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Amp Wattage Measurement - Approximations

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  • #16
    Just decided to read some of the comments under the YouTube video I sent the link to. Sounds like there's a bit of controversy swirling around regarding the video. I don't know him from Adam, so I can't say one way or the other.

    Ok, just thought I could take the easy way out, but I see it's not that simple to measure wattage, but not all that complicated if you are serious about it.

    What I do generally is setup a DB meter one meter on access from my amps and let it rip to record the maximum SPL peaks. But that also has serious flaws in that amps with an upper mid focus have big SPL number, but sometimes very little bass, so it's a mixed bag measuring sound pressure levels as well.

    I've found from gigging with different amps and the same moderately loud drummer, that I need about 105db of midrange to just start to poke a hole in the wall of sound, and have my guitar heard in a solo. A 10 watt amp with the right speaker can do this, if you dime it and the speaker is efficient, but it's easier with a 20 + watt twin speaker amp, for sure.

    Nothing is that simple in this world !
    " Things change, not always for the better. " - Leo_Gnardo

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    • #17
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      • #18
        *IF* you're wanting to 'guess-ti-mate' an approximate output power for a Class-AB push-pull amp by simply looking at its schematic, here's an equation:

        Po ≈ (%)*(Zoo/4)*(gm*Vg)^2

        where:
        Po = estimated output power, watts (avg)
        % = Zo loading effect on tube rp, dimensionless (typical value 0.85-0.9-.95)...see below
        Zoo = OT plate-to-plate impedance, ohms
        gm = output tube average transconductance
        Vg = output tube BIAS voltage, absolute value


        Actual (%) formula: (%) = (rp/(rp+Zo))^2...note: Zo versus Zoo here; rp = nominal output tube plate resistance, ohms.
        Last edited by Old Tele man; 06-19-2019, 04:24 PM.
        ...and the Devil said: "...yes, but it's a DRY heat!"

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        • #19
          Originally posted by Old Tele man View Post
          *IF* you're wanting to 'guess-ti-mate' an approximate output power for a Class-AB push-pull amp by simply looking at its schematic, here's an equation:

          Po ≈ (%)*(Zoo/4)*(gm*Vg)^2
          Doesn't that give peak power as gm*Vg is peak current?
          Average power will be half of that.

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          • #20
            Originally posted by Dave H View Post
            Doesn't that give peak power as gm*Vg is peak current?
            Average power will be half of that.
            No, as Vg is BIAS (DC voltage), not Vgg which is AC peak-to-peak, which I believe you are thinking about.
            ...and the Devil said: "...yes, but it's a DRY heat!"

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            • #21
              One of those commenters is a member here. We know the Great Bandmaster Saga well. That's what turned me off permanently to GW...

              Though I remember poring over the schematics & Trainwreck Pages in his book.

              Justin
              "Wow it's red! That doesn't look like the standard Marshall red. It's more like hooker lipstick/clown nose/poodle pecker red." - Chuck H. -
              "Of course that means playing **LOUD** , best but useless solution to modern sissy snowflake players." - J.M. Fahey -
              "All I ever managed to do with that amp was... kill small rodents within a 50 yard radius of my practice building." - Tone Meister -

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              • #22
                Originally posted by Old Tele man View Post
                No, as Vg is BIAS (DC voltage), not Vgg which is AC peak-to-peak, which I believe you are thinking about.
                The bias voltage is the peak voltage isn't it? That's what I was thinking about. Peak to peak is twice the bias voltage.

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                • #23
                  For power calculation you want RMS current. (gm*Vg) gives peak plate current or amplitude. (gm*Vg)^2 = 2*Irms^2.
                  - Own Opinions Only -

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                  • #24
                    Originally posted by Helmholtz View Post
                    For power calculation you want RMS current. (gm*Vg) gives peak plate current
                    Yes, that's what I thought. When I use the approximation formula on my 15W EL84 amp it give the output as 33W.

