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14th March 2007, 01:49 AM  #1 
diyAudio Member
Join Date: Nov 2004
Location: Midwest Madman

Calculating SE Output Question
Hi Guys,
I am terrible at figuring out the math on these tube circuits, Partly because I can never find the references due to my terrible filing system. But anyway, Im trying to figure out the final output on a geetar amp Im playing with, But I think my math and formula is incomplete. Can Someone enlighten me? Its a common 3 tube amp, 112AX7, 16V6, 1 5Y3 With a single JJ 6V6S, 5K primary load to 8 ohm secondary ( Edcore XSE1585k OT) Hammond PT 3000300 @100ma Rcathode: 500 ohms Voltage across Rcathode = 23.8V Anode @ 391V, Screen @ 386V Pin 8 Cathode @ 23.8V So If I have this correct, 23.8V/R500= .047amps 391V23.8V= 367.2V 367.2 x .047 = 17.26 watts ???? With a 6V6? I do not seem to be overheating and no evident replate issues, No weird screen glows or anything, Longest duration I have used is about 2 hours straight at 5070% lvolume. But How do I determine my final output wattage? Trout 
14th March 2007, 04:56 AM  #2  
diyAudio Member

Re: Calculating SE Output Question
Quote:
Quote:


14th March 2007, 07:44 AM  #3  
diyAudio Member
Join Date: Nov 2002
Location: Athens, Greece

Re: Re: Calculating SE Output Question
Quote:
In general the output power is about a quarter of the dissipated power. For instance the output power of your amplifier will be about 17/4=4.25 Watt. But I have to agree with Miles Prower regarding the calculation from the loadline. 

14th March 2007, 09:32 AM  #4 
diyAudio Member
Join Date: Dec 2003
Location: new zealand

one simple way to do this is to draw the load line on the output curves. Identify the following points. Max Volts on the loadline, min volts, usually where the 0V grid line is, similarly, the max current, and the min current on the load line.
then the power is roughly as follows (VmaxVmin)(Imax Imin)/8 hope that his helps. kind regards 
14th March 2007, 10:00 AM  #5 
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Join Date: Feb 2005
Location: leeuwarden, netherlands

then the power is roughly as follows
(VmaxVmin)(Imax Imin)/8 ????? why divided by 8??? I learned it a different way: draw the load line..... substract the anode voltage at bias point from the anode voltage where ia=0 the current difference is: current at bias point multiply these too and then multiply by 0.5 (so, you take only one of the 'triangles' which interect at bias point) resume: 0,5x delta Va x delta Ia = Pout
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emission over emittor 
14th March 2007, 10:04 AM  #6 
diyAudio Member
Join Date: Nov 2002
Location: Athens, Greece

Have a look over here http://www.valveheart.com/theory/triodeSE.html
You will find it very useful I quess. 
14th March 2007, 01:37 PM  #7 
diyAudio Member
Join Date: Mar 2007

6V6 OUTPUT POWER
As my experience to meassure the actually output power from speaker:
(a) Use a digital Volt Meter. (b) Select meter range to ACV 20V. (c) Connect the meter probe to both output terminals of amplifier ,any polarity can do. (make sure speaker are connected firmly), (d) Slowly adjusting volume control until hearing noisy distorted sound from your speaker, Write down ACV reading , turn down volume at once! (e) then using this ACV meter reading devide your impedance of speaker, you can get the RMS value of watts power output to speaker. (F) RMSWatts times 1.414 will define to Peak power out put We can call this as undistorted output power to specifier speaker. (almost 90% accurcy !) It will better If use signal generator & scope for measuring! Hope it can help for RMS & PEAK power measuring!
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TJL 
14th March 2007, 03:56 PM  #8  
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Join Date: Nov 2004
Location: Midwest Madman

Quote:
Dead Link? Trout 

14th March 2007, 05:35 PM  #9 
diyAudio Chief Moderator
Join Date: Oct 2002
Location: AthensGreece

Works. (Firefox),

15th March 2007, 05:12 AM  #10  
diyAudio Member
Join Date: Dec 2003
Location: new zealand

Quote:
It's the same thing, you are taking half the voltage and current swing that I am taking, therefore, you only have to divide by 2, whereas i have to divide by 8, as my figures are 4 times bigger. make sense? 

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