How to understand the power output in this case? - diyAudio
 How to understand the power output in this case?
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 27th September 2012, 05:35 AM #1 diyAudio Member   Join Date: Jun 2011 How to understand the power output in this case? Hi folks, I learnt in duncanamp that EL34 PP AB1 class can produce power to 100W/ch. Class AB1 P/P Va 775 Vg2 400 Vg1 -39 Ia 50 - 182 Ig2 6.0 - 38.0 Ra S Rk Zout 11,000 Pout 100 THD 5 Notes Shared Rg2 = 750 ohms Since max dissipation power of EL34 is just only 25W/valve, therefore, PP with 2 valves per channel can only produces max 50W/ch. So why the output power is so great? Is this the parallel PP in this case? Thanks!
 27th September 2012, 07:59 AM #2 diyAudio Member     Join Date: May 2002 Location: The great city of Turnhout, BE Blog Entries: 8 The dissipation of a tube is not the same as the max output power. The dissipation is the loss of power that heats the tube, but the output power in class (A)B can be higher. jan didden __________________ Music is dither to the brain; lets me think below the usual chaos - me Linear Audio Vol 12 is out! Check out my Autoranger and SilentSwitcher
 27th September 2012, 10:00 AM #3 diyAudio Member     Join Date: Mar 2008 Class A: Maximum theoretical efficiency is 50%. Class B push-pull: Maximum theoretical efficiency is 78.5%. __________________ Judge: This court appreciates that you invented physics, Mr.Newton, but unfortunately you can't have a patent on it.
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Quote:
 Originally Posted by hungdn Hi folks, I learnt in duncanamp that EL34 PP AB1 class can produce power to 100W/ch. Class AB1 P/P Va 775 Vg2 400 Vg1 -39 Ia 50 - 182 Ig2 6.0 - 38.0 Ra S Rk Zout 11,000 Pout 100 THD 5 Notes Shared Rg2 = 750 ohms Since max dissipation power of EL34 is just only 25W/valve, therefore, PP with 2 valves per channel can only produces max 50W/ch. So why the output power is so great? Is this the parallel PP in this case? Thanks!
The key point is to use Class-B and to max out the tube's Ub: The mean value of the plate dissipation has to be <25W for a whole cycle (EL34). But this is regardless of the exact distribution in this interval of the cycle. So you can go well above the Pa-max hyperbola for some small time interval. This can be proven mathematically quite easy when you assume a sinusoidal signal.

Below you see the EL34 curves for the Operating conditions you published in your post. At the hottest point, the EL34 "burns" over 60W on its plate - not a problem.

I hope, you kind of got the point how this works.
Attached Images
 EL34_800V.png (61.0 KB, 108 views)
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 27th September 2012, 06:50 PM #5 diyAudio Member   Join Date: Jun 2011 Thanks all. Since I am a newbie to electronic and trying to design my first tube amp schematic , so all your information is value to me. A question for the_manta: As I see in the datasheet of TLF EL34, Max anode voltage = 800V Max anode current = .15A So I dont understand that why the load-line is come from the point (0V,320mA)? As your explanation, I understood that in AB class, the tubes are working alternately, so in the given short time, output power of the tube can greater than the max dissipation rate. So is there any formula or something in the datasheet can help me to calculate that? Thanks so much for your help
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But you should know, that this operation principle applies for Class-B operation. Like when you have to squeeze everything out of a tube. Distortion is unimportant in such a case. Nowadays, Class A or warm Class AB operation is prefered.
Also your data is not Class-AB anymore, it is Class-B, as also pointed out in the Telefunken or Philips Datasheet. (The well known 800V/100W/11K-Raa operating point)

Quote:
 So I dont understand that why the load-line is come from the point (0V,320mA)?
I made a small mistake there ! I misread your given value and used 10K-Raa and 800V for the picture.
It's easy. For Class-B operation, Raa = 4*Ra (The total resistance of the output transformer has to be 4 times the value of a single tube. (For Class-A it is 2 times!). Reason: Always only one tube conducting instead of 2 tubes in Class-A)
So Ra= Raa/4 = 11K/4 = 2,75K

Furthermore: Ra=Ub_max/Ia_peak
rearrange: Ia_peak = Ub_max/Ra = 775V/2,75K = ~282mA (instead of 320mA as I drew)

Quote:
 As your explanation, I understood that in AB class, the tubes are working alternately, so in the given short time, output power of the tube can greater than the max dissipation rate. So is there any formula or something in the datasheet can help me to calculate that?
This would take a very long time here. I suggest some good books here:

The concept behind this: Use a power ansatz:

P_anode = P_supply - P_out

So you're left with finding P_supply and P_out. But, you have to find them in a way, that you maximize P_out and that P_anode matches P_anode_max (25W for EL34). So this leads to a derivation of your P_anode function to find the maximum etc....

I don't have classic Litarure in English language, just german. Maybe someone here can give some advice where you can find the proper PDF.

But, again: For a good, low-distortion tube amp you don't need to know this. Class-A Push Pull is unbeaten when it comes to THD and the math behind this is easy.
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 28th September 2012, 01:52 AM #7 diyAudio Member   Join Date: Jun 2011 @the_manta: Thank you so much for your time. So much clear for me.

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