12V regulator IC that can power 6 preamp tubes (900mA) FROM COLD STARTUP?

[Gregg]

I'm regularly using 12V regulated power supply for preamp tubes built around regular 7812 or in case of 6 tubes a 78S12 (just in case). Never had a problem with that. I have also experience with LM2940 and LM1084 powering preamp tubes at 6V for a total of 800-900mA without problems.
Putting a diode at the output (and from input to output) is a recommendation from 78XX series datasheets and it can help when you have high value capacitor/s at the output. Most probably your regulator "sees" a short at the output when powered up and shuts down. I think that diode is recommended there to prevent that.

[Merlinb]

Still, it's all speculation at this point.

Not really. We know it starts with three tubes, we know how heaters behave at start up, and it's a known problem with regulators.

Paul: A variation on the parallel resistor is a zener. When the reg is in shut down there will be full voltage across it -enough to turn on the zener. But once the voltage across the heaters rises enough, the reg should snap back into action and the voltage across it will fall, turning the zener off (more or less), so long term dissipation in the zener might be avoided....

[olddawg]

I don't understand what you think this will accomplish?

First, there is already a diode on the ground pin of the regulator in order to adjust the output voltage from 12V up to something closer to 12.6V. But what I really don't understand is why you think a forward biased diode on the output pin would in any way act to limit current?? Essentially all it will do is cost me a diode drop of voltage.

The regulator will not sense a short condition on it's output at start up with the cold heaters with the diode junction resistance there. The other diode in the regulator ground will increase the output (relative to ground) by .6-.7v to compensate for that .6v voltage drop you are worried about. You could also just source a regulator with a slightly higher output and use just one diode on the output of the regulator.

[loudthud]

Cold filaments of a 12ax7 read roughly 12 ohms for me in series, so each tube will draw 1+ amp for a few hundred milliseconds (assuming it's a 12a series tube).

This didn't sound right to me, until I checked it. I measured 13.2 ohms with a sample of one. So you can expect a surge of 5 or 6 amps at startup. It takes two or three seconds for the current to ramp down to it's nominal value. Some regulators won't like that, especially if they employ some kind of foldback dissipation limiting. The diode in series might trick the regulator into starting up. A thermistor would be better if you can find the right one.

[olddawg]

This didn't sound right to me, until I checked it. I measured 13.2 ohms with a sample of one. So you can expect a surge of 5 or 6 amps at startup. It takes two or three seconds for the current to ramp down to it's nominal value. Some regulators won't like that, especially if they employ some kind of foldback dissipation limiting. The diode in series might trick the regulator into starting up. A thermistor would be better if you can find the right one.

I was going to suggest a thermistor as well but I don't think you will find one that will have a high enough value cold and a low enough value hot.

[Wombaticus]

Not really. We know it starts with three tubes, we know how heaters behave at start up, and it's a known problem with regulators.

Paul: A variation on the parallel resistor is a zener. When the reg is in shut down there will be full voltage across it -enough to turn on the zener. But once the voltage across the heaters rises enough, the reg should snap back into action and the voltage across it will fall, turning the zener off (more or less), so long term dissipation in the zener might be avoided....

I may indeed experiment with a Zener. I had just a brief minute to experiment tonight, so I tried a 5 ohm 5W resistor in parallel. This did indeed make the heaters light up right away and gave me right on 12.6VDC by my meter. But when I looked on the scope, there was about a 70mVrms sawtooth riding on the DC -- probably not good enough. For giggles, I may perhaps try a bit larger value resistor as well, I think I have some 10 ohm ones in the stash.

[Gregg]

I measured 13.2 ohms with a sample of one. So you can expect a surge of 5 or 6 amps at startup.

I always put fuses on my secondaries (except for the bias supply) and usually 1A6 or 2A slow blow fuse is OK for 5-6 tubes. Most of the time a 1A fuse will get blown during startup. You can also use an NTC for inrush current limiting but I usually put mine on the primary for PTs that are more than 200W.

