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Help me with the math....


Coytee

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Over on AK, I told someone they can balance their biamped speaker by using the 20*Log(V1/V2) where V1 and V2 are the respective input sensitivies of each amp.  Calculate the difference and attenuate the more sensitive amp by that many db's.

 

https://www.audiokarma.org/forums/index.php?threads/bi-amping-question.904938/page-2#post-13553036

 

Starts on page 2.

 

I'm wondering if one of us (usually me) is a bit confused...  so at the end I'm asking him for some real numbers so we can do a side by side...  if I'm wrong, I want to know what/where so I don't mislead anyone.

 

 

 

 

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I have a hard time remembering when/whether to use 20x or 10x, but an iteration usually clears it up.  Looks right so far from here with the available info (none).  Didn't follow the link...

 

Add: would be easier to use the overall gain, in dB, of the amps, wouldn't it?

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20*Log(V1/V2) is the equation for voltage ratios, but I think It is a little more complicated.  In addition to input sensitivity, each amp will have its own gain and they are likely different.  So, starting with your equation is fine, but may only get him close.  No substitute for a SPL meter. 

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This is not that complicated. I think we have to be careful what is meant by gain. 

Attached is the gain chart for my SA9500 amp.   

Note the input sensitivity at  Power Amp IN is  specified as 1 Volts  to provide rated output power of 85 watts into 8 ohms.

This is shown on the chart as 1V = 0dBv at the input of the PA 

OdBv is referenced to 0V   so in this case dBv= 20 Log (V PAin/0V) 

 

This gives a Vout of the PA of 26.1V  that corresponds to 85W into 8ohms if you use P= V^2/R

Gain is Vout/Vin.   In dB this is 20xLog (Vout/Vin).    = 20 Log (26.1/1)  = 28.3 dB. 

 

So when I mean be careful of what is meant by gain is that if gain is Vout/Vin to provide maximum rated power you are not going to match your levels. 

What you need to do is find gain for all your amps for an equal Vout given identical Vin. 


For example, you want all your amplifiers to provide the same output voltage for a given input voltage. 

If amp A  (for you HF section for example) provides 10v out for 1V in, it is a voltage gain of 20dB

If amp B (for your LF section for example) provides 20V out for 1V in, it is a voltage gain of 26dB and you would have to use 6dB of attenuation. 

 

Note that Amp B provides 2x the voltage.   You could use a simple voltage divider circuit using 2 10Kohm resiistors in series. 

 

Voltage gain in dB works for any impedance. That is why the chart uses voltage gain because input and output inpedances change in the circuit.

 

If you do not have such a chart then you could determine gain yourself with an audio generator (your laptop or phone and an online audio generater!)

and voltmeter   

 

If you use power gain, which is 10 x Log Pout/Pin   you have to calculate power using voltages and resistance. 

 

In this example, without using any attenuation to equalize the amps: 

Amp A  at 10V out into 8 ohms  gives you  12.5 watts

Amp B at 20V into 8 ohms gives you 50 watts.

the power difference between amp B and amp A is  10 x Log 50/12.5 = 6 dB    the same as the voltage gain! 

 

now all of this assumes equal impedance and sensitivity drivers which is rarely the case!

You would have to take both into account and back calculate the PA voltage into the drivers to provide same SPL, and then provide the right attenuation at the input of the PA to equalize them.  

 

Have fun!
 

E

 

 

 

 

SA9500 level.jpg

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13 hours ago, glens said:

I have a hard time remembering when/whether to use 20x or 10x

 

This will either clear things up, or make them even more confusing:

 

Just remember that dB is always a ratio of powers:

dB = 10 * log10(P2/P1)

Also remember that power is equal to voltage squared divided by resistance:

P = V²/R

So

P1 = V1²/R and P2 = V2²/R

and

dB = 10*log10(P2/P1)

= 10*log10[(V2²/R)/(V1²/R)]

= 10*log10(V2²/V1²) (the "R" cancel-out)

= 10*log10[(V2/V1)²]

Taking the logarithm of an exponent is equal to multiplying the logarithm by the exponent:

10*log10[(V2/V1)²] = 2*[10*log10(V2/V1)] = 20*log10(V2/V1)

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25 minutes ago, Deang said:

"There are three kinds of people in the world. Those who can do math and those who can't."

Actually there are 10 kinds of people in the world, those who understand BINARY and those who don't.

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6 hours ago, Edgar said:

This will either clear things up, or make them even more confusing:

 

Yep.  It either will or it won't.  Actually, the case of voltage vs. power is not terribly difficult (as I'd said, an iteration usually makes it obvious).  Have you got a rule of thumb for sound pressure vs. sound power?  Would it be safe to consider "power" involves "something squared" in all cases, as a rule of thumb?

 

1 hour ago, Deang said:

"There are three kinds of people in the world. Those who can do math and those who can't."

 

Too funny!  Sounds like something Mark Twain would've said.  Who was it?

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2 hours ago, glens said:

Have you got a rule of thumb for sound pressure vs. sound power? 

 

Yeah -- I always have to look it up.

 

Quote

Would it be safe to consider "power" involves "something squared" in all cases, as a rule of thumb?

 

Well, "something times something"; voltage times voltage, voltage times current, current times current, etc.

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1 hour ago, CECAA850 said:

I know it's not but it sounds like a Yogism (Yogi Berra).

My favorite Yogiism quote: "It's amazing how much you can observe just by watching."

 

To keep it relevant to Klipsch speakers..........................the more a cone moves, the more distortion it produces. You just can't watch it when it's in a horn!

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1 hour ago, ClaudeJ1 said:

My favorite Yogiism quote: "It's amazing how much you can observe just by watching."

 

To keep it relevant to Klipsch speakers..........................the more a cone moves, the more distortion it produces. You just can't watch it when it's in a horn!

Why yes you can with a remote camera!! SAOTWU

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2 hours ago, Dave A said:

Why yes you can with a remote camera!! SAOTWU

PWK showed me the Khorn that was built out of plexiglas in the original Hope museum, across the street from the factory, back in 1985. They did Laser Interferometry measurements to show the K-33's motion to be within calculations' predictions of only about 0.060 (1.5 mm) Xmax when the horn was blowing out 120 db of output. This is why I never worry about recommending the Kappa 15C woofer (2.5mm Xmax) for LaScalas, Belles, and my own Quarter Pies bass horn as superior and more affordable than all the rest of the options.

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