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#1 |
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diyAudio Member
Join Date: Jan 2005
Location: Chicago
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How do I determine the power that a given tube with a given transformer, a given B+, into a given load is capable if delivering?
I ask because I hooked up a 6gm8 to some edcor transformers (5K:8) and was surprised to find that it did a pretty good job on my Grados. Gain was a little low, but bass was great and sound was surprisingly good. I am curious what was going on, and how to improve things. |
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#2 |
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diyAudio Member
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I normally do it actual measurement- using a Simpson Power Level meter.
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#3 |
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diyAudio Member
Join Date: Jun 2002
Location: Denver, CO
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It can be estimated using lots of assumptions. Some folks will plot the loadline and determine graphically the power out. Plotting a loadline requires info such as B+ and load impedance. Results are power out of the power tube to the load (transformer). Power to the speaker is reduced by transformer loss (estimated).
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#4 |
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diyAudio Member
Join Date: Jan 2005
Location: Chicago
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I know how to plot load lines, and I am happy with a very rough estimate. So, how do you do it? I found one person saying to muliply change in current by change in voltage and then multiply the whole thing by 8, or divide it by 8, I don't remember. Something like that
As I say, I got a surprising amount of power from a 6gm8, but I'd like to see if it is as surprising as it seems, and how various other tubes might compare so I have a sense what to tinker with. So, even knowing how one tube relates to another, that is, estimating proportional power, would be fine. |
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#5 |
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diyAudio Member
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I would tell you to look at the Radiotron at my site, but for a quicker answer, I suggest you to dig into the articles about loadlines here:
http://members.aol.com/sbench102/po-dis.html |
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#6 |
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diyAudio Member
Join Date: Jun 2005
Location: San Diego, CA
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If you can provide some more detail... I will show an example of first order calculation...
The details of the circuit are important.... Chris |
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#7 |
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diyAudio Member
Join Date: Jun 2004
Location: big smoke
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Remember though headphones are relatively high impedance devices that respond to voltage, not current. Grado rates their efficiency in milliwatts/per severity of tinitus. One watt into Grados is called fireworks.
__________________
Ears aren't microphones. |
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#8 | ||||
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diyAudio Member
Join Date: Jan 2005
Location: Chicago
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#9 | |
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diyAudio Member
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Quote:
Of course, once you start to build the thing, you will probably have to tweak the op-point somewhat to realize the best distortion performance. Fortunately, VTs tend to be quite forgiving concerning op-points, moreso than SS. For more details, see Steve Bench's Site. There are five articles on loadlines under the "Technical Reports" heading. |
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#10 | |
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diyAudio Member
Join Date: Jun 2004
Location: big smoke
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Quote:
__________________
Ears aren't microphones. |
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