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Old 16th December 2008, 02:27 PM   #1
sgerus is offline sgerus  United States
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Default What is the RIGHT way to calculate the Value of the cathode resistor?

Lets say I want to run my triode (2A3-40) at 360V, -65V, and 87mA.

65V/.087= 747R. so 750R on the cathode and where good.

BUT this never lines up on the plot sheet.
Given:
Vb=360V
Iq=87mA
U=4.2
Rp=800

Using this formula: Rk = ((Vb/Iq)-Rp) / (u+1)

Rk=641R. so we use 650R and the plate and grid voltages all line up with the plot sheet, but now there is 100ma flowing through cathode.

So for this example, do we use 750R knowing that where running the valve at 87mA, or 650R to run the valve at the exact plate and grid voltages?
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Old 16th December 2008, 03:00 PM   #2
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Default Re: What is the RIGHT way to calculate the Value of the cathode resistor?

Quote:
Originally posted by sgerus
Lets say I want to run my triode (2A3-40) at 360V, -65V, and 87mA.
First, you can't choose all three. If the plate voltage is 360V and the grid bias is -65V, then the idle current is going to be wherever it falls on the tube's curves.

Second, when computing load lines for single ended output stages you must be careful to remember the output load is an inductor. The actual working load line of the tube will be parallel and above the line you would use for a resistively loaded tube. This site has a good, but easy to understand writeup.

http://www.freewebs.com/valvewizard1/se.html
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Old 16th December 2008, 03:42 PM   #3
Sheldon is offline Sheldon  United States
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Default Re: What is the RIGHT way to calculate the Value of the cathode resistor?

Quote:
Originally posted by sgerus
So for this example, do we use 750R knowing that where running the valve at 87mA, or 650R to run the valve at the exact plate and grid voltages?
Your plate voltage is determined by the power supply, but it can vary depending on the current load. Assuming you have designed the power supply for a given voltage at a chosen current, you will want to select the cathode resistor to deliver close to that current, and let the grid voltage fall where it may, as long as it reasonably close to the predicted value. This value will vary some from tube to tube. Depending on tube type, it can vary by as much as 20% or so. With cathode biased amps, the grid voltage and cathode current both change with the resistor, so you may have to make adjustments to the cathode resistor to get the current you want. Tube variation will make it difficult to get everything exact, so within 10% of the design spec. is fine. A closer match between channels is done by tube selection.

Sheldon
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Old 16th December 2008, 03:43 PM   #4
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When using the curve, are you using anode to cathode voltage as Va, or B+ ?

You shouldn't use the B+ value for cathode biased stages, as the cathode is elevated above circuit common (grid at common). The plate curves are respect to anode-cathode voltage.
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Old 16th December 2008, 04:08 PM   #5
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Zigzagflux is so right, you have mixed up the voltages.

Your formula is wrong. Change Vb to Ua and it will be OK!
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Old 16th December 2008, 04:24 PM   #6
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Default Re: What is the RIGHT way to calculate the Value of the cathode resistor?

Quote:
Originally posted by sgerus
Lets say I want to run my triode (2A3-40) at 360V, -65V, and 87mA.

65V/.087= 747R. so 750R on the cathode and where good.

BUT this never lines up on the plot sheet.
Given:
Vb=360V
Iq=87mA
U=4.2
Rp=800

Using this formula: Rk = ((Vb/Iq)-Rp) / (u+1)

Rk=641R. so we use 650R and the plate and grid voltages all line up with the plot sheet, but now there is 100ma flowing through cathode.

So for this example, do we use 750R knowing that where running the valve at 87mA, or 650R to run the valve at the exact plate and grid voltages?

I took your first set of numbers; 360V anode-cathode, -65V and 87mA
and they do look close on the JJ 2A3-40 plate characteristics curve.
That is a valid operating point for the tube, regardless of how it was
chosen. I'd read it as -67V for 87mA and 360V.

