Calculating SE Output Question

[Trout]

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]

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.

But How do I determine my final output wattage?

That'll have to be figured from the loadline.

 

[satelitis]

Re: Re: Calculating SE Output Question

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.

 

[billr]

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

 

[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

 

[satelitis]

Have a look over here http://www.valveheart.com/theory/triodeSE.html

You will find it very useful I quess.

 

[tjl]

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!

 

[Trout]

Have a look over here http://www.valveheart.com/theory/triodeSE.html

You will find it very useful I quess.

Dead Link?

Trout

 

[Salas]

Works. (Firefox),

 

[billr]

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?