As I understand it, the 'output rating' of an AMP is really at what output wattage you reach 1% distortion.
It's not that you are changing the output power for the same given input... is how much clean output you get from a different inputs.
So, for a xxV input signal, you will always get the same yyV output signal with the ACA. The difference is that, in some monoblock configurations, the AMP can take more input (hence put more output) before reaching the same levels of distortion.
That's my level of understanding right now, I could be completely mistaken.
Rafa.
It's not that you are changing the output power for the same given input... is how much clean output you get from a different inputs.
So, for a xxV input signal, you will always get the same yyV output signal with the ACA. The difference is that, in some monoblock configurations, the AMP can take more input (hence put more output) before reaching the same levels of distortion.
That's my level of understanding right now, I could be completely mistaken.
Rafa.
Yes, and that is absolutely correct.
The reason you are confused is that you need to understand that in a monoblock configuration, we have (4) boards powering 2 speakers. In a monoblock configuration, using the XLR inputs pin 2 has an AC signal in normal phase, and pin 3 has an AC signal that is 180 degrees out of phase. Each of the boards in the monoblock now amplify those signals (in phase and 180 degree out of phase signals) SEPARATELY and then that gets connected to the speaker binding posts (the two black posts on the output). ONLY ONE speaker is connected to this monoblock. The output voltage is not 8V RMS but a little more about 11V RMS, which is right around 15 watts for an 8 ohm load.
So yes, it is 10-11 dB gain for both configurations but the output powers are different. Stereo (single chassis) using RCA's is 8 watts/channel for an 8 ohm impedance and Monoblock (dual chassis) using balanced XLR inputs is 15 watts/channel for an 8 ohm impedance. A stereo amplifier works with a pair of speakers. And the monoblock set works with a pair of speakers too.
Reading more about how bridge tied loads work may help you. In addition, understanding the differences between gain and output power will help as well.
What speakers do you have, what is its efficiency? How big is your room? How far do you sit from your speakers? How much gain is there in your preamp? What is your sources? What is its output voltage in RMS? How loud do you listen?
The answers to these questions will help you elucidate whether you have enough power and enough gain for most of your musical recordings.
Don't feel bad, it's not straightforward in the beginning and asking questions is the only way to learn!
Best,
Anand.
So, to be practical:
- If I connect one stereo ACA to a pair of 8 Ohm speakers and I play music at X volume in the source, I would hear music at Y loudness.
- If I connect two monoblock ACAs to the same 8 Ohm speakers and I play the music in the same source at the exact same X volume (therefore outputting the same volts), would I hear the same Y loudness from the speakers? or would I hear more volume coming out of the speakers? (hence an apparent 'higher gain')
Thanks for any insight,
Rafa.
- If I connect one stereo ACA to a pair of 8 Ohm speakers and I play music at X volume in the source, I would hear music at Y loudness.
- If I connect two monoblock ACAs to the same 8 Ohm speakers and I play the music in the same source at the exact same X volume (therefore outputting the same volts), would I hear the same Y loudness from the speakers? or would I hear more volume coming out of the speakers? (hence an apparent 'higher gain')
Thanks for any insight,
Rafa.
Rafa,
The gains are the same. Therefore at X volume, you will hear the same Y loudness for both. HOWEVER...
For the stereo amplifier, what may happen is that if you turn it up to 2X volume from the source, then the stereo amplifier may clip!
But for the two monoblock ACA's, the 2X volume will give you a higher Y loudness with less chance of clipping.
Put it another way...
2.25V RMS input into the stereo amplifier, will put out about 8 watts into 8 ohms, max.
3.15V RMS input into the monoblock amplifier, will put out about 15 watts into 8 ohms, max.
But if you input 3.15V RMS into the stereo amplifier, you will be well into clipping, and will hear distortion as a result.
Hope that helps,
Anand.
The gains are the same. Therefore at X volume, you will hear the same Y loudness for both. HOWEVER...
For the stereo amplifier, what may happen is that if you turn it up to 2X volume from the source, then the stereo amplifier may clip!
But for the two monoblock ACA's, the 2X volume will give you a higher Y loudness with less chance of clipping.
Put it another way...
2.25V RMS input into the stereo amplifier, will put out about 8 watts into 8 ohms, max.
3.15V RMS input into the monoblock amplifier, will put out about 15 watts into 8 ohms, max.