Your first method of deriving the 65 volts across a 750R at 87mA is
correct. I don't understand why you say it doesn't line up on the plot
sheet. Did you try it like the attached example?

The other formula is incorrect for deriving the static operating point.
Mu is not a DC parameter, but dynamically derived depending on the
operating point and only assumed constant over a small delta. Same
with Rp, so can not be used to derive the DC conditions relative to zero.

Cheers,

Michael

PS as others have implied, the 360V is cathode-anode. With cathode
bias if 65V the B+ will be 425V
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File Type: png 2a3-op.png (61.4 KB, 442 views)
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Old 16th December 2008, 04:58 PM   #7
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Sorry Sgerus,
Didnt look close enough, as I made the assumtion 360V was B+ .

Anyway, dont use that working point IRL as it not in the most linear of the diagram. Between 250-300 will work better.
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Old 16th December 2008, 06:36 PM   #8
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Default Is a JJ 300b more linear than a JJ 2a3?

Here's the same OP plotted on the JJ 300b curves.

Looks a lot more linear, right?

The JJ 300b and 2a3 do in practice have identical plate characteristics,
just plotted a little differently apparently.

I guess linearity is a function of the load impedance also, but I'm
assuming some load line in the 3K5-5K range.

I guess choice of OP is also determined by the desired power output,
and the higher Va offers more Po, all other factors being equal.

With this OP you could get about 8 watts into 3K5, 6 or 7 watts into 5K.

At 250-300V you would be looking at 4-6 watts and could get nice
low distortion also.

I think another valid direction to go is treat it like a 300b and go
400V/75mA which would enable 10W Po into 3K5 and 8 watts into 5K.

The 5K loads have lower distortion at the higher operating voltages
vs 3K5 or 4K. Some folks go down to 2K5 or 3K which can work well
with some speaker types.

I also use the WE 300a/b chart to evaluate OPs for the JJ tubes.

Cheers,

Michael
Attached Images
File Type: gif 300b-op.gif (99.1 KB, 395 views)
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Old 16th December 2008, 07:38 PM   #9
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Hi Michael,
Where do you get the figures from ? The 2A3 curves are incomplete at higher voltages so you can not get all answers from them!

Even though, you can definiately see that the 2A3s are worse at higher voltages, check -100/450V and -120V/500V as examples.

300B and 2A3 does not have identical plate characteristics, as I see it. It would have been nice to have one of each and check them in my AVO.

If wanting to go for higher voltages, 300B must be the way to go.

About JJ I would suggest measuring them before adopting them. I can not speak for 300B or their 2A3 but I started out using JJ-tetrodes when building my first series of guitar amps. Had to leave them due to inconsistency! I even ordered matched pairs from the factory but it didnt help. A total fiasco were 3 out of 5 7591s pairs of where defective.
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Old 16th December 2008, 10:40 PM   #10
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I get my figures on this from running both "types" in amps. I have
used them interchangeably except for the filament transformer.

I have run both JJ 2a3 and JJ 300b in the same amp at the same OP
(400V, 45-60mA range) and identical results. The samples I have
are identical except for filament voltage. The curves they publish are
different but the tubes are the same. There is a different strapping
of the filament, series in the 300b and parallel in the 2a3. That and
the color of the base...

The consistency of the samples I have (6 300bs and 2 2a3s) is
surprising to me, but then there is one bad 300b with an intermittent
anode connection. I would not hesitate to run push-pull with a 300b
on one side and a 2a3 on the other, they are even consistent across
tube number ;-)

I also had really good luck with ECC99s from JJ. The 4 samples I have
are consistent with each other and match within 1 mA side to side
with a common cathode resistor.

I haven't tried a lot of other brands yet except for NOS tubes, which
vary way more than the new JJ.

Cheers,

Michael
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File Type: jpg 2a3vs300b.jpg (59.0 KB, 391 views)
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