Operating Point of 12AT7

[rlyach]

Ok... This was a little more difficult. Using the JJ 2A3-40 datasheet I arrived at the following: The operating point is 59V at 49mA. I ended up making a spreadsheet to model the cathode resistor of 1194 ohms, and a plate load dc resistance of 600 ohms. Looking at the intersection of the resistors and the 2A3 tube I was able to generate a loadline. The peak to peak voltage swing of the 2A3 at full signal should approach 178V. I hope I did all that right. I understand that all these numbers a just paper calculations but it helped me to understand how all this stuff works. I started with a 2V p-p signal and ended up with a 178V p-p swing, giving close to 40db of gain. The 2A3 is an amazingly linear tube. Also the JJ ECC81 is much more linear than the original RCA 12AT7 I was looking at earlier. Now I think I am ready to build this thing. I was surprised at how large the 2A3 tubes were. Thanks for all the help and training. I now know enough to get myself in real trouble. I hope you guys will still be there to bail me out when I plug the amp in and it doesn't work...

 

[jrebman]

Randy,

40 dB?  That sounds somewhat high.

-- Jim

 

[rlyach]

Well.... Thats the theoretical limit, 20 Log10 (178/2) = 38.9 db. At this level the tube is maxed out at 380V. I also assumed maximum gain from the first stage. So this is the best of the best. Since this is the first time I have ever analyzed this type of circuit I could easily be way off. I was following Morgan Jones 3rd ed. At least I have had fun!

 

[4krow]

I really love it when customers ask intelligent questions!
[/quote]

Is this the reason you don't return my calls?

 

[rlyach]

Oh yes... I was calculating the gain at the primary side of the output transformer. If you assume that the output transformer is between 80% and 90% efficient you get a total gain of 37db to 38db. There may be something else I am missing.

 

[Paul Birkeland]

There is a fair amount of gain until the output transformer steps it down (how much depends on how the secondary is configured).

I would have my doubts about the JJ 12AT7 being more linear than an old stock 12AT7.  The only way to know for sure would be to create some curves on your own (we have likely send you an old production 12AT7).

 

[Paul Joppa]

OK, now it gets more interesting  :^)

The load line on your curves is an ~1800 ohm resistance, which is not relevant to the audio signal.

For the audio, the plate load (assuming a resistive speaker load of the appropriate value) is 4000 ohms - that's approximately the speaker resistance (8 ohms on the 8 ohm tap for instance) times the transformer turns ratio squared. Since there is little AC current through the plate choke or the cathode resistor+capacitor, they don't affect the AC load. If you draw a line with a slope of -4000 ohms that goes through the operating point (50mA at 300v, nominal) you can identify the two endpoints at zero grid volts and at -120 grid volts. I get 105mA at 95v, and 10mA at 470v.

From this is is possible to calculate not only the power but also the second harmonic distortion.  I get 4.4 watts and 2.7% second harmonic distortion. You can google "load line triode distortion" or something similar and find articles with more depth.

 

[rlyach]

Paul,

Thanks for the tutorial. I couldn't figure out how to incorporate the AC load and the DC load. So if I understand correctly, You start with the operating point in DC and then draw a loadline based on the AC load and place it on the graph on top of the DC operating point. I will post a new loadline graph with a new gain calculation later. Then I will look up the other calculations you discussed. I really appreciate the help.

 

[rlyach]

OK...  One more try... Here is the updated loadline. I calculated max power of 4.1Watts and distortion of 2.69%. I also added the power rating of the tube. The usable gain still comes in at around 39db. I think I am getting in over my head here. By the way... Can we then use the max usable output and calculate the rated power and distortion from the same graph?

 

[Paul Birkeland]

There's a 22:1 voltage step-down through the output transformer (when wired for 8 Ohms).

 

[rlyach]

I wish I had a completed amp and an oscilloscope before I post this but... Something has been bothering me all night. The output voltage limitation of the B+ seems to only allow for a 30V swing on the plate with respect to GND. Then I remembered something. As part of my job I work with switching power supply circuits, buck and boost regulators. A Boost regulator energizes an inductor and disconnects it from the energizing load. The result is that the inductor voltage rises many times above the input voltage. I think (don't be too harsh on me if I am wrong) that the plate voltage on this circuit will actually go above the B+ voltage for the AC signal. Using a choke as the plate load acts as a boost circuit. The inductor is energized at the DC bias point and when the tube shuts off the inductor continues to supply current and the plate voltage can rise above the B+ supply. If this is true, it is a wonderful reason to use a choke load. It extends the usable range of the tube. Then the loadline will actually get closer to the 4.1W theoretical value. Hence, a 3.5W amplifier. Whew...

Now, for the output transformer. Since the input signal to the first stage is 2 V p-p at 20uA, and the output to the speaker will be several volts at 100's of miliamps, The gain of the amplifier can't be calculated based on the output voltage alone. As the voltage in the transformer is stepped down, the current is stepped up! I think... that calculating the gain at the input side of the transformer may be correct but it will need to be adjusted for transformer efficiency.

