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 14th March 2007, 12:49 AM #1 Trout   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, 1-12AX7, 1-6V6, 1 5Y3 With a single JJ 6V6S, 5K primary load to 8 ohm secondary ( Edcore XSE15-8-5k OT) Hammond PT 300-0-300 @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 391V-23.8V= 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 50-70% lvolume. But How do I determine my final output wattage? Trout
Miles Prower
diyAudio Member

Join Date: May 2005
Location: USA
Re: Calculating SE Output Question

Quote:
 Originally posted by Trout 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 391V-23.8V= 367.2V 367.2 x .047 = 17.26 watts ???? With a 6V6?
That isn't correct. The 47mA is the cathode current, which includes the screen current. The actual plate current is less. Furthermore, that gives the DC input power, not the audio output power. Even so, if those figures are anywhere near correct, that's still pushing a 6V6 pretty hard.

Quote:
 But How do I determine my final output wattage?
That'll have to be figured from the loadline.
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satelitis
diyAudio Member

Join Date: Nov 2002
Location: Athens, Greece
Re: Re: Calculating SE Output Question

Quote:
 Originally posted by Miles Prower That isn't correct. The 47mA is the cathode current, which includes the screen current. The actual plate current is less. Furthermore, that gives the DC input power, not the audio output power. Even so, if those figures are anywhere near correct, that's still pushing a 6V6 pretty hard. That'll have to be figured from the loadline.

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, 08:32 AM #4 billr   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 (Vmax-Vmin)(Imax -Imin)/8 hope that his helps. kind regards
 14th March 2007, 09:00 AM #5 kathodyne   diyAudio Member   Join Date: Feb 2005 Location: leeuwarden, netherlands then the power is roughly as follows (Vmax-Vmin)(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 __________________ emission over emittor
 14th March 2007, 09:04 AM #6 satelitis   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, 12:37 PM #7 tjl   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! __________________ TJL
Trout
diyAudio Member

Join Date: Nov 2004
Quote:
 Originally posted by satelitis Have a look over here http://www.valveheart.com/theory/triodeSE.html You will find it very useful I quess.

Trout

 14th March 2007, 04:35 PM #9 Salas   diyAudio Chief Moderator     Join Date: Oct 2002 Location: Athens-Greece Works. (Firefox),
billr
diyAudio Member

Join Date: Dec 2003
Location: new zealand
Quote:
 Originally posted by kathodyne then the power is roughly as follows (Vmax-Vmin)(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

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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