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#1
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zero dissipation?
Is it theoretically possible for power tube plate dissipation to drop to zero
with large signals? I'm not making this up... I read it on the web. |
#2
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"wsc" wrote in message
... Is it theoretically possible for power tube plate dissipation to drop to zero with large signals? No actually, strictly speaking, since a tube will never fully cut off. In practical terms, yes it often does. Tim -- "I have misplaced my pants." - Homer Simpson | Electronics, - - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting and Games: http://webpages.charter.net/dawill/tmoranwms |
#3
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wsc wrote: Is it theoretically possible for power tube plate dissipation to drop to zero with large signals? I'm not making this up... I read it on the web. It *is* possible for a power tube to sit merilly in its socket, with 500 v applied to its anode, and no current flow, and no signal applied, so voltage x current = zero watts. This occurs in a class C amp at idle. A class B amp at idle also has zero dissipation at idle, but it only exists theoretically, since the tiniest signal causes plate current to flow, and in fact tubes have a gradual cut off so most "class B" amps do have some tiny bias current at idle. On the other hand, it is very ununusual for a tube to have a current flow, but with no voltage between the anode and cathode. This also is a case of zero dissipation. However in solid state mosfet devices for switching, the 'on' dissipation can be very low even with 10 amps in the device because the drain to source voltage is very low. Patrick Turner. |
#4
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Duncan Munro wrote: On 14 Apr 2004 19:22:35 GMT, wsc wrote: Is it theoretically possible for power tube plate dissipation to drop to zero with large signals? Yes, in some cases. Class B and AB tubes do not draw any power at all for part of the cycle. Yes, this is correct. During part of the cycle, the tube current is cut off, even though there is a high voltage rise at the anode. But at idle, all B and AB amps draw a current, and Ea x Ia = the idle power dissipation. Patrick Turner. -- Duncan Munro http://www.duncanamps.com/ |
#5
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No actually, strictly speaking, since a tube will never fully cut off. In
practical terms, yes it often does. Tim To clarify, what I was talking about is was a tube with a large signal where the bias is irrelevant - i.e. a square signal where the tube current goes from zero to the saturation current. If the dissipation were zero in this case, the plate efficiency would naturally be infinite. I think it may boil down to how you compute dissipation. To me its the average plate current times the average plate voltage. But to the guy who's making the claim of zero dissipation, he does the multiplication before the average. Here's the link: http://www.aikenamps.com/Why70percent.html |
#6
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"wsc" wrote To clarify, what I was talking about is was a tube with a large signal where the bias is irrelevant - i.e. a square signal where the tube current goes from zero to the saturation current. If the dissipation were zero in this case, the plate efficiency would naturally be infinite. ** It can never be zero since tubes are not perfect switches. I think it may boil down to how you compute dissipation. ** There is only one correct answer - but various to derive it. To me its the average plate current times the average plate voltage. ** Completely wrong. The product of plate voltage and current has to be computed *at each instant* for a given waveform and a dissipation waveform derived - the average value of this waveform is then the average dissipation. But to the guy who's making the claim of zero dissipation, ** He makes no such claim - he merely gives the result for an imaginary tube that acts like a perfect switch. .......... Phil |
#7
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"wsc" wrote in message
... To clarify, what I was talking about is was a tube with a large signal where the bias is irrelevant - i.e. a square signal where the tube current goes from zero to the saturation current. If the dissipation were zero in this case, the plate efficiency would naturally be infinite. Ah, something like a horizontal output (sweep amp). In this case, the dissipation is in series with the load on the current-drawing half of the cycle. At Pat said, transistors can manage very good class D (either fully on (saturated; by the plate-curve knee) or cutoff) because of their typical Vsat of under 4V for the higher-power devices (equivalent to say, a 4X150's Vsat around 300V). Most stuff is below 2V, with small signal devices getting less than a diode drop (.7V) in some cases. Likewise such a tube would be a 6AU6 with Vsat down below 50V. MOSFETs experience the same characteristic knee curve, but at still lower voltages. Indeed, an IRFZ44 can do what, .02 ohm at 20A? All of 8W peak dissipation! Tim -- "I have misplaced my pants." - Homer Simpson | Electronics, - - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting and Games: http://webpages.charter.net/dawill/tmoranwms |
