Tube Amps / Music Electronics
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|3/6/2006 12:46 PM|
how does one calculate output impedance based on the electrical parameters of an output section? for example, I know my plate voltage is 510V and my total idle current is 140mA. is it possible to calculate impedance with only this data? what other information do I need?
|3/6/2006 12:54 PM|
It's a tough thing to do.
The real way to do it is to compute the anode-to-anode impedance of the output tubes, reflect that through the transformer, then allow for the reduction in output impedance as caused by the feedback of the power stage of the amp.
No, because of the effect of the output tubes and feedback, you really can't calculate it from the power supply and the idle current.
It's far simpler to measure it. Set the amp up running a test signal into a largish load, say 50 ohms resistive. Measure the signal, and check to be sure it's not oscillating or clipping. Now add resistance in parallel to the load until the output voltage is halved. Again, make sure that the signal is not oscillating or clipping. When you're done, the parallel combination of the added load and the original load is a good estimate of the output impedance at that frequency.
The original load makes the measurement harder, but you have to use it because the feedback on tube power stages makes them unstable into open circuits - the old "don't run your amp with no load" stuff.
It's possible and likely that the output impedance varies somewhat with frequency. It's likely not to be one number.
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|3/6/2006 3:52 PM|
the only problem with *measuring* the impedance is that the amp has not yet been constructed. it's still in the planning stages. the amp in question is a 2x100W stereo power amp, running 4x6L6 per side in pentode mode, and no negative feedback. I'm trying to locate an output transformer that will properly match impedances. I realize that a pair of 6L6s usually wants about 5000 ohms, and therefore, a quad would want about 2500, but I was thinking that the higher than usual plate voltage might cause it to want a higher impedance, possibly somewhere above 3k ohms. I have, so far, only been able to simulate it in circuitmaker, and if there is a way that I can measure or calculate impedance based on simulation data, that would work just as well, I guess.
|3/6/2006 5:47 PM|
If your simulator is any good at all, you can run the test on it.
|3/6/2006 7:48 PM|
Ah. I see the problem. I responded to your "measuring output impedance" and gave you a correct answer to a question you didn't mean. Sorry - I was confused.
If I understand you correctly now, what you really want to know is how to specify the impedance ratio/turns ratio (which are two faces of the same thing) for proper load matching.
The output impedance is not the same as load matching. Ignore my earlier response.
Here's the rub - you can't really just look at a tube and a power supply and presuppose what anode to anode impedance will make it work best. What the tube makers did back in the Golden Age was to design a new power tube try to get its characteristics perfect, and then measure what loading the tubes actually worked best with.
Let's ignore the speaker side for a moment. We'll look only at plate to plate loading. What the Golden Age Guys did was to load up a pair of tubes with a transformer that gave a reasonable match, then they would vary the secondary load (which varied the plate to plate load) and they'd measure what came out. For instance, in testing 6L6's, they'd use a garden variety output transformer, but load it with many different load resistors on the secondary, from maybe 1 ohm up to 32, and they would plot the maximum power output, and things like second harmonic distortion, third harmonic, etc.
What universally happened was that power out would have a very broad, gentle peak at some impedance. So would total harmonic distortion. They would test lots of their new tubes, then write data books that told people what impedances to use, based on where the peaks of the power and harmonic generation were.
In general, a tube will NOT produce most power and lowest harmonic distortion at the same loading. For 6L6's IIRC, most power is about 4400 ohms, lowest distortion is about 6K. They would usually specify the lowest distortion point, but the guitar guys figured out that they could have more distorted power, and used the high power loading.
Here's the problem you're facing: the models in circuitmaker are not good enough to do that. Producing a new 100W OPT is a huge task, and what you haven't said in so many words is that you don't yet have an OPT to go with this. So you're either trying to specify one to be built or buy the closest thing you can. Either way, what you have precalculated and simulated in circuitmaker will not be much help. You're going to have to try the actual setup to see what you get.
I suggest that you build a half-power version with only one pair of 6L6's, and only one channel. Diddle the loading by using power resistors like the GAG's did, and pick what loading you like best, in the compromise between power out and clean reproduction. Then either go get one like that, or get one wound up for you. Only then try building one full power channel.
This is very unlikely to be a do-it-right-the-first-time kind of venture.
|3/6/2006 5:58 PM|
I don't exactly understand how this is supposed to work... can you explain how this will help me figure out what primary impedance I need in an OT?
|3/6/2006 6:49 PM|
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