Well, I have been building tube amplifiers for many years and most of the time I could match the theory and actual design/measurements.
For know I am in doubt if my understanding is correct since measurements show something different.
The configuration is based on an e88cc/6922 with a plate- and a cathode resistor, values resp 1k5 and 200.
I changed Rc from 200 to 1k and Rc from 3k and did some measurements but the (for me) strange thing is that I always thought that high Ia/ClassA setting would result in the lowest THD...since your'e moving up in the most linear part of the curves where the distances between the curves along the load line are more equal than at a lower position.
Measurements: 1..2mA...highest plate voltage...lowest THD....
I can imagine that at a higher Ia with a lower Rc the voltage on the cathode is low (2..3V) and that the input signal might be closer to the 2..3V limit than that it is the case with the 2mA and 6V on the cathode (more headroom).
Had measurements, simulation sheets and...yes...curves drawing lines.
What is the correct thinking/approach/explanation?
Arno
For know I am in doubt if my understanding is correct since measurements show something different.
The configuration is based on an e88cc/6922 with a plate- and a cathode resistor, values resp 1k5 and 200.
I changed Rc from 200 to 1k and Rc from 3k and did some measurements but the (for me) strange thing is that I always thought that high Ia/ClassA setting would result in the lowest THD...since your'e moving up in the most linear part of the curves where the distances between the curves along the load line are more equal than at a lower position.
Measurements: 1..2mA...highest plate voltage...lowest THD....
I can imagine that at a higher Ia with a lower Rc the voltage on the cathode is low (2..3V) and that the input signal might be closer to the 2..3V limit than that it is the case with the 2mA and 6V on the cathode (more headroom).
Had measurements, simulation sheets and...yes...curves drawing lines.
What is the correct thinking/approach/explanation?
Arno
With a load that low and plate current similarly low, you're working with almost a vertical load line. Gain follows the equation Vin{mu*Rl/(Rl + rp)}, with rp modified by (mu + 1)*Rk. But you have to determine what the rp and mu are at your operating point- they're not constants.
3k line is really, really steep. Especially when you consider that the plate resistance is way higher than that in your circuit.
Hm...maybe I should add that the two triode-halfs are put in parallel...so when putting the loadlines to the curves (of 1 triode) de value is doubled....sorry for not adding this before
Still, you have ~40k of plate resistance and 3k of plate load. That's fine if you're trying for an effects box.
Why don't you start with defining what you want to do and choosing an appropriate operating point? Then you can derive the correct values.
First of all it's really not possible to say anything definite without knowing the B+ voltage. BUT - if you were going to generalize, for highest linearity you need a horizontal load line, i.e. current source, and the current needs to be pushed well over the bottom curved area of the curves into the part of the curves where the varied Vg1 makes parallel and equidistant lines.
Also, In your configuration the chosen current has nothing to do directly with class A. You are building a single-ended amplification stage and it is really by definition working in class A - to a point where clipping occurs. Now, where this happens depends on how you chose your operating point and load, which in turn depends on what you want to accomplish. With decreasing B+ and required swing, your choice of operating point shrinks ever more, until you reach a combination where a single tube or a tube of that type will not do.
So to sum it up: you need to define what you want, then you can go into how you are going to get it, using the specs and curves, not audiophool 'lore' which gives predefined 'good tube types' and values for Rk and Rc...
Getting back the known track now. The current source and upper part of the curves is obvious. Apparently my misperception came in a split-second where I felt that high current is lowest THD...and this is not as straight forward. But, in case the load line is put where the application needs it....then a higher current/voltage might shift the operation point up the load line, ending up in an area where the spacing between the curves is better. On the other hand it might be that the headroom is lower since you're moving towards the -Vg=0 line....is the latter correct?
Owwwwwwwww......after a good sleep and a decent cup of coffee I took the curves, put in some load lines and wondered what must have got into me when putting in 1k5...🙄🙄😱😀😱 indeed vertical. Bias at -2V..-3V with 10k..20k starts to look more like something that's useful (and some knowledge does seem to get rusty)
The temptation of putting a semicon/FET current source in remains 🙄....But...I would favour to stick with the traditional parts if possible for this configuration.
This feels like a "get up and smell the coffee" moment...thanks for the wake-up-call..🙂
The temptation of putting a semicon/FET current source in remains 🙄....But...I would favour to stick with the traditional parts if possible for this configuration.
This feels like a "get up and smell the coffee" moment...thanks for the wake-up-call..🙂
The major distortion mechanism in triodes is non-linearity in ra. To minimise this you need to load ra as lightly as possible which is why mu followers and CCS anode loads are popular. To get into a low distortion region you should aim for the anode load to be at least 5 times ra at the operating point
Cheers
Ian
Cheers
Ian
Hi Ian,
This stage will have it's output from the cathode since I do need the low impedance (and no extra amplification)
Regards,
Arno
This stage will have it's output from the cathode since I do need the low impedance (and no extra amplification)
Regards,
Arno
OK, so effectively you are using it as a cathode follower. So you really do not need the anode resistor except to manage the anode voltage. For minimum distortion you should decouple the anode. If you do not do this then you do not get the full benefits of the negative feedback in a cathode follower whicj means you will have higher distortion and a higher output impedance.
Cheers
ian
Nice! Put the calculations in practice and guess what...as expected! THD goed way down, gain where I want it etc.
Now an additional question. In case there is only the Cathode resistor of for example 10k and the cathodevoltage is 10V...-Ug= -10V...right? Well if so, what about the -10V curve in the graphs....or is this another situation? Wat if Ucathode is even higher...what about the curves in the tube specs...?
Now an additional question. In case there is only the Cathode resistor of for example 10k and the cathodevoltage is 10V...-Ug= -10V...right? Well if so, what about the -10V curve in the graphs....or is this another situation? Wat if Ucathode is even higher...what about the curves in the tube specs...?
You sure about the many years part? 😉
A cathode follower, with it's massive series feedback, will always be very linear at an amplification of just under 1.
Concernong the cathiode voltage vs. grid curves: As you might recall, in the case of using cathode bias with a cathode follower, you can use two resistors set up as a voltage divider to get the wanted Vgk (remember, it's the difference between grid and cathode that's being amplified). The cathode can be at 100V, using a divider to supply 96V to the grid, giving a bias of 4V. The 'bias resistor' value can be calculated ones the current going through the whole is known.
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Don't know what you mean by that...🙄
Yep. But...preferring not to have an input cap...leaves me with the low value of Rk to set the bias...right?
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