I'm assuming that, since we now know how many millivolts the amp is delivering into a 33 Ohm resistive load, we can then calculate how much power is being delivered into the two 33Ohm load resistors.
That's good to know, so, there's no danger of the cathodes undergoing cathode-poisoning then? or wearing-out prematurely?
Nope. You are fine. See the link below from an expert (a term which I do not use lightly) who posts as MerlinB on this forum.
https://www.valvewizard.co.uk/standby.html
https://www.valvewizard.co.uk/standby.html
Okay, I'm going to see if I can calculate how much power my OTL Tube Headphone Amplifier is delivering into a 33 Ohm resistive load, I know that the signal level being delivered into the load is 1.10V Peak To Peak as shown in the screenshot, we also know that the peak voltage is .585V or 585 millivolts, first thing I need to do is to convert 1.10V Peak To Peak into Vrms, or the Vpeak into Vrms, we do that by multiplying the Vpeak voltage by .707 (because it's a sinewave) if I remember my electronics theory correctly, so:
Vrms = Vpeak X .707
= .585V X .707
= 0.413595Vrms
Or roughly 0.414 Vrms if we round it up.
I did a quick search on Google and this seems to be the equation to use to calculate the AC power into a resistive load:
To calculate AC power into a resistive load, use the formula P = Vrms * Irms where "P" is power in watts, "Vrms" is the root mean square (RMS) voltage, and "Irms" is the RMS current; essentially, it's the same formula as calculating DC power, as in a purely resistive circuit, voltage and current are in phase, meaning there's no phase shift and the power factor is 1.
So, I need to first calculate the RMS current, the load is purely resistive:
According to Ohms Law:
I (Current) = E (Volts)/R (Resistance)
So
Irms = Vrms/R
= 0.414/33 Ohms
= 0.0125454545454545
Or about 12mA RMS
So
P = Vrms X Irms
= 0.414 X 0.012
= 4.968 milliwatts
According to my calculations, my OTL Tube Headphone Amplifier is capable of delivering 4.968 milliwatts RMS into a 33 Ohm resistive load.
Vrms = Vpeak X .707
= .585V X .707
= 0.413595Vrms
Or roughly 0.414 Vrms if we round it up.
I did a quick search on Google and this seems to be the equation to use to calculate the AC power into a resistive load:
To calculate AC power into a resistive load, use the formula P = Vrms * Irms where "P" is power in watts, "Vrms" is the root mean square (RMS) voltage, and "Irms" is the RMS current; essentially, it's the same formula as calculating DC power, as in a purely resistive circuit, voltage and current are in phase, meaning there's no phase shift and the power factor is 1.
So, I need to first calculate the RMS current, the load is purely resistive:
According to Ohms Law:
I (Current) = E (Volts)/R (Resistance)
So
Irms = Vrms/R
= 0.414/33 Ohms
= 0.0125454545454545
Or about 12mA RMS
So
P = Vrms X Irms
= 0.414 X 0.012
= 4.968 milliwatts
According to my calculations, my OTL Tube Headphone Amplifier is capable of delivering 4.968 milliwatts RMS into a 33 Ohm resistive load.
I'm guessing that, since my pair of Audio Technica ATH-M50X headphones are rated at 38 Ohms, the OTL Tube Headphone Amplifier I built is most likely capable of putting at least 5.0 milliwatts into it due to the lighter loading it presents to the amplifier.
I just figured out how I can do a more realistic test of the maximum power output of my OTL Tube Headphone Amp, I could repeat the previous test I did, but instead of plugging in the dummy load with the two 33 Ohm resistors, plug my Audio Technica ATH-M50X headphones in instead, and connect the scope probes to the solder tags of the Headphone Out Socket, something I might do later on today, will be interesting to see if I get different results.