Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: John on July 27, 2017, 05:53:02 am
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Hi Joel, we're taking Sluckey's thread off track; your topic needs it's own.
You've got a push-pull amp (you're using an 18W OT) this means you've got 2 power tubes. IF I'm thinking correctly, your figuring out your pDiss properly using your B+ & current numbers, but you're forgetting that your voltage being dropped across your K resistor is for 2 tubes. You need to take that dissipation and divide it by 2.
IIRC, you've got 350 volts at the plates, and you're measuring 21 volts across the 250 ohm K resistor. 21/250= .084. That's the current. .084 x 329 volts (plate-cathode) is 27.6 watts / 2 tubes = 13.8 diss per tube. So yes, it's a little over the 12 watts dissipation on the data sheet, but your tubes won't melt.
This is assuming that I'm understanding your original question, of course. :icon_biggrin:
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Cheers John.
I'm not following some of your logic. Can you run through the process from the top? Starting with B+ 350V, and 8.4k output transformer. How do you go about drawing the load lines, determining the bias point and hence an appropriate cathode resistor? Then calculate the approximate output power?
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I've never drawn a load line on a power tube, it's way too complicated for me. The only thing I pay attention to is max dissipation, and max plate voltage. Even those figures can be fudged up a little, due to the fact the data sheet are normally a little conservative. And those were drawn for uses other than guitar amps, where we like to push the envelope.
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To draw loadlines on tubes, download a copy of RDH4 from this website.
Practice drawing the loadlines on triodes first, then graduate to the pentodes. chapter 12 and 13 respectively.
I have not seen the method to draw load lines for pentodes operating in UL (ultralinear) mode.
Good luck.
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load lines for pentodes operating in UL
just raise it to 45degrees n ur close enough :icon_biggrin: