Does RMS current matter for Class A

[Mike K]

I know we talk about a Class A amplifier having a constant average current throughout its operation, and of course that makes sense. But I was digging through some stuff and came upon this:



As you might surmise, RMS current at load is not equal to the RMS current at idle. It's 1.22x higher.

Does this have any physical meaning? Or is just a non-practical use of RMS?

[Helmholtz]

Result is correct.
What's the purpose?

E.g., plate RMS current heats the OT primary wire.

[Mike K]

I think it doesn't mean anything. RMS seems to only make sense if you have a sign change but still a source of energy that can do work.

[loudthud]

In the example, the Average current is just I_q, but the Power consumed by the tube and the load (via the OT) is B+ times (I_rms). This is higher than just the Average current times B+ because when the current is near I_p, B+ times the Instantaneous current is Four times the Power when current is just I_q.

[uneumann]

This was a special case where Iq is exactly the AC amplitude.... Wikipedia gives a general form for AC superimposed on DC.
https://en.wikipedia.org/wiki/Root_mean_square Which also shows LT's point.

A special case of RMS of waveform combinations is:^[6]

[Helmholtz]

... but the Power consumed by the tube and the load (via the OT) is B+ times (I_rms). This is higher than just the Average current times B+ because when the current is near I_p, B+ times the Instantaneous current is Four times the Power when current is just I_q.

I have to disagree.
Power consumed by the tube + load is average DC cathode current times B+ (multiply by 2 for total with push-pull).
With cathode biased class A PP using a common cathode resistor, the cathode voltage is essentially constant as is the sum of the cathode currents (i.e. no AC component with matched tubes).
Terman states that in this case no bypass is required.
So here total power consumption is easily calculated from total cathode current times B+.

To get real power, multiplying RMS current by voltage only works if voltage and current have the same waveshape and are in phase (in other words if both are DC or the load is purely resistive with AC).
In all other cases a power factor <1 needs to be included.

[loudthud]

I have to disagree.
Power consumed by the tube + load is average DC cathode current times B+ (multiply by 2 for total with push-pull).

Consider the following. A 1 Ohm resistor has 1 Volt across it continuously. The dissipation is 1 Watt. What is the dissipation if the Voltage is 2 Volts for half the time and zero for the other half of the time ? When the Voltage is 2 volts, the dissipation is 4 Watts. Average that out over the time when dissipation is zero and you get 2 Watts. The Average Voltage is still 1 Volt, but the dissipation doubled. RMS Voltage would be 1.414 Volts so RMS Voltage squared times resistance is 2 watts.

Note: In a 1 Ohm resistor, current and Voltage waveforms are the same.

What is the dissipation when the Voltage is 10 Volts for one tenth of the time and zero the rest of the time ? The average Voltage is still 1 Volt. When the Voltage is 10V, the power is 100W. Spread that out over the whole time period and the average power is 10 Watts. The RMS Voltage is 3.1622 (square root of 10) Volts.

post 44 and post 52
https://music-electronics-forum.com/...imations/page3

post 70
https://music-electronics-forum.com/...n/49229-/page5

Kevin O'Connor takes a bath in post 64 and 66
https://music-electronics-forum.com/...y-so-big/page5

[Mike K]

Perhaps we're missing the point here...

RMS is a time averaging method. The example shown in the OP is a special case, but a very specific special case that was referenced: Full signal of symmetrically biased Class A operation.

The question is does this RMS result have any physical meaning when we compare it to the RMS output of the amplifier or the RMS power supplied by the PT? I made this calculation long ago, but didn't write any notes with it. I believe I was trying to determine if one were to size components based on average current or RMS current. I've since convinced myself that RMS makes sense when we have an AC signal that changes sign, but average is more appropriate when we don't. We could still calculate a RMS current but my thought is the average current of the amplifier corresponds to the RMS current of the PT (because the PT is nominally a symmetric signal about zero, but still can do work in either direction relative to ground).

[Helmholtz]

The general formula for electrical power is P(t) = V(t) * I(t).
In our special case we get P(t) = (B+) * Ib (1+ sin (w*t)) = [(B+) * Ib]+ [(B+) * Ib * sin (w*t)] , with Ib being the idle current.

As the average of a sine is zero, average power is given as Pav = Ib * B+.
No squaring or RMS value needed here.
QED.

The OT wire OTOH gets heated by the copper resistance, so here the total RMS current matters.

[Helmholtz]

.
RMS is a time averaging method. .

RMS value is not the time average.
Rather it's the DC power equivalent of an AC signal at a resistor.
RMS is always positive (due to squaring before averaging), while the time average value can be negative as well.