#8
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wsc wrote: No actually, strictly speaking, since a tube will never fully cut off. In practical terms, yes it often does. Tim To clarify, what I was talking about is was a tube with a large signal where the bias is irrelevant - i.e. a square signal where the tube current goes from zero to the saturation current. If the dissipation were zero in this case, the plate efficiency would naturally be infinite. You're essentially describing a switching situation, at which tubes don't typically do nearly as well as transistors (especially power MOSFETs). No problem at the cutoff end - current is essentially zero, so dissipation is also zero. However, at the saturation end, there will typically be a considerable voltage drop across the tube. Look at a set of plate curves to see this. For the 6L6GC, for instance, with a grid voltage of zero and a resulting plate current of 200 mA, the plate voltage will be some 50 volts -- a 10 watt dissipation (not counting the substantial screen power dissipated under these conditions). I think it may boil down to how you compute dissipation. To me its the average plate current times the average plate voltage. But to the guy who's making the claim of zero dissipation, he does the multiplication before the average. No, it's not the order of the multiplication. Remember that instantaneous plate dissipation equals *instaneous* plate voltage times *instantaneous* plate current. As plate current goes up, voltage goes down -- but not by the same factor. That explains the curves in the link you provided. If you simply multiply average plate current by average plate voltage, you'll get a wrong result. How wrong? Depends largely on the waveshape of the signal. For a "right" result, you have to take each point of the signal, multiply the instantaneous voltage by the instantaneous current, add them up all those instantaneous powers (integrate them), and only then divide by time to get the true average power. In principle, one could derive the power curves for a sine wave signal in conjunction with an approximation of the characteristic plate curves; in practise, however, even this would be quite cumbersome. With "real-life" signal sources - forget it. That's why we tend to use "rules of thumb" based on experience and a safety margin. A practical approach for the home experimenter is to remember that power into the system must equal power out of the system. So, connect a known dummy load to the speaker output, then at various amplitudes, measure the DC input power (average plate current times plate voltage), and subtract the measured output power plus estimated losses in the OPT. The difference obviously has to go somewhere -- that's your plate dissipation at that output power. If you repeat the experiment for a square-wave signal, you'll find the results quite different! Clear as mud? Cheers, Fred -- +--------------------------------------------+ | Music: http://www3.telus.net/dogstarmusic/ | | Projects: http://dogstar.dantimax.dk | +--------------------------------------------+ |
#9
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Remember that
instantaneous plate dissipation equals *instaneous* plate voltage times *instantaneous* plate current. Fred, Thanks for responding. I happened across your cool web site earlier this year and was fascinated by dogzilla. I regularly do the measurements you suggest and the dissipation never goes anywhere near zero. When I've driven tubes way past clipping things always end up in a situation where the input power is roughly evenly divided between the load and the tube (minus some little bits due to heating of the transformer et cetera.) Here's a typical example, where I measured everything except the dissipation -- and figured that out by subtraction. This is just as you said, but I got it from a suggestion in RDH4 on p578. By the time you get to the right side of the graph the output is essentially square. I have a bunch more like this for different bias methods et cetera. http://members.aol.com/wsjcrane/SE_EL84.gif This is mainly why I find the idea of zero dissipation such a conundrum. |
#10
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"wsc" wrote in message
... This is mainly why I find the idea of zero dissipation such a conundrum. Zero *instantaneous* dissipation. In a square wave, half on, half off, the on half is dissipating twice the average power if the off half dissipates none. Simple enough, just follow the watts. Tim -- "I have misplaced my pants." - Homer Simpson | Electronics, - - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting and Games: http://webpages.charter.net/dawill/tmoranwms |
#11
