• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

plotting load-line for OTL amp?

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Following on from my previous loadline thread, I have another question:

How do I plot a loadline for an OTL amp, such as the one shown in http://headwize.com/projects/showfile.php?file=waarde1_prj.htm ????

Would I use the value of R5, which is 4.7K in this example? Or would I use the impedence of the headphones being driven (32R in my case)? If I were to pick my choice from above, then I'd go for the 4.7K, but I could be totally wrong.


R5 & C1 is a filtering/decoupling network for the first stage.

Since the output tube is direct coupled to the first stage. To plot a line for the output stage:

1st Plot a DC loadline for the first stage. B+/(R4+R5+R6)

2nd Find operating plate voltage for the fist stage. This will also be the grid voltage for the output stage.

3rd, Plot a DC loadline for the output stage: Your line will be B+/R6.

4th, find operating point of the output stage: Look at the loadline and find a point where the Vg-c + Vp = B+. But remember that you have a voltage on the grid. So that means if you find your point to be at Vg-c to be -10V and you have a grid voltage of 50V that means the cathode voltage will be 60V

5th find an AC REFERENCE loadline of output stage: Your AC load is R6//R7//Headphone impedance. Use any voltage/AC load to plot a reference AC loadline.

6th, Make the AC loadline. To draw an actual AC loadline, make a line parallel to your reference, except do it so that it intersects the DC operating point.

Note: that the real life AC behavior will be more linear than your plotted line because the stage is a cathode follower, which by nature uses %100 negative feedback. The AC loadline is just used to give you an idea of optimal operating point and how much output voltage to expect.

B+ = Positive supply voltage
Vg-c = Voltage between the grid and cathode
Vp = Plate voltage
// = Parallel

Check out the links on the thread Online tube learning for newbie’s they may give you a better idea.

OK does this sound right?

1. Load line for E88CC would be drawn between 150V at Ia = 0 and 0V at Ia=17mA.
2. The operating point for cathode resistor of 820R would be at -3.2V @ 4mA. I got this figure by plotting Ia for each value of Vg/820. Where this line (shallow curve) intersects with the loadlne is the operating point. Vp = 115V, Ia = 4mA
3. OP tube loadline is plotted with Ia = 0 at 150V and Ia = 45mA (150/3K3) at 0V.
4. OK, so it seems that an operating point of -50V with a Vp of 100V looks good, but how do I calculate the voltage on the grid? Is this value the same as the Vp for the input tube? If so then it is around 115V, so the cathode voltage would actually be 165V??
5. AC reference loadline 3K3//10K//32 = 31.5R or 32R

So, for 2V p-p on the input, I would get a voltage swing from 90V to 140 on the plate of tube 1, which is 50V p-p. This would give something like voltage 70V p-p on the output? In other words a total amplification factor of 35???

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