Could someone please help me understand this equation? - diyAudio
Go Back   Home > Forums > Amplifiers > Tubes / Valves

Tubes / Valves All about our sweet vacuum tubes :) Threads about Musical Instrument Amps of all kinds should be in the Instruments & Amps forum

Please consider donating to help us continue to serve you.

Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
Reply
 
Thread Tools Search this Thread
Old 30th April 2003, 08:53 PM   #1
diyAudio Member
 
Join Date: Sep 2002
Location: Oregon, USA
Default 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:
The formula for determining the cathode resistor's value (in the absence of a plate resistor) is a simple one:

Click the image to open in full size.

where, Rk equals the cathode resistor; Vb, the B+ voltage; Iq, the desired idle current; rp, the plate resistance; and mu, the triode's amplification factor.
How do we get to that expression? Vb seems to be a constant, so the voltage at the cathode, Iq.Rk = Vb - (drop inside the tube), which is Iq.rp (assuming rp is for AC as well as DC resistance, which he says it's sometimes used interchangably for). This gives me Rk = Vb/Iq - rp, with no mu + 1. I've seen mu + 1 come in when the grid is grounded (either for real or my shorting the voltage source driving the grid while "Thevenising") - so if we're considering AC/dynamic conditions, for a change /\Vk (delta Vk), the grid-cathode difference also goes up by /\Vk, so the plate voltage changes by (mu + 1)./\Vk. But Vb is a constant. And that's where I get lost and don't know how to proceed.

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 Vg-k above the (grounded) grid, so Rk = Vg-k/Iq.

Any help would be most appreciated.

Thanks,
Saurav
  Reply With Quote
Old 30th April 2003, 09:28 PM   #2
diyAudio Member
 
Join Date: Jun 2002
Location: Denver, CO
Default 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
  Reply With Quote
Old 30th April 2003, 09:35 PM   #3
EC8010 is offline EC8010  United Kingdom
diyAudio Moderator Emeritus
 
EC8010's Avatar
 
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 single-line 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 spot-on.
__________________
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...
  Reply With Quote
Old 30th April 2003, 09:39 PM   #4
SY is offline SY  United States
diyAudio Moderator
 
SY's Avatar
 
Join Date: Oct 2002
Location: Chicagoland
Blog Entries: 1
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.
__________________
You might be screaming "No, no, no" and all they hear is "Who wants cake?" Let me tell you something: They all do. They all want cake.- Wilford Brimley
  Reply With Quote
Old 30th April 2003, 09:43 PM   #5
EC8010 is offline EC8010  United Kingdom
diyAudio Moderator Emeritus
 
EC8010's Avatar
 
Join Date: Jan 2003
Location: Near London. UK
The curves are by far the most useful thing...
__________________
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...
  Reply With Quote
Old 30th April 2003, 09:58 PM   #6
diyAudio Senior Member
 
fdegrove's Avatar
 
Join Date: Aug 2002
Location: Belgium
Default DO I LOVE CURVES, OR WHAT?

Hi,

Quote:
The curves are by far the most useful thing...
Oh yes...which is why I plot them for every tube that goes into my own circuits.
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,
__________________
Frank
  Reply With Quote
Old 30th April 2003, 10:11 PM   #7
diyAudio Member
 
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:
Your penultimate paragraph is spot-on.
OK, thanks, that's reassuring

Quote:
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
And I think I know how to do that. rp is the slope of the plate curve at the operating point, gm is... ok, change in plate current by change in grid voltage for constant plate voltage, so a vertical line intersecting plate curves, measure the current change against the Y axis grid and the grid voltage change by the values on the curves I cross. Does that sound correct?

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
  Reply With Quote
Old 30th April 2003, 10:25 PM   #8
EC8010 is offline EC8010  United Kingdom
diyAudio Moderator Emeritus
 
EC8010's Avatar
 
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 over-simplified 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 non-linear with Ia.
__________________
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...
  Reply With Quote
Old 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.
  Reply With Quote
Old 30th April 2003, 10:32 PM   #10
SY is offline SY  United States
diyAudio Moderator
 
SY's Avatar
 
Join Date: Oct 2002
Location: Chicagoland
Blog Entries: 1
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.
__________________
You might be screaming "No, no, no" and all they hear is "Who wants cake?" Let me tell you something: They all do. They all want cake.- Wilford Brimley
  Reply With Quote

Reply


Hide this!Advertise here!
Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
Feedback equation Aidan135711 Tubes / Valves 10 16th January 2009 03:47 PM
Suspension Sag Equation hooha Subwoofers 19 22nd February 2008 12:24 PM
Dipole Model Equation tg3 Multi-Way 1 13th October 2003 03:21 PM
Which port equation to use? Yoda Multi-Way 3 10th November 2001 04:31 AM


New To Site? Need Help?

All times are GMT. The time now is 12:55 AM.


vBulletin Optimisation provided by vB Optimise (Pro) - vBulletin Mods & Addons Copyright © 2014 DragonByte Technologies Ltd.
Copyright 1999-2014 diyAudio

Content Relevant URLs by vBSEO 3.3.2