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                    • #25
                      First, there is no such thing as WATT(RMS)!

                      Why? Because RMS is just the mathematical-equivalency operation (0.707*pk) that's applied to the peak value of sinusoidal VOLTS and/or AMPS to convert them to their equivalent AVERAGE values:

                      SINUSOID(avg) = 0.707(rms)*SINSUSOID(pk).

                      POWER(average) = VOLT(rms)*AMP(rms)...where RMS*SINUSOID(pk) = DC.heating(average).

                      When working with sinusoidal volt and amp waveforms: WATTS(avg) = VOLT(rms)*AMP(rms)...where AVG watts is the "time-averaged" DC-heating equivalent of an (assumed) sinusoidal waveform.
                      ...and the Devil said: "...yes, but it's a DRY heat!"

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                      • #26
                        Originally posted by Old Tele man View Post
                        First, there is no such thing as WATT(RMS)!
                        True, that's why I said average power in post #19

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                        • #27
                          Originally posted by Old Tele man View Post
                          First, there is no such thing as WATT(RMS)!

                          Why? Because RMS is just the mathematical-equivalency operation (0.707*pk) that's applied to the peak value of sinusoidal VOLTS and/or AMPS to convert them to their equivalent AVERAGE values:

                          SINUSOID(avg) = 0.707(rms)*SINSUSOID(pk).

                          POWER(average) = VOLT(rms)*AMP(rms)...where RMS*SINUSOID(pk) = DC.heating(average).

                          When working with sinusoidal volt and amp waveforms: WATTS(avg) = VOLT(rms)*AMP(rms)...where AVG watts is the "time-averaged" DC-heating equivalent of an (assumed) sinusoidal waveform.
                          The average (or mean value) value of a pure sine is zero. RMS is not average. The unit Watt (W) is defined as the product of the units Ampere (A) and the unit Volt (V).

                          The output power of an amp is usually measured using a load resistor. This reflects to a resistive (real) plate load Rp (=Zoo/4). Consequently Rp*Irms^2 gives the real (as opposed to apparent) power. Nothing wrong with calling the result RMS power in a steady state situation.
                          Anyway your formula gives peak power which corresponds to twice the value you would get from (Vout^2)/Rload (Vout being RMS output voltage).
                          Last edited by Helmholtz; 06-20-2019, 07:18 PM.
                          - Own Opinions Only -

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                          • #28
                            Originally posted by Helmholtz View Post
                            Rp*Irms^2 gives the real (as opposed to apparent) power. Nothing wrong with calling the result RMS power in a steady state situation.
                            Are you sure, isn't that average power?
                            RMS power would be the RMS value of the power waveform which isn't useful.

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                            • #29
                              Class-AB1 PP amps do not conduct 360-degrees, only about 190-205-degrees, or more generally 'slightly more' than 50% of a FULL sinusoid, so referencing a 'pure' sinewave is only valid prior to the PI and output tubes...or, at the load.

                              The ONLY place the two quasi-halves of the full pre-PI sinewave become a FULL 360-degree sinusoid again is *inside* the OT -- specifically in the secondary (output) winding(s) -- because anything prior to that is (basically) pulsing alternating ± half waves (PI, output tubes, each primary half of OT).
                              Last edited by Old Tele man; 06-20-2019, 11:19 PM.
                              ...and the Devil said: "...yes, but it's a DRY heat!"

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                              • #30
                                Originally posted by Dave H View Post
                                Are you sure, isn't that average power?
                                RMS power would be the RMS value of the power waveform which isn't useful.
                                You're completely right. It's correctly mean power or average power. But I am used to call Vrms*Irms "effective power" (which would translate into RMS power), knowing that it's actually apparent power and that the RMS value of time-varying power is of no use. I know it's sloppy.
                                Last edited by Helmholtz; 06-20-2019, 09:15 PM.
                                - Own Opinions Only -

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