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Old 16th January 2008, 08:26 PM   #1
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Default working out Z out for tube with IXYS IXP10M45S plate load

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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Old 16th January 2008, 08:30 PM   #2
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It's just ra (at the operating point). As simple as that.
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Old 16th January 2008, 09:07 PM   #3
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Default so if we use e180f as triode

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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Old 16th January 2008, 09:11 PM   #4
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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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Old 16th January 2008, 09:51 PM   #5
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Default Re: so if we use e180f as triode

Quote:
Originally posted by duderduderini
e180f ... working at 5.5 mA and 153V
Probably a little low on the current. More like 12-16mA is probably a good point to shoot for.


http://www.triodeguy.com/Triodeguy%2...asurements.pdf
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Old 16th January 2008, 09:56 PM   #6
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Default At the risk of sounding completely stupid

Hi Again
Ok mu=50
Gm.. is that transconductance or is it derived by operating point (150V @ 5.1mA) ?
Thanks
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Old 16th January 2008, 10:28 PM   #7
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Default Thanks for the triode datasheet

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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Old 17th January 2008, 01:39 AM   #8
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Default So is this the Ra

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?
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Old 17th January 2008, 07:15 AM   #9
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Default Re: So is this the Ra

Quote:
Originally posted by duderduderini
using Ec's formula ra(triode) = gm x mu(g1-g2).
That's because the twit EC had his brains out of gear. He meant to say ra = u/gm. If he had spent less time on formatting and more on thinking, he wouldn't have confused you. My apologies.
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Old 18th January 2008, 07:13 AM   #10
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Default I am always confused

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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