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Old 4th May 2006, 01:56 AM   #1
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Question Determining Output Power

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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Old 4th May 2006, 02:17 AM   #2
arnoldc is offline arnoldc  Philippines
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I normally do it actual measurement- using a Simpson Power Level meter.
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Old 4th May 2006, 05:09 PM   #3
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Default estimated

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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Old 4th May 2006, 05:37 PM   #4
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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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Old 4th May 2006, 06:36 PM   #5
Giaime is offline Giaime  Italy
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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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Old 4th May 2006, 08:17 PM   #6
cerrem is offline cerrem  United States
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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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Old 4th May 2006, 10:03 PM   #7
rdf is offline rdf  Canada
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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.
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Old 4th May 2006, 10:23 PM   #8
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Quote:
Originally posted by Giaime
I suggest you to dig into the articles about loadlines here:
http://members.aol.com/sbench102/po-dis.html
That's excellent. Thanks.

Quote:
Originally posted by cerrem
If you can provide some more detail... I will show an example of first order calculation...
The details of the circuit are important....
It is just a transformer with one side connected to the plate and the other to B+, and a resistor and cap from the cathode to ground.


Quote:
Originally posted by rdf
Remember though headphones are relatively high impedance devices that respond to voltage, not current.
That's true for sennheisers and other high impedence phones, but Grados (at 32 ohms) are all about current.

Quote:
Originally posted by rdf
Grado rates their efficiency in milliwatts/per severity of tinitus. One watt into Grados is called fireworks.
Now that probably is true.
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Old 4th May 2006, 10:43 PM   #9
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Quote:
Originally posted by dsavitsk
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
It's NBD, actually. Here's a loadline I did for a SE 6BQ6GTB project I'm working on. Given the Pd of this tube (12W) it looks like Vpkq= 250Vdc is a good place to start. For this voltage, assume that the Vpp`= 500Vdc, since the xfmr primary represents a "phantom" Vpp rail. The primary inductance does this. Once you have that voltage determined, it's simply a matter of fishing for a loadline that gives the best linearity without busting the Pd spec. For this, I print the plate curves out to make it easier to play with them. In this case, the THD= 0.3%. Of course, this is just an estimate since the equation used assumes that the second harmonic is the only component present. That never occurs in the real world, but it is a useful estimate. Once you have that loadline determined, simply read off the maximum and minimum voltages to figure power out: Po= [(Vpmax - Vpmin)/(Ipmax - Ipmin)]/8.0. The divide by eight turns P-P values into RMS values.

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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Old 4th May 2006, 11:24 PM   #10
rdf is offline rdf  Canada
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Quote:
Originally posted by dsavitsk


That's true for sennheisers and other high impedence phones, but Grados (at 32 ohms) are all about current.

In the world of headphones 32 ohms eats power, in the world of OPTs meant for speakers it's nothing. I've run headphones directly off the output of devices ranging from a single stage 6C45 5-kohm SE to a Crown D-75 (deaf sportscaster.) Power really isn't important until your start paralleling.
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