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Speaker amps for headphones - Page 129

post #1921 of 3116
As for DACS. That is a good one. And if you want to improve it or most others get an Audiophilleo 1 or 2 with pp. Find it used. It will make an improvement in the dac.

The second is wav files this helps a lot to. That will help you save up the money faster for the stax.

Al d
post #1922 of 3116

Al:

 

I'm currently doing a comparison test of 15 or so DACs.

 

http://www.head-fi.org/t/689783/december-2013-mid-level-dac-comparison

 

I think I have plenty of good candidates. :)

post #1923 of 3116
Funny wow some list. I can only add that if you would try the Audiophilleo they would all pretty much sound the same .


That would save you a lot of time. Google it. And reviews it is why I bought it and it does work.

Have fun wow very impressive man .

Al D
post #1924 of 3116
Quote:
Originally Posted by Gary in MD View Post
 

Are you using one of the Stax transformers, or something else?

Yep a modded SRD-7sb mk2. Not based on spritzers board but I had some left over and fairly expensive PIO polypropylene caps that I put to good use to replace the existing ERO MKP coupling caps.

post #1925 of 3116

"Preferred Headphone Resistor Network" Circuit Analysis

or How the Resistor Network Works

Using my HE-500 headphones we'll analyze the "Preferred Resistor Network" using Ohm's Law. Inputting the HE-500's specs of 38 ohms impedance and 89 dB/mW sensitivity into the Headphone Power Calculator spreadsheet shows:

Headphone_Resistor_Network_Preferred_AnalysisPwr1.jpg

So with my HE-500s connected directly to an amplifier providing 0.389 volts RMS you get 10.24 milliamps (0.01024 amps) of current flowing through the headphones and 95 dBSPL Vrms of loudness. That loudness level would be too loud for most.

 

Using our Resistor Network Calculator on this webpage we input Headphone Impedance = 38 ohms, and the standard "Preferred Network" resistor values of R2 = 6 ohms, R3 = 2 ohms:

Headphone_Resistor_Network_Preferred_AnalysisCalc.jpg

The calculated "Effective Speaker Load" is the total Headphones + "Preferred Network" circuit impedance/resistance of 7.9 ohms and Attenuation is shown as -12.37 dB.

 

Now lets plug the 0.389 volts from the amplifier and the 7.9 ohms of total circuit resistance into our "preferred network" and analyze it using Ohm's Law.

Headphone_Resistor_Network_Preferred_Analysis.jpg

First we calculate the total circuit current using Ohm's formula: Current = Volts / Resistance = 0.389v / 7.9ohms = 0.0492 amps of current (or 49.2 milliamps). We can also calculate the total circuit Power (watts) using:  Watts = Volts^2 / Ohms = 0.389v^2 / 7.9ohms = 0.0192 watts (19.2 milliwatts). So the amplifier is putting out 0.389 volts, 0.0492 amps and 0.0192 watts.

 

Now we can calculate the voltage drop caused by resistor R2's 6 ohms. All 0.0492 amps of current run through R2 so we can use the formula Volts = Current * Resistance = 0.0492amps * 6ohms = 0.2954 volts (295.4 millivolts).

We can also calculate the watts dissipated by R2 using the formula Watts = Amps^2 * Ohms = 0.0492amps^2 * 6ohms = 0.0145watts.

 

Since our Headphones are in parallel with resistor R3 they will both get 0.2954 volts but the 0.0492 amps will be divided between them.

 

To determine how the current is split between R3 and the Headphones we must first calculate the resistance of R3 + Headphones in parallel. The formula is 1 / (1 / R3 + 1 / Headphones) = 1 / (1 / 2ohms + 1 /38 ohms) = 1.9 ohms.

Now we can calculate the voltage drop across R3 and the Headphones. The formula is: Volts = Current * Resistance = 0.0492 * 1.9ohms = 0.0936 volts (93.6 millivolts).

 

Knowing the voltage drop across R3 and the Headphones we can calculate the current split through them. Lets start with R3. The formula is Current = Volts / Resistance = 0.0936v / 2ohms = 0.0468 amps flowing through R3. For the headphones it is 0.0936v / 38ohms = 0.0025 amps (2.5 milliamps) flowing through the headphones at 0.0936 volts (93.6 millivolts).

 

Now we can also calculate the watts dissipated by R3 using the formula Watts = Amps^2 * Ohms = 0.0468amps^2 * 2ohms = .00438 watts.The watts dissipated by the headphones is: 0.00246amps^2 * 38ohms = 0.00023 watts (0.23 milliwatts).

