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There is 26.5 V across the 12.6 V heater, but the current is not too excessive, so, yes, the model is suspect. But there are other issues, some related to the heater perhaps. But it's a weird circuit anyway, including a valve/tube/ a FET and an IC. Is it some sort of demonstration of those mixed-up technologies?

Have you yet run simpler simulations, with just the 12AU7, to verify that its model is good?

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From where I stand, having done only a cursory check, it might not be a good model.? It seems to be generating energy in its heater that comes out of the heater pin.? How is it doing that?

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Andy

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For me the simulation is correct.

8.6V at the LM317 input, 5.8V at the output.
Gain of about 15.

I run version x64 24.1.2

On 2/17/25 11:54 AM, John Woodgate wrote:

There is 26.5 V across the 12.6 V heater, but the current is not too excessive, so, yes, the model is suspect. But there are other issues, some related to the heater perhaps. But it's a weird circuit anyway, including a valve/tube/ a FET and an IC. Is it some sort of demonstration of those mixed-up technologies?

Stepping back and looking at it...

The IRF530 appears to be connected as a class A amplifier with the LM317 as a constant current load. Input to the IRF530 is from the tube which also has a current source as its load. From there, things get weird.

24V is very low for a 12AU7. Data sheets show a range of plate voltages from 100V to 300V.

The heater drive is extremely weird. Instead of a 12V supply, the IRF530 load current gets connected here. Which would be bad in almost every way I can think of. It will drive the source voltage on the IRF530 way high. Meaning that the output voltage range is very asymetric. The current at least is about what the tube wants.

Then there is the voltage at the heater. In excess of the available 24V drive so there is a problem in the model.

Gee, I wonder if this is part of the problem:

* Heater model
*
* Can be operated from AC or DC power sources.
* NB: When operating from DC power sources, "Skip initial transient
* solution" must be checked, to make use of this model.
*

Perhaps one of the heaterless models would be better. I really like that heater warm up time parameter of 10.5 seconds. I bet that works really well with a 10 millisecond simulation time. :-)




--

David Schultz
"The cheeper the crook, the gaudier the patter." - Sam Spade

Following up on that. I added a 6V source just for the heater and terminated the LM317 current source into ground. Low and behold I see a +/150mV sine wave at the output. I don't know if a real 12AU7 would be happy with a plate voltage of 12V (the DC operating point) but the model works with it.

--

David Schultz
"The cheeper the crook, the gaudier the patter." - Sam Spade

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Forgot to mention I use the alternate solver.

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How come that I can run the simulation, which gives credible results?

On 2/18/25 9:16 AM, Andy I via groups.io wrote:

IIRC, a 12AU7's nominal heater voltage is supposed to be 12.6 volts (hence the "12" in the part number), so this is way below normal, and I doubt that the part works right.? Was your circuit supposed to provide the 12AU7's heater with the right voltage (12.6 V)?

I dug up the data sheet. It is a dual triode with two heaters. 6.3V in parallel or 12.6V in series.

The model has several dependent voltage sources and appears to have been designed to work when driven by a voltage source. In this circuit it is driven by a current source which it clearly doesn't like. I am not going to dive deeper into its netlist to figure out why.

--

David Schultz
"The cheeper the crook, the gaudier the patter." - Sam Spade

Actually, it is not "driven" by a current source.
It is rather "biased" with a current source, which is a perfectly legit way of operating a 12AU7.
The actual signal source is a voltage source.
Since only half of the dual triode is used, it's rather normal to use only one-half of the heater, hence a 6V connection. Again a quite common arrangement.

On 2/18/25 12:40 PM, Carlo wrote:

Therefore, do you think that the difference w.r.t. .TRAN without UIC comes down to the fact that in the latter the initial ITS step picks as stable operating point a nonsensical one for the 12AU7's heater (as you said where the 12AU7's heater generates energy from (very) thin air) ?

The heater model seems to dislike the current source drive. Driving it with a dedicated voltage source produces much different results.

6VDC produces a constant current even over long time scales. I used 10 seconds.

The comments in the 12AU7heater model indicate that it works with both DC and AC so I gave that a whirl. 6VRMS (4.2V peak) actually gives time varying results. Current starts at a little over 500mA peak but settles down to around 200mA peak after a second or so.

Oh, the signs of current and voltage match the expectation that the voltage source provides power while the heater consumes it.

