Amp Power Output Calculations
PP18 Power Calculations
Using the Aiken amp info from the Q+A page here:
It is possible to check accurately, the power output of your amp with a dual beam oscilloscope, and knowing the speaker impedance you are connecting to. The rest comes from calculation as stated in the above article.
The small differences from the article in these following examples will come from not using a fixed 16 or 8 ohm fixed resistance at 400Hz, but a 1 kHz signal across a 16 ohm 66W speaker. This signal does not necessarily drive the amp at its maximum current output.
For the PP18 example, I connected one channel to the anode of the LTP pair, as this is where distortion is generated at volume, so I set the volume level at just below the point the sine wave distortion starts to occur. This is important to understand, as there is a big difference between the power used in an amp at distortion and a pure sine wave as explained in that article, as the tubes may be on near full saturated conduction for each half cycle with a square wave for example, so pass much more current for much longer than they would for a non distorted signal.
Channel 2 is connected across the speaker load. Note that these signals are peak to peak values.
For the PP18 I measured 40Vpp at the LTP anodes at just below distortion, and 48Vpp at the 16 ohm speaker jack connection, with my new global NFB pot set at minimum NFB.
The power consumed in the speaker is calculated using P = Vsquared / R
This equation works for RMS signal values only, so first convert the speaker peak to peak values to one peak, by halving it then divide this by the square root of 2 or 1.414 to give the peak RMS voltage:
48V/2 = 24Vpeak.
RMS peak = 24/root2 = 24V / 1.414 = 16.97V RMS.
Using P = Vrms squared / R = 16.97V x 16.97V / R = 288 / 16 ohms = 18 Watts exactly.
Seems the name PP18 is chosen well…
In my amps case, I also took readings for maximum NFB and got only 5Vpp undistorted at the LTP anodes at max volume, and 2.4Vpp at the speaker. This is a giant reduction of 40V/5V = 8 times – heavy attenuation.
With this pot control up full it drops the power output to a tiny:
2.4V/2 = 1.2Vpeak = 0.848Vrms.
P = 0.848V x 0.848V / 16 ohm = 0.72 / 16 = 0.045 Watts.
This is 400 times less power at full volume than with no NFB at full volume. This is still quite audible which shows again the remarkable but very non – linear relationship in human ear sensitivity to sound energy.
Another point of academia here is the turns ratio relationship of the output transformer.
At max power I read the voltage at the power tube anode also = 640Vpp when the OT sec 16 ohm tap reading was 48Vpp, so a voltage ratio of 640V/48V = 13.33:1 from primary to secondary, or a gain ratio (attenuation) of 1/13.33 = 0.075 from primary to secondary.
The quiescent HT for this amp with no signal sits at 347V DC. The cathode is biased at 16.5V.
This means that at full volume, the voltage swing across the OT primary is a high proportion (320V/347V) of the available DC, but just for a short time. This is partly why there can be more than double the power output for a Push Pull design over a Single Ended design:
“Not only this, but Class AB also allows us to operate the valves at much higher anode voltages and the load line can actually go beyond the maximum power dissipation curve. In this way the peak power level can actually exceed the combined maximum anode dissipation power of the two valves!”
Each tube is only conducting for half the cycle time but can be at higher power. One switches off when the other is on full.
If you know your HT voltage and OT primary impedance, a very rough power estimate can be made for a PP amp using Blencowe’s:
P = 2 * (HT-50)^2 / Rload
Which in my case gives 2 x (297) ^2 / 8000 ohms OT primary = 88209 x 2 / 8000 = 176418 / 8000 = 22.05225W. Not 18W but fairly close.
Each tube is using about 9W RMS each for each half cycle, which for a JJ 6V6 in a Push Pull design is within spec:
Typical characteristic: Class A1
singl tube Push-Pull
Ua = 250 V 250 V
Ug2 = 250 V 250 V
Ug1 = -12,5 V -15 V
Ia = 45 mA 70 mA
Ig2 = 5 mA 13 mA
Ra = 50 k??–
Ra-a = – 10 k?
N = 4,5 W 10W
From Blencowe p190, the turns ratio of an OT is related to the square root of the (primary impedance / the secondary tap impedance).
This Danbury OT has a primary impedance of 8000 ohms, so the turns ratio to the 16 ohm secondary is:
square root (8000 / 16) = SR 500 = 22.36 times less OT sec turns
The ratio at the 8 ohm tap would be:
square root (8000 / 8) = SR 1000 = 31.62 times less OT sec turns
This ratio value between taps is 31.62/22.36 = 1.414 = SR of 2 times larger.
This means that the voltage found at the 8 ohm tap would be SR of 2 times less, or about 48V / 1.414 = 34V.
This is approximately what I measured at the 8 ohm tap though it was not under load.
Looking at this in terms of current, then more current would be pulled from the 8 ohm tap with an 8 ohm speaker attached by this SR of 2 proportional amount as the impedance is now halved for the same frequency but power is constant.
Doing the calculation for these values should show that the power used remains the same if I had connected an 8 ohm speaker instead. The current increases for this SR 2 proportional drop in voltage.
34Vpp = 17Vpeak = 12Vrms.
P =Vsquared / R = 144 / 8 ohms = 18.0625 Watts
Just for completion and get an idea how much current actually flows in a speaker, the power equation can be re-arranged (from Ohms Law, V=IR) and using P = I squared x R for both taps.