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"wsc" wrote in message ... Remember that instantaneous plate dissipation equals *instaneous* plate voltage times *instantaneous* plate current. Fred, Thanks for responding. I happened across your cool web site earlier this year and was fascinated by dogzilla. I regularly do the measurements you suggest and the dissipation never goes anywhere near zero. When I've driven tubes way past clipping things always end up in a situation where the input power is roughly evenly divided between the load and the tube (minus some little bits due to heating of the transformer et cetera.) Here's a typical example, where I measured everything except the dissipation -- and figured that out by subtraction. This is just as you said, but I got it from a suggestion in RDH4 on p578. By the time you get to the right side of the graph the output is essentially square. I have a bunch more like this for different bias methods et cetera. http://members.aol.com/wsjcrane/SE_EL84.gif This is mainly why I find the idea of zero dissipation such a conundrum. ** There was no claim of zero dissipation in any real situation. You really need to go figure what was meant by the phrase "taken to the extreme... " Plus read all the posts in reply to yours. .................. Phil |
#12
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Zero *instantaneous* dissipation. In a square wave, half on, half off, the
on half is dissipating twice the average power if the off half dissipates none. Simple enough, just follow the watts. Tim Tim, Correct me if I'm misreading this, but I think Mr. Aiken is saying that with a large square wave the average dissipation approaches zero -- his reasoning being that when the current is high the voltage is low and when the voltage is high the current is low, and that the switching part can be ignored? I think his graph for push-pull EL34s shows the average dissipation. |
#13
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wsc wrote: Zero *instantaneous* dissipation. In a square wave, half on, half off, the on half is dissipating twice the average power if the off half dissipates none. Simple enough, just follow the watts. Tim Tim, Correct me if I'm misreading this, but I think Mr. Aiken is saying that with a large square wave the average dissipation approaches zero -- his reasoning being that when the current is high the voltage is low and when the voltage is high the current is low, and that the switching part can be ignored? I think his graph for push-pull EL34s shows the average dissipation. I think that the key word here is "approaches". This is true for any switching situation, an ideal switch will never dissipate any power; either the current is zero, or the voltage drop across it is zero - either eventuality producing zero dissipated power. Some switches are more ideal than others; vacuum tubes tend to be pretty close to ideal in the off-state (zero current), but not great in the on-state (maximum current) because of the substantial voltage drop that still exists across them under this condition. In the example I gave in a previous post, a 6L6GC in full saturation still dissipates about 10 watts. Averaged out over a 50% duty cycle square wave, this would still be a dissipation around 5 watts -- and whilst that "approaches zero" it's still not even "approximately" zero compared to many solid-state devices under similar situations. Cheers, Fred -- +--------------------------------------------+ | Music: http://www3.telus.net/dogstarmusic/ | | Projects: http://dogstar.dantimax.dk | +--------------------------------------------+ |
#14
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I think that the key word here is "approaches".
Fred, It took a while, but I finally convinced myself you guys are right about the power calculation. I guess the reason I've never seen anything remotely close to zero dissipation does have to do with the word "approaches." The biggest hurdle, as you said, would seem to be getting the voltage to approach zero. This is a rather monumental task involving a robust driver circuit. In such a case the plate dissipation in the power tube is just a small fraction of the power losses which now include grid dissipation in the power tube as well as plate dissipation in the driver tube. I think a meaningful statement about dissipation would probably have to include that stuff? |
#15
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"wsc" wrote in message
... In such a case the plate dissipation in the power tube is just a small fraction of the power losses which now include grid dissipation in the power tube as well as plate dissipation in the driver tube. I think a meaningful statement about dissipation would probably have to include that stuff? Absolutely! Unless you're trying to market of course, in which case you'll include only plate efficiency. Grid ratings be damned! Also the other reason why MOSFETs are so superior to just about anything. Until you break the grid insulator. #$@^. Tim -- "I have misplaced my pants." - Homer Simpson | Electronics, - - - - - - - - - - - - - - - - - - - - - - --+ Metalcasting and Games: http://webpages.charter.net/dawill/tmoranwms |
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