But if you input 3.15V RMS into the stereo amplifier, you will be well into clipping, and will hear distortion as a result.
Hope that helps,
Anand.
It does. A lot.
I have a follow up question though:
If I put 1V into the input of a stereo ACA into channel A, it will output a 3.16V differential between the + and - speaker terminals (10dB of voltage gain).
Now, if I feed the same 1V into EACH of the channels, but one of them 180 out of phase (balanced connection), the differential between channel A output and gnd would be 3.16V. But the differential from channel B output and gnd would be -3.16V.
Is this correct? If so, the differential between both 'hot terminals' (which are the ones now connected to the speakers) would be 6.32V. Wouldn't this mean that there is more 'gain' by combining both channels? Or is that not how this works at all?
Thanks again,
Rafa.
I have a follow up question though:
If I put 1V into the input of a stereo ACA into channel A, it will output a 3.16V differential between the + and - speaker terminals (10dB of voltage gain).
Now, if I feed the same 1V into EACH of the channels, but one of them 180 out of phase (balanced connection), the differential between channel A output and gnd would be 3.16V. But the differential from channel B output and gnd would be -3.16V.
Is this correct? If so, the differential between both 'hot terminals' (which are the ones now connected to the speakers) would be 6.32V. Wouldn't this mean that there is more 'gain' by combining both channels? Or is that not how this works at all?
Thanks again,
Rafa.
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Thank you very much Anand. It is confusing. A possibly stupid question comes to mind, can the pin 3 signal from the XLR that is 180 degrees out of phase be inverted to be in phase with the signal out of pin 2? Is this possible? Yes I have to do a fair amount of reading to understand this.
Rafa, what you say is also my question. Using basically the same setup, the same music, and the same source volume, but comparing two different modes of wiring the amp (monoblock or stereo), does the speaker volume and sound remain the same or does it change?
Since I built my own speakers I have never measured their sensitivity.
Rafa, what you say is also my question. Using basically the same setup, the same music, and the same source volume, but comparing two different modes of wiring the amp (monoblock or stereo), does the speaker volume and sound remain the same or does it change?
Since I built my own speakers I have never measured their sensitivity.
It could be done, but why would you want to?
The whole idea behind a balanced scenario is that you generate your music signal, and then you send two 'versions' of the same signal. One is the '0 phase' signal, the other one the 'reversed' version of the first (180º out of phase). Then you send both signals inside two different cables.
As the signal flows, those cables will pick up hum, radio frequency noise, jitter... any type of disturbances. The genius part is that these noises are not 'reversed' in the different cables. They are 'in phase'.
When you get to the amp, you 'only' amplify that which is opposite from one signal to the other. Whatever is 'in phase' gets 'subtracted' (added really, but that is beyond this explanation
) and therefore eliminated. In that way you get a much cleaner signal!So the 'out of phase' between pins 2 and 3 is very much desirable and something you pay handsomely for on equipment to be able to have a real balanced scenario.
Best,
Rafa.
Edit: If you were to put two identical in-phase signals into both channels of the ACA and hook the speakers to both black outputs, you would actually ONLY hear the hum and noise and nothing of the music.
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Gain and power are not the same thing, and it's very common to be confused by that, so don't feel bad.
Power - as Rafa says, basically the amount of voltage the amplifier can swing into the load, which is your speakers. As the impedance of the speakers changes, and is referred to as an ohmic value, volts and ohms all mix together and make watts. There's a bit about how the current required goes up as the load impedance goes down to get those volts into the speaker, but for the time being just remember volts into the load.
So amplifier power is commonly referred to in watts. (Which, just like the RCA or "cinch" connector becoming the standard is one of the great tragedies of audio, but that' a discussion for another day...)
Or to put it another way, power is how many electrons the amp can push into your speakers, and this is not infinite - bigger amps can push more, smaller amps less.
Gain - Is the multiplier that the input sees in the amplifier. If you put in a signal to the ACA, it will be multiplied by a fixed ratio, this is gain.
The ACA has a gain of 11db. A signal on the input will come out the speaker posts 11db louder. 11db is also a voltage multiplier of 3.5x, so a 1v signal on the input will come out as 3.5v on the output, a 2v signal will be 7v on the output, 3v will be 10.5v, etc...
