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Old 17th April 2003, 04:03 PM   #1
Morse is offline Morse  United States
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Default Help! Dumb questions about a spec sheet....

Hi everyone and sorry to pester you with dumb questions, but here goes. Thanks in advance for your help!

On the Philips data sheet at http://frank.pocnet.net/sheets/010/e/ECL82.pdf

1. Is the "supply voltage" the raw B+ or is it the voltage of the plate relative to the cathode?

2. The pentode figures imply that 9.5V peak to peak on G1 will drive the pentode section completely. Using the recommended Rk=650r for the pentode section, and 37.8mA current at the cathode, it looks like the cathode is floated 24.6V. Is this right? It looks like the cathode is going to be floated way over the grid at all points of operation. Is this right?

3. If I'm reading all this correctly, then if I use B+ =272 + Vdrop due to OPT + Vcathode, and Vpeak to peak out from triode = 9.5V, then I should be able to get 3.5wpc at 10%THD.

Sorry for the dumb questions, but I'm having trouble finding useful characteristic curves on the 6BM8 (already got the Svetlana one and it only goes down to -10V on G1). This and the RCA tube manual are the best data I've got so far and if I'm reading this right, I like this operating point better than the one in the RCA manual (only gives 1.8w output at 10% THD IIRC).

Thanks again,
Morse
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Old 17th April 2003, 06:51 PM   #2
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Default 6BM8/ECL82

Hi,

Quote:
1. Is the "supply voltage" the raw B+ or is it the voltage of the plate relative to the cathode?
Yes, this is the voltage measured from the top of the anode resistor to ground.

2+3. Looks alright at first glance.

Cheers,

No such thing as a dumb question, Morse.
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Old 17th April 2003, 07:06 PM   #3
Joel is offline Joel  United States
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Frank, that's not correct actually!
"Supply Voltage" is always the voltage before the load - the raw B+. "Plate Voltage" is measured at the plate.

A 12AX7 would have a supply voltage of 300V, but an anode or plate voltage of only 150V or so.
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Old 17th April 2003, 07:14 PM   #4
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Hi,

Joel,

To clarify, some European tubemanufacturers such as Philips give an extra piece of data stated as Vb, which Philips calls "supply voltage".

I think what you mean is Va, which is measured directly at the anode.

Am I missing something too?
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Old 17th April 2003, 07:19 PM   #5
Joel is offline Joel  United States
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Quote:
Originally posted by fdegrove
...Am I missing something too?

Yes. Check the "maths", and everything should fall into place....
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Old 17th April 2003, 07:26 PM   #6
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Default ALRIGHT,ALRIGHT...

Hi,

LOL.

Quote:
Yes, this is the voltage measured from the top of the anode resistor to ground.
O.K. So let's rephrase this:

No, it is not really the raw B+ in that it normally isn't just the filtered voltage value.

So, Vb is the anode voltage measured between the top of the anode resistor and ground.

Va is the voltage measured at the tail end of the anode resistor and ground.

In the 12AX7 example Vb = 300VDC and Va = 150VDC.

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Old 17th April 2003, 07:27 PM   #7
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Default Plate Voltage

I thought the plate voltage specified on the data sheets is measured across the tube only--not necessarily to ground. That is, measure at the plate and cathode pins (above any cathode resisters, etc.).
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Old 17th April 2003, 09:47 PM   #8
Morse is offline Morse  United States
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Default Thanks and another question....

Thanks Frank and Joel; I think I get it now.

Okay, here's another question:

Given the following conditions (RCA given operating point), what power output can I reasonably expect? Could it support the 3.5watts that the Philips spec claimed?

For the pentode section of a 6BM8:

1. V plate to cathode = 200V, Ip = 35mA
2. V grid 2 (referenced to cathode) = 200V, Ig2 = 7mA
3. cathode floated 16V to provide bias
4. 5kr OPT, 8r load
5. mu = 9.5

If I feed g2 with 13.5Vrms, I get 13.5*9.5 = 128Vrms and Vout(rms) = 128*((8/5000)^.5) = 5.13V

Okay, with Vout = 5.13 I get 5.13^2/8 = 3.3 watts rms.

Can a 6BM8 support this under these conditions? If not, I can increase the B+ up to around 300V without too much hassle (I'm using a Hammond 269JX 500VCT@60mA trafo in a fullwave configuration with SS rectification).

Assuming that those numbers look okay, I'll set the triode section for a gain of about 11 to keep from overdriving the pentode. Sound like a (good) plan?

Thanks again for all your help and understanding!
Morse
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Old 17th April 2003, 10:42 PM   #9
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Default ECL82

Hi,

Quote:
what power output can I reasonably expect? Could it support the 3.5watts that the Philips spec claimed?
The tube can certainly handle this dissipation, no problem there.

If you consider a SE amp with this type I think an effective output around 2 W is more realistic.

In case you care for a couple examples you can have a look at these:

ECL82

Try to work through the diagrams and you'll probably see that the Philips datasheets are pretty useful stuff to have.

Quote:
I'm using a Hammond 269JX 500VCT@60mA trafo in a fullwave configuration with SS rectification
That is going to give you way too high a B+ with a FWB using diodes...unless you need that voltage for something else, I dont see much use for it.

Cheers,
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Old 17th April 2003, 11:15 PM   #10
Morse is offline Morse  United States
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Thanks Frank;

>>>....That is going to give you way too high a B+ with a FWB using diodes....<<<

Sorry, my fault - I'm going fullwave, but not a bridge. Hmmm, I've just checked it again with PSUD II, and using anything from 75mA to 90mA idle current I'm right in the ballpark around 250VDC for raw B+. That's with a 100uF->390r->100uF filter, by the way.

Anyway, thanks again - it's great having knowledgeable people to ask!

By the way, this one's purely a "what I have on hand" type of amp - hopefully I won't have to buy anything ('cept maybe another pair of 6BM8's as spares...) for it. That's one reason for the trafo - I've got one and it looked good on paper anyway...

All the best,
Morse
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