I hope I did not just make a fool of myself.

Randy

 

[Paul Birkeland]

You are correct about the plate choke and that the AC signal on the plate can pop up above B+, as there is magnetic energy stored in the coil and the entire piece of iron functions in the DC realm much like a resistor, but in the AC realm the impedance presented is much higher, but also depends a bit on frequency (inductance on the low end, capacitance on the high end).


If you know the power output at the speaker terminals and the impedance that this power is made into, then you have your current number as well, and you could actually work backwards to solve for the unknowns.

-PB

 

[rlyach]

Now this design is starting to make sense. The whole thing fits together incredibly. The 12AT7 at the front end (mu 60) gives approximately a +/- 60v swing which feeds the 2A3 biased at -60V, utilizing the entire operating range of the 2A3 in AC (0 to -120v). I know you can never really get to 0V because of grid current. This design is amazing. It looks so simple but there is so much behind it. In order to do more calculations with the output load I will make use of the 22:1 turns ratio. Also knowing that at 8 ohm output I get 4K ohm input, I should be able to derive all the specs of the transformer (at least for an 8 ohm hookup). I am still having trouble figuring out how to represent the gain of the entire amp. Thanks for all the patience in helping me walk through this design in detail. All the analysis is allowing me to get so much more out of this build.

 

[jrebman]

Randy,

Yeah man -- good stuff.  The magnetics are also very impressive (even more so when you consider that they also work in the s.e.x. amp).

And not at all unlike the Orcas -- deceptively simple, but far more to it than meets the eye.  No wonder they work so well together.

You're in for a real treat when it all comes together.

-- Jim

 

[Paul Joppa]

The magnetic effect of the plate load (allowing instantaneous voltage to exceed the power supply voltage) works exactly the same in series feed, and in push-pull, where the inductance of the output transformer supplies the energy.

As for gain, audio amplifiers are usually specified for voltage gain, and separately for maximum power into a given load. On the Community page, I have a white paper on the subject of signals and noise, which illustrates the problems with that approach...  :^)

 

[rlyach]

Paul,

That make sense now that I think about it. Now for the final chart. Using the transformer ratio of 22:1 I calculated the actual load at 3872 ohms. Then I got picky and added subdivisions to the graph for better estimations. Plotting the new loadline against the JJ published datasheet for the 2A3 you get the following:

Operating point: Vgk=-59, Ip=49mA, Vp=299V.
Max power: 5.28W
Max distortion: 3.68%

Now, if we use the rated power of 3.5 watts and draw new limits we get:

Max power: 3.5W
Max distortion: 1.85%

This also shows that at 3.5 watts there is a 10 volt guardband on the grid of the 2A3. Very safe operation! Now all we have to do is bump the numbers up for the losses in the transformer. If I assume 3.5W out of the transformer I get 5.29vrms at .661 amps. That converts to 7.5V peak. Then using the winding ratio the input needs to swing 165V plus losses, and we have 310V from the plate. There is an impedance divider formed by the parafeed cap and the transformer which will divide this down, and this is also frequency dependent. If I had the resistance, capacitance and inductance of the output transformer I could probably calculate the efficiency of the transformer. Still too many unknowns.

UPDATE: Changed graph to reflect Paul's comment.

 

[Paul Joppa]

Yes, transformers are amazingly complex beasts, given how simple they appear to be!

The actual turns ratio is smaller, about 20 for the OT-2 if I recall correctly. The losses due to resistance of the wire are around 0.5dB - again, if I recall correctly - and those resistances add to the reflected load resistance to get the effective actual primary impedance. There are losses due to eddy current in the steel laminations, and losses due to hysteresis of the magnetic properties as well. The last two appear as a shunt resistance, lowering the actual effective impedance.

I'll run out of room if I try to get into leakage inductance, winding capacitance, primary inductance (varies with both signal level and frequency), skin effect in both wire and laminations, insulation dieectric properties, etc. So I won't.  :^)

Incidentally, your power number seems a bit high. Because the output signal is distorted, the positive peak is different from the negative, so you have to look at the peak-to-peak voltage (or current) to get a good estimate. I see 460v-92v = 368v pk-pk. Divide V-squared by (8 times load impedance) to get power, closer to 4.5 watts.

 

[rlyach]

Paul,

Yes, If I use the full distorted signal Vp-p I get 4.37W. The website I used to tutor myself on triode curves only used half of the wave to calculate max power. I think the reasoning is that it is the worst case since the high current side is larger in magnitude than the low current side, but as you point out, it is a little unrealistic. I think I will save myself a lot of time and not dig into transformer theory any deeper. I am already dreaming about this stuff. By the way... Are the iron upgrades (nickle) at magnaquest for the paramour applicable to the Stereomour? I don't really need to spend more money but maybe in the future???