Hi Johnny,
Thanx for your reply. I do use data sheets as a guide but I wanted a formula of sorts so I can know what to expect with new builds and to study and learn what variables affect the output power and why. I basically have a new amp I built and although I can calculate the disibation of each tube by measuring the Plate voltage and cathode current I dont know what power output to expect.
Thanx for your reply. I do use data sheets as a guide but I wanted a formula of sorts so I can know what to expect with new builds and to study and learn what variables affect the output power and why. I basically have a new amp I built and although I can calculate the disibation of each tube by measuring the Plate voltage and cathode current I dont know what power output to expect.
A quick edit......My new amp uses one half of a ecc83 as a line stage amp and a kt88 on each side SE UL. The screens are connected to the OT from tabs close to 45-50% shared winding with the anodes.(the output tranformer is not designed for distributed loading I improvised). Screen grid resistors are 1k. B+ is 395v Plate Voltage is 375v and the primary load resistance is 8k(high I know for kt88's). Bias is fixed and the idle plate current is 40ma. I would like to know how much output power to expect.
Thanx again!
Mario.
Thanx again!
Mario.
Of course:
Pout(RMS) = Vout(RMS) * Iout(RMS)
Just multiply the two bits you read off the diagram and that's it !
It would take a set of curves with a load line portraying the particular loading you are using to generate an accurate estimate of the design. The next best thing would be to perform a power output test and measure it. Lacking that capability, here is another approach:
1. Your plate B+ is 375 vdc.
2. Assume that the tube can draw that down to 60 v at full power output.
3. That represents a peak voltage of 315 volts developed across the OPT pri.
4. Divide that by 1.414 to obtain an rms value, and then multiply the result by 90% to account for OPT losses.
5. That means that around 200.5 vac will developed across the OPT winding.
6. Applying the standard power formula of E x E / R to your design yields:
200.5 x 200.5 / 8000 or 5.0 watts RMS into an external load.
You can also estimate power output from the dc power consumed by the stage as well, but for lack of any better information, 5 watts will be a pretty accurate estimate of what you can expect your design to produce.
Dave
1. Your plate B+ is 375 vdc.
2. Assume that the tube can draw that down to 60 v at full power output.
3. That represents a peak voltage of 315 volts developed across the OPT pri.
4. Divide that by 1.414 to obtain an rms value, and then multiply the result by 90% to account for OPT losses.
5. That means that around 200.5 vac will developed across the OPT winding.
6. Applying the standard power formula of E x E / R to your design yields:
200.5 x 200.5 / 8000 or 5.0 watts RMS into an external load.
You can also estimate power output from the dc power consumed by the stage as well, but for lack of any better information, 5 watts will be a pretty accurate estimate of what you can expect your design to produce.
Dave
A question for dcgillespie (Dave). The 315 V you calculated - is that the peak-to-peak voltage or the peak voltage?
I'm used to solid-state, OTL (output transformerless) designs, and those the voltage you calculated would be the peak-to-peak output voltage (not the peak voltage). I'm guessing, however, that using a transformer allows the voltage at the plate to swing both above and below the B+ voltage, doubling the output voltage swing for a given supply voltage? That would make your calculation correct, and quadruple the output power compared to a transformerless design.
In the fast-moving world of electronics, it's hard to believe the best technology we have for amplifying an electric guitar is now over a century old. Lee De Forest received his patent for what we now call the triode in 1908!
-Flieslikeabeagle
I'm used to solid-state, OTL (output transformerless) designs, and those the voltage you calculated would be the peak-to-peak output voltage (not the peak voltage). I'm guessing, however, that using a transformer allows the voltage at the plate to swing both above and below the B+ voltage, doubling the output voltage swing for a given supply voltage? That would make your calculation correct, and quadruple the output power compared to a transformerless design.
In the fast-moving world of electronics, it's hard to believe the best technology we have for amplifying an electric guitar is now over a century old. Lee De Forest received his patent for what we now call the triode in 1908!
-Flieslikeabeagle
Hi Flies --
You are exactly correct. The complete load line of a SE output stage extends both ways from the quiescent B+ operating point, in which case, it would then represent a peak to peak voltage value.
I was just trying to keep my presentation simple, and therefore discussed only the half of the load line represented by a positive grid voltage swing, and then treated it as a peak value only.
Of course as you surmised, whether you treat 1/2 of the load line as a peak value and divide by 1.414, or use the full load line's peak to peak value and divide by 2.828, it gets you to the same place, and same power output. Thanks for asking!
Dave
You are exactly correct. The complete load line of a SE output stage extends both ways from the quiescent B+ operating point, in which case, it would then represent a peak to peak voltage value.
I was just trying to keep my presentation simple, and therefore discussed only the half of the load line represented by a positive grid voltage swing, and then treated it as a peak value only.
Of course as you surmised, whether you treat 1/2 of the load line as a peak value and divide by 1.414, or use the full load line's peak to peak value and divide by 2.828, it gets you to the same place, and same power output. Thanks for asking!
Dave
They get the same result, just Ohm's Law. It does not matter whether it's pentode or triode...
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