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Old 9th February 2007, 09:02 AM   #31
EUVL is offline EUVL  Europe
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> How come no output coupling cap?
No need. That was the whole point for using JFETs.

> Am I missing something?
Yes.
Vgs is -ve for a JFET.


Patrick
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Old 9th February 2007, 09:11 AM   #32
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Quote:
Originally posted by EUVL


Yes.
Vgs is -ve for a JFET.

Shame, I missed it

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Old 9th February 2007, 09:23 AM   #33
steenoe is offline steenoe  Denmark
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Quote:
Shame, I missed it
Thats not a shame I am sure many of us learned a little from that.
I asked this:
Quote:
What is the output impedance of this circuit? A standard headphone expects to see around 120r.
Steen
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Old 9th February 2007, 09:27 AM   #34
EUVL is offline EUVL  Europe
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> I asked this:

Was not in your original post.

Well, I would say the output impedance would be the source resistor plus 1/Yfs for the JFET at the working point, which is about 1 ohm. So in total 4 ohms. If you what lower impedance, you could lower the source resistor to say 0.68R (a la ZV9) with corresponding increase in bias current (and heat).




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Old 9th February 2007, 09:48 AM   #35
steenoe is offline steenoe  Denmark
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Quote:
Was not in your original post.
Thats right. It was the usual case of being overlooked when editing.
If the output impedance is so low and headphones build to international standards should expect 120r, wouldnt it be a good idea to put a 120r, 5watter in series with the output?

Steen
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Old 9th February 2007, 10:03 AM   #36
EUVL is offline EUVL  Europe
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> If the output impedance is so low and headphones build to international standards should expect 120r, wouldnt it be a good idea to put a 120r, 5watter in series with the output?

Are you sure ?

I mean I saw a few audiophile headphones with around 40 ohm impedance (Sennheiser HD595, Grado SR80, .....). Surely you do not want to drive them with 120R Rout ?

Grey's circuit has even lower output impedance, IMO.


Patrick
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Old 9th February 2007, 10:12 AM   #37
EUVL is offline EUVL  Europe
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> http://www.epanorama.net/circuits/he...ttenuator.html

They argue that you should put a 120ohm in series.
This is essentially an output attenuation.
You have to remember that the follower circuit has close to unity gain.

But I think you should ask Grey. I am not into headphones at all.


Patrick
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Old 9th February 2007, 10:22 AM   #38
steenoe is offline steenoe  Denmark
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I am "sort of" sure.
Take a look at this: http://sound.westhost.com/project70.htm
It says that a headphone should want to see 120r, regardless of its own impedance. Its easy enough to test though, so its not a big deal. Just thought I would mention it
If someone knows, please tune in.

Steen
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Old 9th February 2007, 10:38 AM   #39
EUVL is offline EUVL  Europe
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Quote from the same webpage :

" ..... but some 'phones seem to prefer lower source impedance."




Patrick
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Old 9th February 2007, 11:51 AM   #40
EUVL is offline EUVL  Europe
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For the technically minded, the load line with 3V Vds at 200mA bias, and 3 ohm source resistor.


Patrick
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