 

We can use Kirchhoff's Voltage Law to double-check our voltage calculations. The law states that the sum of all voltages around a closed circuit is equal to zero. Sure enough we add the output voltages of R2 + (R3 & Headphones) = 0.2954v + 0.0936v and subtract that from the amplifier output of 0.389 volts and you get 0.

 

So our amplifier puts out 0.389 volts, 0.0492 amps and 0.0192 watts

 

And the resistor network drops that to 0.0936 volts (93.6 millivolts), 0.00246 amps (2.46 milliamps) and 0.00023 watts (0.23 milliwatts) through the headphones.

 

We can plug our calculated 0.0936 volts at the headphones into the Headphone Power Calculator spreadsheet:

Headphone_Resistor_Network_Preferred_AnalysisPwr2.jpg

With 0.0936 volts RMS getting through the resistor network to the headphones you get a nicely matching 2.46 milliamps and 0.23 milliwatts and a loudness of 82.64 dB. If we compare this loudness to the 95.0 dB of the headphones connected directly to the amplifier we get an attenuation of: 82.64dB - 95.0dB = -12.36 dB which matches the attenuation calculated by the Headphone Network Calculator on this webpage.


Edited by robrob - 12/19/13 at 11:49am
post #1926 of 3116
Quote:
Originally Posted by robrob View Post
 

I was working up a circuit analysis of the "preferred resistor network" and everything worked out except I was surprised by the watts dissipated by resistor R2. I've been using an R2 of 6 ohms and rated at 5 watts. But when an amp putting out 10 volts RMS into the "preferred network" with 38 ohm impedance headphones the wattage calculated for R2 comes out to 9.6 watts.

 

Watts = Amps^2 * Ohms = 1.266amps^2 * 6ohms = 9.6watts

 

Why is a 5 watt resistor R2 adequate?

 

 

Oy, bit of a long way around to do a simple calculation, but I guess it works...

 

answer to your question: Because you're not running your amp at maximum, and you're not playing sine waves or cranking DC through your headphones.

post #1927 of 3116
Quote:
Originally Posted by Armaegis View Post
 

 

 

Oy, bit of a long way around to do a simple calculation, but I guess it works...

 

answer to your question: Because you're not running your amp at maximum, and you're not playing sine waves or cranking DC through your headphones.

Please don't take him literal and crank DC through your headphones people :o

post #1928 of 3116

One of the secrets to golden ear listening is learning how to properly bias your eardrums into class-A...

post #1929 of 3116
Quote:
Originally Posted by Armaegis View Post
 

One of the secrets to golden ear listening is learning how to properly bias your eardrums into class-A...

Lol! You got a chuckle out of me on that one :D

post #1930 of 3116

If Rob likes to play his music at a "typical" 123.2dB, he's probably done irreparable bias already...

post #1931 of 3116
Quote:
Originally Posted by Armaegis View Post
 

If Rob likes to play his music at a "typical" 123.2dB, he's probably done irreparable bias already...

Yeah, I took notice of that...

post #1932 of 3116
Quote:
Originally Posted by Armaegis View Post
 

If Rob likes to play his music at a "typical" 123.2dB, he's probably done irreparable bias already...

What?

post #1933 of 3116
Quote:
Originally Posted by Armaegis View Post
 

 

answer to your question: Because you're not running your amp at maximum, and you're not playing sine waves or cranking DC through your headphones.

 

Thanks Armaegis, that makes sense. I figured if 130dB is the pain threshold then how loud could 123.2dB be? ;)

 

My Mjolnir at 16 Vrms into 38 ohms impedance puts out a theoretical max volume of 127.3dB.

 

Quote:
 Please don't take him literal and crank DC through your headphones people 

But the noise floor for DC is so low. . .:blink: 

 

I did the circuit analysis to show, in very basic steps for a non-electronic technician type, what's going on in the circuit and how it ties into amplifier power and headphone loudness.


Edited by robrob - 12/16/13 at 12:32pm
post #1934 of 3116
Quote:
Originally Posted by robrob View Post
 

 

Thanks Armaegis, that makes sense. I figured if 130dB is the pain threshold then how loud could 123.2dB be? ;)

 

My Mjolnir at 16 Vrms into 38 ohms impedance puts out a theoretical max volume of 127.3dB.

 

But the noise floor for DC is so low. . .:blink: 

 

I did the circuit analysis to show, in very basic steps for a non-electronic technician type, what's going on in the circuit and how it ties into amplifier power and headphone loudness.

OK i have to clarify something here. 120dB at over 30 seconds in duration is going to damage your hearing. Look up any audiologist-approved chart.

 

EDIT: I have a general rule of never exceeding 110dB, in any duration.


Edited by brunk - 12/16/13 at 1:04pm
post #1935 of 3116
How loud is a jet flying over your head? Low flying at the airport .
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