# 12AX7 Plate Current vs. HT Voltage

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#### ITPhoenix

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.

#### DF96

'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.

#### counter culture

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.

#### ITPhoenix

'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.

Thank you.

#### funk1980

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

#### Palustris

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.

#### ITPhoenix

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!

#### ITPhoenix

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.

#### Palustris

"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.

#### ITPhoenix

...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.

#### DF96

A 25VA transformer will provide about 8-10W after rectification and smoothing. At, say, 400V that means 25mA so your 9mA will be fine. If AC heaters are included too they will use 4x6.3x0.3=7.5W=7.5VA so you get only about 17mA for HT. Still enough.

Your descriptions may suggest some confusion between current and power, but this is unclear.

#### ITPhoenix

A 25VA transformer will provide about 8-10W after rectification and smoothing. At, say, 400V that means 25mA so your 9mA will be fine. If AC heaters are included too they will use 4x6.3x0.3=7.5W=7.5VA so you get only about 17mA for HT. Still enough.

Your descriptions may suggest some confusion between current and power, but this is unclear.

The main reason to use another torroid is to avoid the winding labor, which was decidedly undesireable. The +-35 is not too bad, and it was decided to put the 6.3v winding there since the core is already being modified. It would be the same wire gauge, just another much more distributed winding--only a few more minutes--about 36 turns.

True, I am not always careful with my distinctions of circuit terms and parameters. In order to derive the power dissapation or consumption of a circuit, we must know at least two of the following values: voltage, current, and resistance--in DC terms at least. If there is a duty cycle, that too may also be significant in the design or diagnosis of a circuit.

At any node (junction) in a circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.

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#### DF96

Be as clear as you can. We can't help if we don't know what you are thinking.

#### ITPhoenix

Be as clear as you can. We can't help if we don't know what you are thinking.

Agreed. I had posted a question concerning electret microphone elements. I could not understand how the AC was generated from a varying DC supply. It was amazing how difficult it was to get a straight answer. It appears veterans "just know" some things without needing to explain them.

Finally, after many different attempts from several members, it was seen the DC blocking capacitor was responsible for generating the negative part of the audio waveform. But even this was implied rather than stated.

#### DF96

One of the unfortunate frustrations in the early days of learning a new subject is when you reach the stage that you can ask questions but lack the knowledge to understand the answer. Sometimes a grossly simplified answer can help, provided that all realise that this is what it is; otherwise it can store up trouble for later. Some people develop their own 'understanding' which helps them get through this stage but this probably has to be discarded or corrected later on.

Be patient and persistent. To get a reasonable working knowledge of electronics takes years (then decades to get really good). A common mistake nowadays is for people to expect to become an expert after a few months playing, or reading a couple of books.

#### ITPhoenix

Agreed. For example, I was worried about the power supply for the SS power section. So I did some studying and asked a question (you answered in fact), and tried to read the lengthy thread on cap value. Long story short, something similar to what happened with my "how does the AC get generated from an electret?" except the asker was seeking a "standard" formula.

It appears there really isn't one, but excellent tips and guidelines were established here and there...

By chance, looking at this last night The Strange Effects of AC Ripple on a Class AB Power Amp I finally realized what I have gotten myself into, but also a good place for a beginner to learn about tubes.

The effect of power supply design for a SS amp will have some effect on the output. In the link above, although it discusses Class A-B power bottles, it appears the power supply can greatly effect the quality of the output.

So it hit me. My SS power amp is A-B, and I was worried about droop (PS sag) and AC ripple, assuming zero sag and zero ripple must be good.

Now that is dispelled, and I may have to add a droop resistor, and remove some filtering caps so it doesn't sound like a 1960s, 9v hand-held transistor radio. Or maybe not--who knows?

Moreover, my tube preamp was designed to work with the standard inverter and push-pull bottles, like the one in the link (Fender Bassman 5F6-A). So now I may really be in for it. That may have to be modified as well.

It's ok though, because learning and problem solving is more advantageous for me than a good sounding guitar amp. In fact, it's been decades since I played. I do have to hold one again just to test this thing. I will use a Gibson with Humbucking pickups, and my friend is still currently playing and exclusively uses a Strat with those kind of pickups. We both know what a good amp sounds like. Getting it there is quite another story.

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