Conrad Johnson used a simple 78L12 to power 6 preamp tubes in its preamps for years

Maybe you meant 7812? 78L12 has a rating of 100mA.

[R.G.]

I think you're hoping for a miracle. You're almost certainly in for some redesign instead of an IC swap.

You have correctly identified the start current issue, and I believe that the refusal to start is caused by the filaments never heating enough with the regulator limit to start bootstrapping the current down. The regulators and tube filaments' current/resistance characteristic could well be conspiring to not let the voltage rise by generating enough volts*amps in the heater wires.

You want simple solutions or elegant solutions?

The simple solution is to use two of the regulator you already use. You know from testing that one regulator will start and regulate three tubes. Use two regulators, one for three tubes, one for the other three.

Elegant solutions will require different approaches, like perhaps a kickstarter to allow the currents to be unregulated or semi-regulated until current drops below X. There are some others.

But from the design limitation you've added - make it work, but don't change anything but the regulator - I'd say that splitting the load and adding one more regulator would be expeditious.

Upon further thought:

You *might* get by with just paralleling the regulator with another identical one. Current sharing is always an issue with hard-paralleling things, but in this case the regulator is able to stand the normal current all by itself. It's only the startup surge that's an issue, and this will be over and done with before they burn up.

The lowest-voltage regulator will eat all the current it can, start lowering its voltage, and then the second and slightly-higher voltage regulator will cut in and provide more current. This should be enough to get the heater temps started up and the current started down, and once that happens, the situation resolves itself.

It's worth a try, and uses parts you already have.

Another thought is that low dropout regulators like these are really fussy about their load impedance. They must be used with a capacitance on the output that's not too high, not too low. If you're missing that capacitor, your regulator may be oscillating. Have you looked for that?

[Merlinb]

Another thought is to put a thermistor in series with the regulator output..

You would want the thermistor in series with the input of the regulator, so that it doesn't interfere with the voltage across the heaters.

[Mick Bailey]

You would want the thermistor in series with the input of the regulator, so that it doesn't interfere with the voltage across the heaters.

My thought was that if it's in series with the input, the regulator would still 'see' a near-short and not kick in. Depending on the characteristics of the particular device the drop across it shouldn't be more than about 90mV on-load.

[Wombaticus]

The regulator will not sense a short condition on it's output at start up with the cold heaters with the diode junction resistance there.

I'm slightly gobsmacked, but credit where credit is due... This actually works. I confess I still don't fully grasp why, as the junction resistance should be very low at the moment of activation as the initial surge current is quite high. Certainly it must be an order of magnitude below the 2 or 3 ohm resistance of six tubes in parallel... Can someone explain what I'm missing here?

[soundguruman]

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[Dave H]

I'm slightly gobsmacked, but credit where credit is due... This actually works. I confess I still don't fully grasp why, as the junction resistance should be very low at the moment of activation as the initial surge current is quite high. Certainly it must be an order of magnitude below the 2 or 3 ohm resistance of six tubes in parallel... Can someone explain what I'm missing here?

Perhaps it's the 0.7V drop of the diode and not its junction resistance that fools the regulator?

[Gregg]

What I meant in my previous post was a diode from ground to the output pin as seen in this datasheet (fig.21):

http://www.makershed.com/v/vspfiles/...ages/l7805.pdf

[R.G.]

I'm slightly gobsmacked, but credit where credit is due... This actually works. I confess I still don't fully grasp why, as the junction resistance should be very low at the moment of activation as the initial surge current is quite high. Certainly it must be an order of magnitude below the 2 or 3 ohm resistance of six tubes in parallel... Can someone explain what I'm missing here?

It worked!?! KEwl!!

I think the explanation is simply that as OD said, the regulator senses that its output voltage is bigger than 0.000 - by about 0.6V, and that's enough to convince it that it's not trying to power a dead short. That makes sense to me.

Neat trick. I've not seen that before.