Got some money, going to buy OPTs for PP GU-50

[TG]

Well, if the curves are that straight (which they never are), the tangent matches the curve.
But for your calculations in post #18 you should have taken all the deltas from the orange line:

[rongon]

Why? Aren't we interested in the delta of the current and voltage of the actual device, and not some straightened tangent derived from measurements of the device?

I feel like I'm missing something big here.

Figuring the rp from the orange line I get 0mA at 300V and 98mA at 380V.

380 - 300 = 80
0.098 - 0 = 0.098 = 816.33

Really??

I'd figure if you were going to draw a tangent, you'd draw it so that it intersects both points on your plate curve down to cutoff, so that you incorporate changes in plate resistance that happen as the tube approaches cutoff (where rp goes up and gm goes down). That's what I tried to accomplish with the thin red tangent I drew from the high voltage/high current point on the curve to the low voltage/low current point.

I'd better go look at the RDH4 again.


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[rongon]

OK, I checked the books. It does seem you want to 'sample' the grid curve in a 'central' part of it, where the curve is fairly straight. That will tend to give you a lower value of rp.

rp = mu / gm , so if you know those two other constants you can figure out rp. Solving for mu first:

gm = δVp / δVg

- I found that starting from an operating point of Vp = 340V and Ip = 98mA, the grid bias is -60V.

- Grid bias is -70V at Vp = 380V
- Grid bias is -50V at Vp = 304V

δVp = 380-304 = 76V
δIp = 20V

76V / 20V = 3.8 <-- That's the mu
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Now for gm

gm = δIp / δVg

- I found that at Vp = 320V and Ip = 120mA, Vg = -50V
- At Vp = 320V and Ip = 40mA, Vg = -70V

δIp = 80mA
δVg = 20V

0.08 / 20 = 0.004 <-- gm = 4mA/V
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rp = mu / gm

3.8 / 0.004 = 950 ohms
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In the latest scrawling on the plate curves, I drew a tangent from 320V-40mA (-70V) to 380V-98mA (-70V). That gives rp = 1035 ohms.
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I'm going to split the difference and say rp = ca. 1000 ohms.

rp is not truly constant, so isn't a bit of a judgment call? Yes, at a particular operating point there is one value of plate resistance. But the plate resistance will vary once the triode starts swinging out of its linear window of operation.

So yeah, I think 1000 ohms works for me.
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[TG]

This whole load line method is just an approximation, you're absolutely right.
The load line is not a line in real life either, so there's no point in ultra-precise calculations

[rongon]

Thank you TG!

A few well-chosen 'scope grabs with real life loads speaks volumes. That's a spot-on illustration.

You got me to go back to the books on this. I'd read the chapters on calculating rp in the RDH4, MerlinB's and Morgan Jones' books, but I missed that crucial detail of how to choose a section of the grid curve to make the calculations. Yes, they talk about using the straight part of the curve. I see clearly now how that is just an approximation, an educated guess -- so there's no use nitpicking.

I learn better when I have a real life problem to solve, and this did the trick.

So, thanks again!
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[TG]

I didn't really do anything, but you're welcome

[nerdorama]

I've thought about a project like this for a while. I have a nice pair of 6.6k OPT's set aside and plenty of GU-50's but it seems too many other things have taken priority. Haven't decided what sort of driver circuit to use but probably either a fet follower like the Baby Huey project or interstage PP transformers. I think you could use 300b operating points as a good starting design since the triode curves are fairly similar. You might also consider a pentode design with local feedback. It can sound very triode like but with better power out.
Good luck and keep us posted,
John

[rongon]

Mostly I'm looking to start building things out of parts I've collected over the years. I have too much stuff!

I have a pair of OPTs from an Eico ST40. They're rated 8.2k pri:4,8,16 secondary, 20W, no screen taps so only pentode or triode operation feasible. The thing I worry about is if a 20 watt-rated OPT can take the 90mA quiescent plate current from each GU-50 triode. In theory the two tubes should be balanced, so the standing current nulls. But in practice? What happens when the amp gets driven into Class AB?

Perhaps I should leave those OPTs for a nice PP EL84 amp and purchase a pair of 50W rated OPTs for this project.

I calculate I'll get 13W per channel, Class A triode. Long ago, I built a PP 300B amp that made about that. It was pretty whompous, but it did hum some from its AC filaments.

The plan right now is to use a Williamson-style driver circuit employing 6SN7 voltage amp-6SN7 cathodyne to 6V6-triode LTP, then to the GU-50s. Open loop sensitivity looks like about 250mV to 1W out, approx. 1V to full power. That seems about right. I could add 6dB of gNFB if that ends up being desirable. Looks good on the screen, at least.
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[nerdorama]

What is the resistance of the ST-40 primary? 90mA shouldn't cause too much heating in watts. Seems like I saw Mike of Magnequest make a similar calc and offer some guidelines about how much is too much for an unknown transformer.
John

[TG]

From the ST40 manual I can see that the tubes were originally biased at 56mA, and the OPT's primary DCR is just under 200 Ohm (each half).
So running GU-50 at 90 mA will add almost 2W of extra heat per OPT. Not optimal, but probably tolerable.
I wouldn't really be bothered about DC imbalance though, especially with self bias (and even more so with individual self bias).