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Old 27th December 2017, 04:15 PM   #1
tommydoan84 is online now tommydoan84  United States
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Tubelab SSE - Cathode resistor understanding
Default Tubelab SSE - Cathode resistor understanding

hi all,

I'm a 1st time SSE builder and I hope you can help me shed the light on choosing the right cathode resistor R17 and R27 for different tube types.

I have read several threads in this forum about cathode resistor and also trying to build a switch to select different resistors for different tubes like member Russ "rkinze".

I look over George's tube& application table.
"choose a cathode resistor value for a given amount of tube dissipation" Tubes and Applications | Tubelab

My build include
Power Transformer: Edcor XPWR059
Output Transformer: Edcor CXSE25-8-5K
Output tube: KT88 genalex gold lion & EL34 - Electro Harmonix

My build's B+ voltage is around 450V. Looking at EL34 & KT88 data sheet and George's table for KT88, EL34 450V, I still don't know which value of Rk to use for these tubes.

Can someone show me what values of dissipation on the 2 data sheet I need to look at and compare with George's table ? I check every value of dissipation in the 2 spec sheet but they are not matching with George' Pwr out values

EL34 Electro harmonix's data sheet:
https://www.newsensor.com/pdf/electro-harmonix/el34eh.pdf

KT88 - Gold lion data shee
http://www.jacmusic.com/KT88/kt88_spec_sheet.pdf

I really appreciate your inputs.
thanks,
Tom
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Old 27th December 2017, 04:36 PM   #2
Treybal is offline Treybal  Chile
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Default Tubelab SSE - Cathode resistor understanding

Hello, in a similar build 440-460 V, i have use 560 ohms 10 W Mills. Over 5 having no pcf problems with JJ EL34, Svetlana 6550B and KT88 Gold Lions (new production).

Cheers

Last edited by Treybal; 27th December 2017 at 04:39 PM.
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Old 27th December 2017, 05:06 PM   #3
scott17 is offline scott17
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Yes, same as Treybal I've used 560R 10W with a similar B+ 440V. No problems.
To run the EL34 at a little less dissipation you could use a 680R-10W resistor and then install a switch to parallel a 2.2K-5W when you run the KT88. Effectively ~520 ohms. Use a wirewound for both.

To answer your question, the values you look at on the data sheet are Plate Dissipation (Pa) and Screen (grid 2) Dissipation. If you are running the tubes as triodes, since the plate and G2 are tied together, you can add the plate and screen dissipation.

Last edited by scott17; 27th December 2017 at 05:20 PM. Reason: add info
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Old 28th December 2017, 05:04 AM   #4
tommydoan84 is online now tommydoan84  United States
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Tubelab SSE - Cathode resistor understanding
Thank you Treybal and scott17 for the reply and explanation.

Instead of using both wirewound resistors, can I use 680R-10W wirewound and the rest of resistors in // with 680R to be 3W metal oxide ?

Thanks
Tom
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Old 28th December 2017, 10:52 AM   #5
scott17 is offline scott17
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If you already have the metal oxide resistors then use them. I've read that metal oxide are noisier than wirewound or metal film, but I'm not sure it will matter in this case. Someone else may want to comment. I don't have any personal experience using metal oxide in this application.

You said "resistors", does that mean you are going to use more than one to make the 2.2k? When using the KT88, the 2.2k in parallel with the 680R is going to need to dissipate close to 1 watt.
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Old 28th December 2017, 03:08 PM   #6
tommydoan84 is online now tommydoan84  United States
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Tubelab SSE - Cathode resistor understanding
Thanks Scott. In this case, I'll stay with wirewound all the way. I didn't know metal oxide is noisier. I have not bought all the resistors for this just yet. I only use 1 resistor in // with 680R. I thought putting resistor in // will increase power. So if 680R is already 10W, we only need 3W or 5W of the other resistor right ?

Thanks,
Tom
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Originally Posted by scott17 View Post
If you already have the metal oxide resistors then use them. I've read that metal oxide are noisier than wirewound or metal film, but I'm not sure it will matter in this case. Someone else may want to comment. I don't have any personal experience using metal oxide in this application.

You said "resistors", does that mean you are going to use more than one to make the 2.2k? When using the KT88, the 2.2k in parallel with the 680R is going to need to dissipate close to 1 watt.
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Old 28th December 2017, 03:31 PM   #7
scott17 is offline scott17
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Hi Tom,
If the resistors were both the same value then yes, each one would dissipate half the wattage. They both have the same voltage across them so if you do the math you'll see. P=E^2/R. I would use a 5W 2.2k. For anode and cathode resistors I like to take the actual wattage dissipation and multiply by 4.
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Old 28th December 2017, 04:47 PM   #8
tommydoan84 is online now tommydoan84  United States
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Tubelab SSE - Cathode resistor understanding
Thank you Scott. Really appreciate your inputs and explanation!

Regards,
Tom

Quote:
Originally Posted by scott17 View Post
Hi Tom,
If the resistors were both the same value then yes, each one would dissipate half the wattage. They both have the same voltage across them so if you do the math you'll see. P=E^2/R. I would use a 5W 2.2k. For anode and cathode resistors I like to take the actual wattage dissipation and multiply by 4.
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