SO, if you have a small signal on the input, it's only going to get 3.5x louder, yet then a big signal will also only get 3.5x louder, until the amplifier runs out of power...
This is an incomplete answer, but covers the basic ground.
Power - as Rafa says, basically the amount of voltage the amplifier can swing into the load, which is your speakers. As the impedance of the speakers changes, and is referred to as an ohmic value, volts and ohms all mix together and make watts. There's a bit about how the current required goes up as the load impedance goes down to get those volts into the speaker, but for the time being just remember volts into the load.
So amplifier power is commonly referred to in watts. (Which, just like the RCA or "cinch" connector becoming the standard is one of the great tragedies of audio, but that' a discussion for another day...)
Or to put it another way, power is how many electrons the amp can push into your speakers, and this is not infinite - bigger amps can push more, smaller amps less.
Gain - Is the multiplier that the input sees in the amplifier. If you put in a signal to the ACA, it will be multiplied by a fixed ratio, this is gain.
The ACA has a gain of 11db. A signal on the input will come out the speaker posts 11db louder. 11db is also a voltage multiplier of 3.5x, so a 1v signal on the input will come out as 3.5v on the output, a 2v signal will be 7v on the output, 3v will be 10.5v, etc...
SO, if you have a small signal on the input, it's only going to get 3.5x louder, yet then a big signal will also only get 3.5x louder, until the amplifier runs out of power...
This is an incomplete answer, but covers the basic ground.
Rafa,
Those 2 scenarios are actually different. You have described a dissimilar scenario
The equivalent scenario would be 1V into the input of a stereo ACA into channel A is the same as an input of 0.5V into pin 2 of the XLR and 0.5V into pin 3 of the XLR. Remember the two opposite phases combine to make a full AC signal. So 0.5+0.5 = 1. The splitting of the phases helps in eliminating/cancelling noise issues. You will see this in Nelson Pass' measurements of the ACA monoblock using XLR inputs. The distortion is much lower in the <1 watt region because of this.
Make some sense I hope!
Another way to understand it is that the monoblock can accept HIGHER input voltages as it can swing more output volts, and hence have HIGHER output power, which is precisely what I said in my previous post.
We can illustrate the math with the power equation but you might get more confused then!
Best,
Anand.
No, no, I'm not getting confused. I'm getting a grasp of things.
This is really interesting, so I will keep on putting out examples until I get it, if you are so generous as to keep with me.
Lets take balanced out of the equation and do it with an RCA Stereo vs. RCA S/E monoblocks scenario:
- My pre connected to a stereo RCA outputs 1V to the left channel, 1V to the right one. So, channel L will output 3.16V, channel R will output 3.16V. I think we all agree this is true, right?
Now switch to monoblocks:
- my pre outputs the exact same 1V into the left channel, which is amplified (and inverted) into -3.16V in the L output (we have -3.16V on the left black post).
- Those -3.16V are then re-channeled to the R channel through a 39K resistor which would turn that signal back into -1V (inverted original value)
- Now that -1V into channel R is amplified (and inverted) to +3.16V on the R output (we have +3.16V on the R black post).
Again, wouldn't that give a 6.32V differential between the binding posts?
Thanks so much for bearing with me.
Rafa.
This is really interesting, so I will keep on putting out examples until I get it, if you are so generous as to keep with me.
Lets take balanced out of the equation and do it with an RCA Stereo vs. RCA S/E monoblocks scenario:
- My pre connected to a stereo RCA outputs 1V to the left channel, 1V to the right one. So, channel L will output 3.16V, channel R will output 3.16V. I think we all agree this is true, right?
Now switch to monoblocks:
- my pre outputs the exact same 1V into the left channel, which is amplified (and inverted) into -3.16V in the L output (we have -3.16V on the left black post).
- Those -3.16V are then re-channeled to the R channel through a 39K resistor which would turn that signal back into -1V (inverted original value)
- Now that -1V into channel R is amplified (and inverted) to +3.16V on the R output (we have +3.16V on the R black post).
Again, wouldn't that give a 6.32V differential between the binding posts?
Thanks so much for bearing with me.
Rafa.
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This voltage discussion is intriguing and i have a question about my pre-amp that maybe can be answered here.
I use the Schiit saga, and their website reads its max output is ">10V RMS (active stage)"
I know this pre-amp to be passive, as in it only attenuates the signal that goes into it. So what is this 10V output? Does the voltage change the louder i turn it up (less it attenuates)?
I use the Schiit saga, and their website reads its max output is ">10V RMS (active stage)"
I know this pre-amp to be passive, as in it only attenuates the signal that goes into it. So what is this 10V output? Does the voltage change the louder i turn it up (less it attenuates)?
What they mean is though the active buffer is engaged, being a buffer it does not do anything to the voltage being applied to the input. It does, however, add current if needed.
Anyway, that circuit has limits as well, and if you put an 11V signal on the input and turn it all the way up, it's going to stop doing anything on the output at 10vThat said, a 10V signal is pretty rare.
6L6 this is fascinating. I did a preliminary impedance measurement of my speakers, they came at 5.9 Ohms, however, the system had been on most of the day. I turned it off before measuring and will keep it off until tomorrow and then retake the reading. The point is that my speakers have low impedance.
I am not sure if I took off the lower frequencies from the driver because they would be sent to the subwoofer, which is self-amplified, I have to check my notes. However, it does appear that my speakers will consume more current when needed than the standard 8 Ohm ones. So, what is there to do, before I build the amps.
As you probably realize I am not proficient in doing circuit math. I am planning to build the standard ACA 1.6 monoblock balanced inputs, and everything I have purchased already maintains the values in the schematic. Mine will be standard ACA 1.6 boards only modified by changing the electrolytic caps to more musical ones.
So, let's say that I may, at least occasionally, need more current flowing to my speakers, maybe beyond what the standard circuit config may provide. This is the question: Extreme Boky posted his system changes and he has C1, the output coupling cap, beefed up to 10,000 uF in parallel with a 2.2 film cap. I can get a suitable 10,000 uF cap in the musical (series II Gold Tone KG) Nichicon series, and I can get a nice and not gigantic 2.2 uF metal film cap to replace the circuit's C1 as specified. Because of the parallel arrangement, the overall capacitance at the C1 spot will be low, but the capacitors will be able to hold a lot more current than the 3300 uF cap specified in the schematic. Would this be an adequate solution to the, so far only potential, issue of not having enough current available at amplifier level to cover the needs of my 5.9-ohm speakers? Please let me know if this is a viable solution to cover a potential and occasional need for increased current availability from the amplifiers. This is something I can do with no problems, is it a reasonable thing to do?
Thank you
I am not sure if I took off the lower frequencies from the driver because they would be sent to the subwoofer, which is self-amplified, I have to check my notes. However, it does appear that my speakers will consume more current when needed than the standard 8 Ohm ones. So, what is there to do, before I build the amps.
As you probably realize I am not proficient in doing circuit math. I am planning to build the standard ACA 1.6 monoblock balanced inputs, and everything I have purchased already maintains the values in the schematic. Mine will be standard ACA 1.6 boards only modified by changing the electrolytic caps to more musical ones.
So, let's say that I may, at least occasionally, need more current flowing to my speakers, maybe beyond what the standard circuit config may provide. This is the question: Extreme Boky posted his system changes and he has C1, the output coupling cap, beefed up to 10,000 uF in parallel with a 2.2 film cap. I can get a suitable 10,000 uF cap in the musical (series II Gold Tone KG) Nichicon series, and I can get a nice and not gigantic 2.2 uF metal film cap to replace the circuit's C1 as specified. Because of the parallel arrangement, the overall capacitance at the C1 spot will be low, but the capacitors will be able to hold a lot more current than the 3300 uF cap specified in the schematic. Would this be an adequate solution to the, so far only potential, issue of not having enough current available at amplifier level to cover the needs of my 5.9-ohm speakers? Please let me know if this is a viable solution to cover a potential and occasional need for increased current availability from the amplifiers. This is something I can do with no problems, is it a reasonable thing to do?
Thank you
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Absolutely the right train of thought!
Using the RCA input bridge configuration does add approximately 6db of gain because the voltage has been multiplied in the manner you describe.
Oh that’s interesting. So my modi dac which has an output of 2V is just passed along. So basically I have a 2 volt output from my saga when using that dac?What they mean is though the active buffer is engaged, being a buffer it does not do anything to the voltage being applied to the input. It does, however, add current if needed.
That said, a 10V signal is pretty rare.