channel )... is 1/50 sufficient? For 60w, and a 6 ohm load, that would require a ~1ohm, ~1 watt resistor.. but that would sort of hose the damping, huh?
Please, READ my post fully before giving your opinion!
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Your neighbours, obviously, and you when you get tinnitus.
tonyEE: (1) The speakers are nominally 6 ohms... so if I put a half ohm on the (+) of each amp, a half watt should be enough?
I'd say at least 2 W.
Do you have any information about the gain differences between the channels and the DC offset differences? If so, you can calculate the peak output voltage difference between the channels and divide it by the series value of the two 0.25 ohm or whatever resistors.
Assuming it's 400 related to 8 ohm: 8 ohm/400 = 0.02 ohm.
Yes, because the current distribution resistors will be much greater than the amplifier output impedance, at least at low frequencies.
There will only be a small increase in power, because the power supply rail voltages will limit the output voltage.
Yes.
Pano once did an interesting test, see the thread https://www.diyaudio.com/community/...h-voltage-power-do-your-speakers-need.204857/
It's a long thread, but you only have to read the first couple of posts to be able to interpret the test results. About half the respondents never ever use voltage levels higher than those equivalent to 30 W into 8 ohm.
Like ejp wrote, paralleling makes sense when you have loudspeakers with an unusually low impedance. It's pointless otherwise.
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(1) Damping factor = 8/output impedance.
(2) Incomprehensible.
(3) No change.
(4) Only if you reduce the load impedance. Otherwise how can the current increase? Ohm's law. If you want more power into existing speaker loading you have to increase the voltage, i.e. bridge.
(5) Yes.
(2) Incomprehensible.
(3) No change.
(4) Only if you reduce the load impedance. Otherwise how can the current increase? Ohm's law. If you want more power into existing speaker loading you have to increase the voltage, i.e. bridge.
(5) Yes.
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Putting the amp in parallel increases the current, not the voltage, but you still deliver more power.
If the output impedance is 0.02 ohms, wouldn't it be better to put maybe a lot lower than 1/8th of an ohm? That seems like a lot of power will be lost in there. What is the lowest practical resistor I can find that is still accurate? I used to play with calibrated 0.01 ohm power resistors that we used for measuring current but those are expensive. More expensive than the amps I want to play with.
I'm using both channels of a stereo amp, so they should be as well matched as any could be.
Thanks. let me know if you can find that thread. I looked and couldn't find any such.
Oh, yeah, I'm not planning on doing this to tube amps. <:-0
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I updated the post and included the link.
That 2 W is just the power dissipation I calculated for the 0.5 ohm resistors you proposed, rounded up to an integer number of watt. I never recommended 2 ohm.
That higher current will never flow through your 6 ohm load, because the amplifier output voltage will clip before reaching the voltage needed to push that current through your 6 ohm load.
That 2 W is just the power dissipation I calculated for the 0.5 ohm resistors you proposed, rounded up to an integer number of watt. I never recommended 2 ohm.
That higher current will never flow through your 6 ohm load, because the amplifier output voltage will clip before reaching the voltage needed to push that current through your 6 ohm load.
(1) Thanks
(2) Yeah, you replied while I was editing the error in my reply to you. Sorry.
(3) I still don't get it why voltage have anything to do it. We are going for adding the current from two channels, the power is gained by more current, the voltage stays the same.
Yeah, I'm afraid of that... but the speakers I plan on using only go down to 45Hz or so.
You aren't doubling the current. How do you magically stop Ohm's Law from applying?
You'll need to halve the speaker's impedance in order to double the current into it.
Because you are not looking at the POWER equations..... which is what I'm after, MORE power.
If I double the current, then the power quadruples. ( Jeez... I guess that whole year doing ElectroMagnetism did pay for something after all).
To find the Power (P)
[ P = V x I ] P (watts) = V (volts) x I (amps)Also:
[ P = V2 ÷ R ] P (watts) = V2 (volts) ÷ R (Ω)
Also:
[ P = I2 x R ] P (watts) = I2 (amps) x R (Ω)
No, no... I know it's late.... but THINK. I'm doubling the CURRENT.
It's a PARALLEL arrangement of TWO AMPLIFIER CHANNELS.
"(4) Bridging adds its own problems, mostly the output impedance."
The output impedance will be increased a lot more by adding series resistors to paralleled amps than it will by bridging them. So long as the amps are happy driving 3 ohms you'll get four times the power by bridging than you would with a single amp, or a pair in parallel.
The output impedance will be increased a lot more by adding series resistors to paralleled amps than it will by bridging them. So long as the amps are happy driving 3 ohms you'll get four times the power by bridging than you would with a single amp, or a pair in parallel.
You aren't doubling the current. You are doubling the available current. That extra current cannot be drawn unless you increase the voltage or reduce the load. Ohm's Law. 200th anniversary coming up.
No need to shout. I = V/R. You're not proposing to alter V or R, so where's the extra I?
(it's 9:42AM here and I'm fresh as a daisy ;-))
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