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# HifiMAN HE-6 Planar Magnetic Headphone - Page 963

Quote:
Originally Posted by Armaegis

Or just not use a ridiculously overpowered speaker amp to begin with

that's no fun.

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+1

Quote:
Originally Posted by SilverEars

Not true.  In parallel configuration, the current will flow more through the less resistive path, and drop more power there.  P=IV.  V is the same for both loads(the resistor and the headphone).

If the output V is the same, then the headphone will see the same amount of power regardless of the parallel resistors. The total power put out by the amp is the summation of the paralleled loads.

Say V=10, headphone Z=50, then with only the headphone you drop P = IV = VV/Z = 2watts

Say you have an R=8ohm dummy load, so with no headphones connected you have P = VV/R = 12.5watts

if you put have both the dummy load and the headphone connected in parallel, the equivalent resistance seen by the amp is ~6.9ohm

stuff that into the power equation and you get P = 14.5watts, which is 2 + 12.5

Yes

Quote:
Originally Posted by Armaegis

If the output V is the same, then the headphone will see the same amount of power regardless of the parallel resistors. The total power put out by the amp is the summation of the paralleled loads.

Say V=10, headphone Z=50, then with only the headphone you drop P = IV = VV/Z = 2watts

Say you have an R=8ohm dummy load, so with no headphones connected you have P = VV/R = 12.5watts

if you put have both the dummy load and the headphone connected in parallel, the equivalent resistance seen by the amp is ~6.9ohm

stuff that into the power equation and you get P = 14.5watts, which is 2 + 12.5

Yes, precisely.  Headphone sees the 2W, dummy is seeing 12.5W, but both dummy and the phone are seeing 10V as you have just shown.  So each dissipate power in different amount, but the V drop is the same is what I meant.

Quote:
Originally Posted by SilverEars

Not true.  In parallel configuration, the current will flow more through the less resistive path, and drop more power there.  P=IV.  V is the same for both loads(the resistor and the headphone).

I don't understand the purpose of even tapping into the speaker amp terminal and using resistors to attenuate. Headphones don't require more than minimal power.  Headphone jack does exactly what the resistor does by attenuating for the headphone input.  Even the series resistor is at a high risk unless everything is calculated out correctly.  I don't agree with all this big wattage amp is better argument now that I understand what it is essentially doing and the risks involved.

Yes, but it doesn't matter if the amp can supply the voltage (doesn't hit the current limit) with the parallel resistor in place, the headphone sees EXACTLY the same power no matter what the parallel resistor value you use. ONLY  if it pulls so much current from the amp that the amp can't provide the voltage does it affect the power delivered to the headphone!

Some amplifiers don't have a headphone out.  My First Watt F1 doesn't.  Neither does my Parasound HCA1000A.  Neither does my Krell...  I don't use any resistors on any of them.  Straight to the taps, baby!

Edited by potterma - 8/2/14 at 3:06pm
Quote:
Originally Posted by potterma

Yes, but it doesn't matter if the amp can supply the voltage (doesn't hit the current limit) with the parallel resistor in place, the headphone sees EXACTLY the same power no matter what the parallel resistor value you use. ONLY  if it pulls so much current from the amp that the amp can't provide the voltage does it affect the power delivered to the headphone!

But there are two parallel paths for the current, the current has to be split from the supplied.  Each load will dissipate inversely proportional to the resistance from the amount that is being drawn.

Edited by SilverEars - 8/2/14 at 3:06pm
Quote:
Originally Posted by SilverEars

But there are two parallel paths for the current, the current has to be split from the supplied.  Each load will dissipate inversely proportional to the resistance from the amount that is being drawn.

Yes, that's correct.  But, it the voltage across both resistors is unchanged, you can change R1 all day long and the power dissipated by R2 will be unchanged.

Interesting discussion. Moar illumination please...and ice-cold southern sweet tea if you've got any.

Quote:
Originally Posted by Silent One

Interesting discussion. Moar illumination please...and ice-cold southern sweet tea if you've got any.

Sorry, fresh out of sweet tea...

Its simple.  V= I * R

If V stays constant and R (headphone) stays constant what is I going to do?  Nada!

Quote:
Originally Posted by potterma

Yes, that's correct.  But, it the voltage across both resistors is unchanged, you can change R1 all day long and the power dissipated by R2 will be unchanged.

Didn't know there was another resistor.  Must have skpped that section, sorry for the confusion.  Yes, the added resistor would split another path, but it's just a parallel value of both, if it's both 8 ohms, the effective should be 4.

Anyway, thanks for helping me refresh DC electronics.

Quote:
Originally Posted by potterma

Almost, correct.  Don't forget to add the series resistor.  Looks like a 24 ohm in the pic?  If so then, assuming nominal 50 ohms for the HE-6, 10||(24+50) or 8.8 ohms...

You are right. I forgot to add the resistor in series because I have never used them in my system.

Now the discussion has HE-6 clarity. And I'm no longer confused.

I guess this is why rating go by SPL, but do speakers have SPL?  When I looked at max headphone SPL, the biggest value I ran across was 136dB.  Although, the speakers have 8ohm or 4 ohm load, it needs lots of power and has much greater power handling ability than a headphone.  This of course depends on the speakers, number of cones etc..

Edited by SilverEars - 8/2/14 at 3:26pm
Quote:
Originally Posted by SilverEars

Didn't know there was another resistor.  Must have skpped that section, sorry for the confusion.  Yes, the added resistor would split another path, but it's just a parallel value of both, if it's both 8 ohms, the effective should be 4.

Anyway, thanks for helping me refresh DC electronics.

Sorry...  Let R1 be the resistor across the speaker taps and let R2 be the headphone resistance.

Now, change R1 to whatever value you want, keep V across R1 and R2 the same (no current limit) and current through R2 cannot change, hence, the power dissipated by R2 cannot change.

Quote:
Originally Posted by SilverEars

I guess this is why rating go by SPL, but do speakers have SPL?  When I looked at max headphone SPL, the biggest value I ran across was 136dB.  Although, the speakers have 8ohm or 4 ohm load, it needs lots of power and has much greater power handling ability than a headphone.  This of course depends on the speakers, number of cones etc..

Yes, speakers are rated by their efficiency, just as headphones are.  Normally an SPL at 1 meter, dB/W

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