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Speaker amps for headphones - Page 115

post #1711 of 3116
Quote:
Originally Posted by robrob View Post
 

Is that log base 20?

Is that dBi or dBl?

 

Great suggestion, I will add that feature. Right now a 0 value gives a DIV 0 Error.

 

Armaegis, could you take a look at the equations and verify you agree with how I'm calculating the Effective Speaker Load and Attenuation?

 

VOLTAGE Attenuation or gain =  20 log(x).

Try a few numbers out, you should get this:

 

if attenuation is 1/10, then attenuation = -20 dBV or -20 dB Volts

 

if gain is 10, then gain = 20 dB

 

no gain or attenuation = o dB.0

post #1712 of 3116
Quote:
Originally Posted by Chris J View Post
 

 

I think you may have hosed up one concept.    :o

 

If resistor 2 = 6 Ohms and you assume the amplifiers output impedance is 0.1 Ohms, then the amplifier's output impedance is ACTUALLY 6 + 0.1 = 6.1 Ohms.

 

But you want to calculate "Effective Headphone Output Impedance", i.e from the headphone's point of view, what is the apparent output impedance?

 

This number will be 2 Ohms in parallel with 6.1 Ohms, i.e. 1. 506 Ohms.

If you neglect the amplifier's assumed output impedance (i.e. 0.1 Ohms), then you get 1.5 Ohms.

Hence, the amplifier's output impedance may be neglected in a SS amp in this configuration, especially since the output impedance will probably be much less than 0.1 Ohms.

 

But if it's a tube power amp AND the output impedance is as high as an Ohm or two, then it starts to get more significant.

As you can see in the formula I posted I do add R2 (6) + AmpOut (0.1) then take the reciprocal to add it to 1/R3 (1/2). I left in the AmpOut variable at your suggestion so it can be used if it's a significant number. Am I missing something?


Edited by robrob - 12/4/13 at 8:46am
post #1713 of 3116
Quote:
Originally Posted by Chris J View Post
 

 

VOLTAGE Attenuation or gain =  20 log(x).

 

That's what I have in the Attenuation formula I posted.

post #1714 of 3116
Quote:
Originally Posted by robrob View Post
 

 

That's what I have in the Attenuation formula I posted.

 

Ahhhhhhhh.......so it is!  :o  I feel shame.

 

Should have a negative sign, but that's getting picky.

post #1715 of 3116
Quote:
Originally Posted by robrob View Post
 

As you can see in the formula I posted I do add R2 (6) + AmpOut (0.1) then take the reciprocal to add it to 1/R3 (1/2). I left in the AmpOut variable at your suggestion so it can be used if it's a significant number. Am I missing something?

 

 

Attenuation:

Oh, now I see it.    (Smacks head!)

in the brackets, the stuff on the bottom goes on the top, the stuff on the top goes on the bottom.

 

For the formula Eff HP Out:

Something is hosed up in excel because that formula looks correct, but when I put in numbers, they don't work out.:confused_face_2: 

For example:  

Amp output impedance = Res2 = 2 Ohms

Res3 = 4 Ohms

Eff HP Out should be approx. 2 Ohms, but spreadsheet calculates 0.8 Ohms,

I think it calculates Res2 in parallel with Res3 in parallel with Amp output impedance.

 

Try putting brackets around (AmpOut+Res2).

So:   1/(AmpOut+Res2)


Edited by Chris J - 12/3/13 at 4:15pm
post #1716 of 3116

Chris, could you download my latest version and check again. I checked the numbers using a calculator (thank god for the reciprocal key!) and they were correct. I'll add that missing minus sign ;)

 

http://robrobinette.com/images/Audio/Headphone_Resistor_Network_Calculator.xls

post #1717 of 3116
Quote:
Originally Posted by robrob View Post
 

Chris, could you download my latest version and check again. I checked the numbers using a calculator (thank god for the reciprocal key!) and they were correct. I'll add that missing minus sign ;)

 

http://robrobinette.com/images/Audio/Headphone_Resistor_Network_Calculator.xls

 

Ah yes.

I see the mistake.

 

This is the corrected version:

 

=IF(Res2=0,"N/A",1/((1/(AmpOut+Res2))+IF(Res3=0,0,1/Res3)))

post #1718 of 3116

So many great minds... churning out great stuff for the rest of us to use!  

post #1719 of 3116

It's like high school physics class all over again with all the kids trying to solve a problem, lol.

post #1720 of 3116
Quote:
Originally Posted by Chris J View Post
 

 

Ah yes.

I see the mistake.

 

This is the corrected version:

 

=IF(Res2=0,"N/A",1/((1/(AmpOut+Res2))+IF(Res3=0,0,1/Res3)))

Thanks Chris, I found it too, it's corrected now. I also realized the JavaScript version was doing some string addition in the Attenuation formula.

post #1721 of 3116

Something wonderful is about to happen...

post #1722 of 3116

I think Mike is about to have a mathgasm...

post #1723 of 3116
Quote:
Originally Posted by robrob View Post

 

Thanks Chris, I found it too, it's corrected now. I also realized the JavaScript version was doing some string addition in the Attenuation formula.





 



deleted!;


Edited by Chris J - 12/4/13 at 9:33am
post #1724 of 3116

I'm still trying to figure out which one is Paul, and which one George...

post #1725 of 3116

I did find another error in the formulas and fixed it. I also added the other 2 resistor network.

 

The latest is available here as always: http://robrobinette.com/images/Audio/Headphone_Resistor_Network_Calculator.xls

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