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30th April 2003, 08:53 PM  #1 
diyAudio Member
Join Date: Sep 2002
Location: Oregon, USA

Could someone please help me understand this equation?
I got my Morgan Jones book yesterday, and between that and the Tubecad site (and trying to remember Thevenin's theorem from my college classes), I'm pretty confused. The equation is here: http://www.tubecad.com/articles_2003...ier/page5.html. It calculates a cathode bias resistor, and it says:
Quote:
I thought you picked a cathode bias resistor by looking at your operating point. You want current Iq, to get that you need the cathode to be a specific voltage Vgk above the (grounded) grid, so Rk = Vgk/Iq. Any help would be most appreciated. Thanks, Saurav 
30th April 2003, 09:28 PM  #2 
diyAudio Member
Join Date: Jun 2002
Location: Denver, CO

I'll try to help
but you must realize that I am a total newbie.
Pick your operating point, say 250v at 40 ma. That point will be somewhere on your loadline at some intersection with a grid bias, say 10v. If you put 250v on the plate (between the plate and cathode) it will conduct 40 ma, for this hypothetical case. The grid is biased at 10v using a series resistor and, in this case, would be 250 ohm resistor (10v/.o40=250 ohm). This simply raises the cathode above the grid 10v (i.e. grid is 10v). As Joel puts it, "getting it?" Don't know if this is what you were looking for, I don't want to do a derivation...I'll let others do that. Joel, John, Brett, Frank and others here can do better at explanation. Rick 
30th April 2003, 09:35 PM  #3 
diyAudio Moderator Emeritus
Join Date: Jan 2003
Location: Near London. UK

The equation assumes a theretically perfect triode, and considers DC conditions only. Real valves are different, and plotting a loadline on reliable curves gives the best answer. It's worth remembering that access to full data sheets was quite limited during the valve heyday, and singleline data was common, so theoretical approximations were actually quite useful. You have the benefit of detailed data, so you can do it properly...
Your penultimate paragraph is spoton.
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30th April 2003, 09:39 PM  #4 
On Hiatus

Even if the formula is correct, you still need to go back to the characteristic curves to see what rp and gm are at your actual desired current, unless you pick the one current for which the tube manual gives you those values. Bah, it's easier to calculate the resistor from the curves directly or determine the value empirically.
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30th April 2003, 09:43 PM  #5 
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The curves are by far the most useful thing...
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30th April 2003, 09:58 PM  #6  
diyAudio Senior Member
Join Date: Aug 2002
Location: Belgium

DO I LOVE CURVES, OR WHAT?
Hi,
Quote:
A time consuming business, I tell you. Maybe I should put a Sophia on my list for Santa. They invariably differ from the characteristic curves as presented by the manufacturer, often not for the better. Which also goes a long way to explain as to why people do hear differences between identical type numbers from different manufacturers, batches etc. Cheers,
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Frank 

30th April 2003, 10:11 PM  #7  
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Join Date: Sep 2002
Location: Oregon, USA

OK, assuming a theoretically perfect triode with constant rp and gm, could one of you explain how that formula works for DC conditions? I thought it would be DC, I didn't see why I would need to consider AC conditions to find the value of Rk. I thought mu is the ratio of the change in plate voltage to the change in grid voltage, both required to produce the same change in plate current. How does one then bring mu back into DC analysis? Using mu = gm/rp, I guess, and assuming rp is valid for DC. Even then, I don't understand how the (mu + 1) divisor comes around, even for a perfect triode.
Quote:
Quote:
It's embarassing (and frustrating) when a book treats something as so basic that it doesn't need explanation, and I spend most of the evening trying to figure it out. Maybe I should have started with Rozenblit's book 

30th April 2003, 10:25 PM  #8 
diyAudio Moderator Emeritus
Join Date: Jan 2003
Location: Near London. UK

All (valve or transistor) amplifiers are a mix of DC and AC conditions. AC conditions vary with DC conditions. Thus, DC conditions must be found first.
Early theorists tried to predict DC conditions using grossly oversimplified physical models. Hence the equation you presented. They aren't terribly good, but they are better than nothing. You have curves available, so you can ignore these poor models. Yes, that is how to predict ra and gm. However, it's usually better to predict mu and use it to predict gm (by calculation with ra), than attempt to predict gm directly. This is because gm is very nonlinear with Ia.
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30th April 2003, 10:31 PM  #9 
diyAudio Member
Join Date: Sep 2002
Location: Oregon, USA

Thanks, I think that answers my question. mu would be calculated by drawing a horizontal line. I can see how that would have less error than drawing a vertical line.

30th April 2003, 10:32 PM  #10 
On Hiatus

rp = dVp/dIp at fixed Vg
gm = dIp/dVg at fixed Vp But, as I pointed out, if you're going to go to the trouble of determining these derivatives from the graphs, you may as well determine the operating point directly. A nice simplifying assumption is that Vk = 0, which is fairly true for most practical plate and cathode voltages. I'm out drinking in your town tonight. If you hear some really loud, obnoxious people, that's probably us.
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