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David Schultz
"The cheeper the crook, the gaudier the patter." - Sam Spade

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In real life, the problem with current drive is that it may result in thermal runoff. As the temperature of the filament increase, the resistance increases, resulting in voltage (and power) increase, which in turn increases temperature, and so on.Heating time is excessive, because at start-up, the resistance (and power) is low.
Not a good idea.

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That is what happens with heaters. The resistance is much lower at room temperature than at (red-hot) operating temperature. The heater voltage is 12.6 V or 6.3 V, not 6.0 V.

The comments in the 12AU7heater model indicate that it works with both DC and AC so I gave that a whirl. 6VRMS (4.2V peak) actually gives time varying results. Current starts at a little over 500mA peak but settles down to around 200mA peak after a second or so.

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That's why the old incandescent? 'fairy lights' were so unreliable. From the point of view of any one lamp, the much higher resistance of all the others in series means that it is current-driven, so the lamp with the highest resistance is in danger of thermal runaway.

In real life, the problem with current drive is that it may result in thermal runoff. As the temperature of the filament increase, the resistance increases, resulting in voltage (and power) increase, which in turn increases temperature, and so on.Heating time is excessive, because at start-up, the resistance (and power) is low.
Not a good idea.

Le 18/02/2025 à 19:40, Carlo a écrit?:

Yes, exactly. The 12AU7 model includes only one-half of the heater (pin H1 & H2 in subckt model). Therefore the current through it should be designed to be about 150mA. Indeed the ratio 1.25/7.5 for LM317 in "current source" configuration is 166 mA.

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On Tue, Feb 18, 2025 at 08:53 AM, Jerry Lee Marcel wrote:

Actually, it is not "driven" by a current source.
It is rather "biased" with a current source, which is a perfectly legit way of operating a 12AU7.
The actual signal source is a voltage source.
Since only half of the dual triode is used, it's rather normal to use only one-half of the heater, hence a 6V connection. Again a quite common arrangement.

UIC flag for .TRAN analysis skips the initial DC operating point solution step (aka ITS). All energy storage components (i.e. capacitors & inductors) by default start transient with 0 stored energy.

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Therefore, do you think that the difference w.r.t. .TRAN without UIC comes down to the fact that in the latter the initial ITS step picks as stable operating point a nonsensical one for the 12AU7's heater (as you said where the 12AU7's heater generates energy from (very) thin air) ?

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On Tue, Feb 18, 2025 at 07:16 AM, Andy I wrote:

With UIC, and no other changes:

Normal solver -> 1.165 V @ 186 mA into the heater.? That's somewhat better.? But still not right.

Alternate solver -> 1.165 V @ 186 mA into the heater.

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That is what happens with heaters. The resistance is much lower at room temperature than at (red-hot) operating temperature. The heater voltage is 12.6 V or 6.3 V, not 6.0 V.

This is often the cause of filament failure. On the baby computer replica we run the heaters at around 3v for a short while before switching to 6.3v to limit the initial "shock".
I can't remember the last heater failure...

Dave

On 2025-02-18 18:57, David Schultz via groups.io wrote:

The comments in the 12AU7heater model indicate that it works with both DC and AC so I gave that a whirl. 6VRMS (4.2V peak) actually gives time varying results. Current starts at a little over 500mA peak but settles down to around 200mA peak after a second or so.

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The heater voltage is 12.6 V or 6.3 V, not 6.0 V.

Preamp tubes are very often heated at reduced voltage, for noise optimization.
Particularly in condenser microphones, where secondary emission (grid current) is to be avoided at any cost. E seies (6.3V nominal) are often run at 5.5-5.8V.

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Indeed; it's not safe to report current direction through? a passive component. I think that the direction of current measured through a 'wire'? is always correct. To measure it may require temporarily modifying the schematic to include a 'wire' long enough. Press ALT and hover over the wire. Left-click to plot the current. I can't find that in the Help, though.

I haven't checked your schematic, but did you know that current through a 2 port device always flows into port 1 and out of port 2. The standard symbol has no visual indication which is port 1. Try rotating the resistor 180deg.

--

Regards,

Tony?

On 17 Feb 2025 17:00, Carlo <carlo.cianfarani@...> wrote:

Hello,

I uploaded the project "Bravo_Ocean_ampli" in Files->Temp. As you can check the .tran analysis returns a weird DC operating point (aka ITS/Initial Transient Solution). Namely the current through R4 connected to LM317 Out pin is negative -165.5mA. Of course physically it can't be the case (note that it isn't a problem related to how R4 pins are arranged in the schematic).

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What could actually be the problem with this circuit simulation ? Thanks.