Current draw limitations in the AC load line?

[Bergheim]

Hey all

I'm trying to get a better understanding of the current draw in the AC load line. Below is the AC and DC load lines of a 12AT7 gain stage with 18k plate and 470R cathode, fed from a B+ of 310V. The AC load is a 600 ohm (@ 1kHz) reverb tank. First of all, please ignore the fact that the AC load line grossly exceeds the max dissipation line.
According to Merlin and a few others, the extra current draw that the AC load represents is sourced from the coupling cap, but I don't understand to what extent this applies to. As far as I can see, Ohm's law is working perfectly fine as long as the AC load line crosses the 0V grid line at or below the max current that the plate resistor will pass at full B+ voltage: I = U/R = 310/18k = 17.2mA.

Now, what happens when the AC load line crosses the 0V grid line at 24.5mA? The required voltage for the 18k plate resistor to pass 24.5mA is 18k*24.5 = 441 volts. The theoretical plate resistor's resistance at the point where the AC line crosses the 0V line equals U/I = (310-210)V/24.5mA = 100V/24.5mA = ~4.1k. None of this makes any sense to me, since all current in the circuit must pass through the plate resistor, and the B+ is only 310V.
Will the tube run out of headroom and clip the output at 17.2mA no matter what the grid voltage is?

[mhuss]

It'll actually start to clip before you get to 17ma, because (unlike a BJT), a tube has significant internal resistance even when in saturation. This is part of the reason why preamp tubes are normally not biased so that the plate is at 1/2 of the supply voltage - more like 2/3, as a compromise between saturation and cutoff.

On a characteristic curves chart, only some subset of all the values displayed are possible given a fixed set of external factors such as load and supply voltage.

[Bergheim]

Yes, the internal resistance is easy enough to see from the slope of the grid curves. If the tube had no internal resistance when the grid hit 0V, that grid line would've rosen vertically from the 0Va point, and the plate current swing would only be limited by the plate resistor. Looking at the DC LL only, the 12AT7's plate with a 18k plate load and 310V B+ can never be pulled to a lower voltage than ~120V (@ ~11mA), because of the internal resistance.

In normal preamp circuits, we're mostly after voltage swing rather than current swing. Thus, the current capacity of the circuits is something we rarely have to deal with. In the following graph, I've drawn an orange line at 17.2mA that represents the maximum current through the 18k plate resistor with a B+ of 310V. As shown in the 600 ohm AC LL (green line), the plate current is supposed to rise to 24.5mA when the grid hits 0V (circled in black). If the orange "plate resistor current limit" line is legit, it clearly shows that 24.5mA can never be reached, and that the plate resistor runs out of headroom @ 17.2mA when the grid is pulled below ~0.7V.

Does this make sense or am I completely lost?

[mhuss]

Yes, your analysis is correct. In your second example, the entire region above the orange line cannot reached given those constraints.

[blackeye]

Hi, first post on this forum. Wanted to explore this question a little deeper.

IMHO, if all current has to pass through the load resistor, then why is the AC load line any different then the DC load line? I think the whole point of an AC load line is that now, with an AC signal, you have two paths for current to flow. One is through the load resistor, and the other is through the coupling cap and whatever input impedance the following stage shows. In this case, most of the AC current is actually coming from that small 600-ohm load.

I think that the valve is happy to conduct 24mA of peak AC current, until you take it too far above that max wattage line and start melting.