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#1 |
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diyAudio Member
Join Date: Mar 2004
Location: Albury NSW Australia
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Hi There
I would like some guidance as to how to work out Z out of a tube with its plate load being supplied by the IXP10M45S. I have read the article at maarten@platenspieler.com and I figure It probably should be MU follower formulas given the output is taken from the anode of the driven tube. Given the Current regulator does not have the plethora of caps and resistors though, I am wondering if it may be accurate. I would appreciate any help Thanks Nick
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"Better to say nothing and keep them guessing than to speak and remove all doubt." |
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#2 |
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diyAudio Moderator
Join Date: Jan 2003
Location: Near London. UK
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It's just ra (at the operating point). As simple as that.
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The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference... |
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#3 |
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diyAudio Member
Join Date: Mar 2004
Location: Albury NSW Australia
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Hi Ec
Hope your xmas woas good and that 2008 is fine for all. The datasheet for the e180f seems to only provide data in pentode op. I will look for more info At this point i have the setip working at 5.5 mA and 153V so is it as simple as V/I? Thanks
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"Better to say nothing and keep them guessing than to speak and remove all doubt." |
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#4 |
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diyAudio Moderator
Join Date: Jan 2003
Location: Near London. UK
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If you look on the E180F data sheet, you should find it gives a value for mu(g1-g2). This is the same as the triode mu. You can find gm at your operating current using the pentode data, and from that use ra(triode) = gm x mu(g1-g2).
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The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference... |
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#5 | |
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diyAudio Member
Join Date: Jan 2005
Location: Chicago
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Quote:
http://www.triodeguy.com/Triodeguy%2...asurements.pdf
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http://www.ecpaudio.com |
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#6 |
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diyAudio Member
Join Date: Mar 2004
Location: Albury NSW Australia
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Hi Again
Ok mu=50 Gm.. is that transconductance or is it derived by operating point (150V @ 5.1mA) ? Thanks
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"Better to say nothing and keep them guessing than to speak and remove all doubt." |
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#7 |
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diyAudio Member
Join Date: Mar 2004
Location: Albury NSW Australia
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I will try increasing current.. i find with other tubes that they can sound a bit unpleasant at higher anode currents.. Is that my imagination?
The 180F at 5.1 doesnt sound objectionable at this point but i will double the current and see for myself Once agin thanks for the datasheet. Nick
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"Better to say nothing and keep them guessing than to speak and remove all doubt." |
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#8 |
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diyAudio Member
Join Date: Mar 2004
Location: Albury NSW Australia
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Hi
using Ec's formula ra(triode) = gm x mu(g1-g2). refering to the triode sheet for the 180F mu=60 gm =17.6 so rA=1056 on the triode data sheete refered to in this post a valus for Rp is given of 3.4k Ra is not the same as rp i take it? Thanks
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#9 | |
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diyAudio Moderator
Join Date: Jan 2003
Location: Near London. UK
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Quote:
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The loudspeaker: The only commercial Hi-Fi item where a disproportionate part of the budget isn't spent on the box. And the one where it would make a difference... |
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#10 |
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diyAudio Member
Join Date: Mar 2004
Location: Albury NSW Australia
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Hi EC
dont worry.. i sppend much of my life in a confused state these days as I find it helps with the vagaries of modern life. Thanks to all for your help. I plan to construct an riaa stage based on the e180f in triode mode and the ixys current regulator.. Will see how it works. Thanks Nick
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