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Old 30th June 2004, 02:10 PM   #1
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Default Load line for CF

Hi folks;

I am new to the forum (see post in introductions) and so why not jump right in.

This tube stuff is new to me and totally fascinating. I have been studying the determination of load lines using plate curves. Steve Bench has a very nice write up on loadlines in CC stages on his website which I have poured over. I got the data sheet for Svetlana's ECL32 and ran through the process a couple of times. But I am left with a question.

How does the process differ when designing a CF stage? Since the currents are essentially the same do you use the same process?
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Old 1st July 2004, 04:12 PM   #2
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Yes, you do. A cathode follower is simply a stage with 100% feedback. If you use the loadline to determine the open loop gain, you can use the feedback equation to determine the closed loop gain. Mind you, most of the time, it's sufficient to consider the gain to be unity.
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Old 1st July 2004, 07:07 PM   #3
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Since plate current equals cathode current in class 1 (no grid current) amplifiers, you can place the load in the plate circuit for your analytical example. Then you have to connect output voltage in series with input voltage, in such a way that plate remains constant and cathode (and B+ as it would be in a normal plate loaded stage) move instead... but that's complicated so we'll ignore that for now. And we can, since it doesn't have anything to do with how the tube sees things.

Say you have a 12AU7 triode running from +300V and the grid is biased at +150V from an external source (perhaps the plate of a preceeding stage) and there is a 10k cathode resistor. The first assumption you can make is that cathode voltage will be near grid voltage, and you can eyeball that there will be about 15mA flowing (150V/10,000ohms).

(If the tube were cutoff, which is impossible because we know current is flowing, there might be 20V between grid and cathode. That places the cathode at 150 + 20 = 170V, which means there are 17mA unaccounted for. So the tube cannot be cut off. Of course, cathode isn't equal to grid voltage, as the tube would then be saturated, fully on; but that requires a load line to find what voltages and currents would be present because a triode has an indefinite saturation point.)

For further analysis (and an assumption like this is often sufficient!), we'll have to refer to the plate curves. Operating point is around 300-150 = 150V plate to cathode and 15mA. You can mark this point if you want.

Now, say current drops to zero: cathode voltage will disappear. 300 - 0 = 300V will be across the tube, so mark a point at 300V, 0mA. If it were possible for the tube to turn fully on and become a short, Vp-k = 0V and so 300V are across the load resistor. 300/10,000 = 30mA, so mark 0V, 30mA. Connect these two dots with a straight line. This is the load line.

If you marked the earlier point, you'll see it falls exactly on the line. That's because we used ohm's law to find the current, so of course it's going to be on the line! It's not the exact operating point, but it's close enough.
Now, you can see it's pretty darn close to the Vg=0V curve. If you move over to the intersection, you'll see you can only get +15V swing before grid current sets in. Compared to the available -20Vg swing, which would produce almost -150V output, you can see this is a pretty trashy operating point. Because of the high current, it'll drive a line pretty well, but you don't have the voltage output available for more than line level output (so no good if you are trying to say, drive output tubes into class 2 (grid current)).
Anyway, that point at 150V on the loadline is around -1.5Vg. So if grid is 150V, cathode is 151.5V and actual current is 15.15mA. See how little difference that makes to the assumtion? If you must, you could bump the point over a pixel or two to represent 151.5V but gee, it's not worth it, and your tube is less accurate (to the data) than that!

Within the available swing of +/-1.5Vg, the extremes of 0Vg and -3Vg land at approx. 132V and 170V, for +18 and -20V. Average is 18+20 / 2 = 19V (peak) for 1.5V peak input, so gain is 13. I don't remember the equation for distortion but it'll probably be like, 3% 2nd harmonic.

Now, at last, we can apply NFB. Output voltage appears 'under' the signal, hence Vin = Vo + Vtube, where Vtube is the grid voltage. We said a swing of 19V for 1.5V in, so Vin = 19 + 1.5 = 20.5V. This is for the same output voltage of 19V, so what happened? NFB happened, and it reduced our gain (to less than unity!). But it pulled distortion and Zo with it. How much? By the change in input voltage: 20.5 / 1.5 = a factor of 13.7, reducing gain and distortions. If 2nd H was 3%, it is now .22%.

There ya have it, that's pretty much it. Really should edit this monster of a post though, it's far longer than necessary... oh well!

Tim
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Old 1st July 2004, 11:52 PM   #4
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I appreciate the responses. Sch3mat1c, I will need to study your reply in detail (time to warm up the printer )
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Old 2nd July 2004, 05:13 AM   #5
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Hehe, yeah. If you need to find it easier it's on my website --v
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