P= 18W, so I squared = 18W/16 ohms = 1.12 so the current flow is SR of 1.12 = 1.067 Amps at the 16 ohm tap, and SR of 18W/8 = 1.5 Amps at the 8 ohm tap.
If I had a 4 ohm tap it would be SR of 18W/4 = 2.12 Amps.
Just to point out here also how much difference there can be to say a big 100W JMP Marshall with a 4 ohm cab, this would be about:
SR of 100W / 4 ohms = 5 Amps.
Maggie366 Power Calculations
This was a bit inaccurate because of undefined scope traces. I have an EL34 in this amp at present.
This amp design starts distorting at the second triode of the 12AX7 which occurred at about 230Vpp of the HT of 354V DC.
The 16 ohm tap voltage measured 35Vpp. This gives an RMS of 17.5 / 1.414 = 12.4V
P = 12.4 x 12.4 / 16 ohms = 9.57W
This is well within the anode dissipation for this EL34 tube of course. I just prefer the tone of EL34s to 6V6s.
Ua = 250 V
Ug3 = 0 V
Ug2 = 265 V
Ug1 = -10 V; -13,5 V
-13,5 V; -16,5 V
Ia = 100 mA
Ig2 = 14,9 mA
S = 11 mA/V
Ri = 15 k?
µg2/g1 = 11
Iaz (Ug1 = -30 V) < 7 mA
Ua0 = 2000 V
Ua = 800 V
Wa(max) = 25 W
Ug20 = 800 V
Ug2 = 450 V
Wg2(max) = 8 W
Ik = 150 mA
Uk/f = 100 V
Rk/f = 20 k?
So, about half the power of the PP18 as expected for a single valve.
Another important difference between these designs is that an SE amp is only 50% efficient
as it conducts even when no signal is being amplified, unlike Push Pulls that draw minimal quiescent current and conduct fully when amplifying so are closer to 80% efficiency.
If you don’t have a scope, you can probably get a rough idea what power your amp uses just by taking the RMS voltage reading across the speaker with your DDM – send a sine tone in and take the AC voltage reading at the speaker. When distortion occurs with a sine tone you may hear a different tone appear with the original – a harmonic –indicating the onset of distortion but you will probably not be able to hear distortion. You can at least get a max voltage figure for full volume with distortion as below.
Use the RMS equation above with your speaker value.
Marshall Mercury 2060
Just for interest, I’ll try that method with the Mercury as its speaker wires are easily accessible, and I know to expect about 5W from its single EL84 power tube from web info.
It was not possible to hear the sine distortion, but knowing the position of the volume control roughly, where distortion comes in on guitar for this amp, it was around a 4Vpeak RMS value across the speaker that the DDM read. I know this is RMS because this DDM reads AC mains 240V RMS which is a peak value NOT a peak to peak value (that would be 480Vpp RMS).
So P = 16 / 4 = 4W very roughly.
From my prior Post on this amp the speaker appears to be a:
Celestion (Rola) G12 “Small” Green Back 4Ohm. Model: G12 | Date Code: BJ27X Dates Feb 1976, | Frame Code: T1692 | Cone Code: 1777?
It measures 4.4 Ohms DC resistance, so would be a 4 Ohm AC impedance speaker.
Or as mine looks more grey – possibly :
Four Creamback G12H 55hz speakers inside a 1973 Marshall 1982B cab
“By 1972-1973, they started to use cream or grey-ish colored plastic back covers instead of the famous green ones. A bit off-putting, maybe (unless you prefer cream or don’t care about the colour – like me), but it didn’t change the sound at all; at least not in these early Creambacks.”
So, P = V squared /R = 16 / 4 ohms = 4W
This sounds about right to me and is well within the max anode dissipation limits of an EL84:
Class A1 amplifier:
Ua = 250 V
Ug2 = 250 V
Rk = 135 ?
Ia = 48 mA
Ig2 = 5,5 mA
Ra = 5,2 k?
Ug1eff (50mW) = 0,3 V
Ug1eff(N) = 4,3 V
N (10%)1) = 5,7 W
N 2) = 6 W
1) Ug1 fest fixed grid bias
2) Ig1 +0,3 ?A
Ua = 300 V
Wa = 12 W
Ug2 = 300 V
Wg2 = 2 W
Ug1 = -100 V
Ik = 65 mA
Rg1 = 1 M??for automatic bias
Rg1 = 0,3 M??for fixed bias
Uk/f = 100 VN 2) = 6 W
These 2060s are very under-estimated amps and I don’t think they deserve that bad press I read about them being in the top 5 most hated Marshalls. On the contrary, for a small practise amp they have character, a Marshall history, a very good rock/blues tone and overdrive, a cutting high end if you need it, a tremolo and are damn loud for their valve size, with a good quality 25W Celestion speaker if it’s still the original. Most importantly for the tone – they come in bright orange or red!!
I have a theory on the misleading serial number badge wattage of 25W on these amps too – I think someone originally misread an EL34 spec. sheet instead of an EL84, OR from pics above, someone just read the speaker label at 25W?:
Ua0 = 2000 V
Ua = 800 V
Wa(max) = 25 W
My speaker label is missing unfortunately but I’ll assume its 25 Watts from the research above.