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Old 20th March 2013, 03:22 PM   #1
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Default 12AX7 Plate Current vs. HT Voltage

I tried reading the Army's Basic Theory and Application of Electron Tubes to find my precise plate currents, but have been left more confused.

According to the chart, the curves have these negative numbers I presume are bias voltages, but I am not sure how to find them with preamp tubes. I know the bottles have their own bias supply, but the pres do not. I presume it has to do with resistor values with the pres.

At any rate, I have the desired voltages and the resistor values. All of the tubes are 12AX7.

The purpose of this question is to determine the correct dropping resistors for the HT supply under load, by simulating the tube loads with dummy resistors.
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Old 20th March 2013, 03:33 PM   #2
DF96 is offline DF96  England
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'Load lines' are (probably) your answer. The valve curves are usually marked with the negative grid bias, which is usually supplied via a bypassed cathode resistor.
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Old 20th March 2013, 03:45 PM   #3
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Why not try downloading LTspice from Linear Technology and running a simulation yourself? It's useful when trying to get a feel for what's going on in circuits. I'm pretty sure it comes with a few tube models, but they're easy enough to find and install anyway.

It's best not to rely too much on sims though, or try to design purely in the simulator by throwing parts on the screen, it leads to a poor understanding in the end. Do what DF96 suggests, try to get a grasp of designing using load lines and then cross-check your ideas in the simulator.

Just google 'tube load lines'
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Old 20th March 2013, 03:45 PM   #4
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Originally Posted by DF96 View Post
'Load lines' are (probably) your answer. The valve curves are usually marked with the negative grid bias, which is usually supplied via a bypassed cathode resistor.
True. Some have no cap, but looking closer at the diagram, cathode voltages are given, like, 1.4-4.3v.

Back to the load chart...

Thank you.
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Old 20th March 2013, 05:48 PM   #5
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The load lines present values for the grid being more negative than the cathode, or in other words, the cathode being more positive than the grid. The grid in preamp tubes is usually at 0 volts (through a high value grid leak resistor). The cathode is made more positive by means of a cathode resistor. The value of the cathode resistor can be found in through use of the load line
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Old 20th March 2013, 08:08 PM   #6
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Originally Posted by ITPhoenix View Post
looking closer at the diagram, cathode voltages are given, like, 1.4-4.3v.
If you have cathode voltages then you can compute the current using ohm's law: divide the voltage by the cathode resistor for the current.
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Old 20th March 2013, 08:13 PM   #7
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funk1980: Thanks. That's the part in the Theory paper that was buried somewhere. I will have to go back to it again...

Turns out I forgot to see some series resistors in the circuit, thinking the three values on the HT PS were going in. No wonder things weren't lining up in the chart! Those charts are for normal operation, not for high gain amps as well.

I was more concerned with blowing the tubes because I am modifying the amp to use a SS power section, so the original design with all its RC values and the removal of the giant inductor (with its considerable resistance), would be off. Two of the triodes do not have series plate resistors and are fed with 384v, but the cathode resistors are larger on these: 100k instead of ~2k. There are also large value "grid stoppers" I believe they call them; the ones in series with the incoming signal.

A closer look at the schematic also provides the voltage after the resistor (if there is one) presumably under load.

The free version LT Spice does not have tubes out of the box, as far as can be seen, even in the "hidden" example library within the program folder.

What I will do is start the four tubes under load and adjust the voltages as necessary. Open circuit supply is about 460, but there are series resistors with the diodes, plus the unkown transformer regulation drop, plus the three series resistor that were already valued at the three supply voltages.

Thank you everybody!
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Old 20th March 2013, 08:34 PM   #8
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Originally Posted by Palustris View Post
If you have cathode voltages then you can compute the current using ohm's law: divide the voltage by the cathode resistor for the current.
Sorry, I did not see your post. Thanks, you saved me countless hours in the manual.

With this information, adjustable load resistors may be placed in the circuit to accurately simulate the load, without using the series feed resistors, which go on the preamp circuit board. It will be several weeks before everything is in; hence the final adjustments now.

I read somewhere my particular circuit would have a total load of 35mA, which obviously would include the wasted, total series resistance power. When I tried to apply that load before the RC sections, the supply voltage took a deep dive. I had wound my own secondary (yes, 720 turns on toroid) but I had forgotten to compensate for the huge wire resistance for 420 feet, so the regulation was very poor.

Doubling or even tripling the free-air wire size to compensate for it being under insulation is not the only concern!

Now I am using a dedicated 25va toroid for just the HT supply. This should be more than enough, even at a .85 power factor.
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Old 20th March 2013, 08:48 PM   #9
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"circuit would have a total load of 35mA, which obviously would include the wasted, total series resistance power."

Kirchhoff is rolling in his grave! At any node in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node. You need to re-read your text on AC and DC circuit theory.
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Old 20th March 2013, 09:17 PM   #10
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...Kirchhoff is rolling in his grave! At any node in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node. You need to re-read your text on AC and DC circuit theory.
I know the sum of currents and ohms law. I calculated the load of each triode and found the total to be 9mA. But this does not include the power wasted in the diode dropping or RC resistors in order to get these currents to be realized.

I have not calculated everything yet, but the parasitic losses must be significant. Either way, they would be required to properly size the transformer, unless you go overkill. I believe I am well covered with 25va.
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