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

[Alan0354]

Gents,

Thanks for the many helpful replies. There is evidently at least one misunderstanding here though. To be clear, I am NOT powering the 12V regulator from a 48V supply. The 48V supply is an independent circuit. And since it's based on a voltage quadrupler, there is absolutely no way that I could source enough current from that to power a heater string. It's intended solely for providing a maximum of 20mA or so for a couple of phantom-powered mics. The 12V regulator is being run from a full wave bridge rectified 12.6VAC supply.

EDIT: Added schematic [ATTACH]27271[/ATTACH]

To be honest, I don't think it's the over current that burn the regulator. Now that I saw the schematic, regulators have current limit and it just doesn't burn like you describe. If it is not over power then it must be something else. In general, regulators needs to have protection diodes from output back to input in case the input collapse and the voltage get below the output voltage.

[olddawg]

Sorry, no, I think there's far too much current involved for such a crude solution to be practical.

It's just a filament supply? 1.2 amps? What are you concerned with. I've seen supplies like that in stereo equipment for 35 years. I thought filament supplies were not that critical. What "crudeness" are you worried about?

[Alan0354]

It's just a filament supply? 1.2 amps? What are you concerned with. I've seen supplies like that in stereo equipment for 35 years. I thought filament supplies were not that critical. What "crudeness" are you worried about?

Problem is you cannot cover different scenarios. Yes if you have all tubes plugged in, you calculate so the zener take a minimum amount of current to regulate the 12V. BUT what if someone unplug the tubes? All of a sudden, the zener has to take up all .9As and that's 11W dissipation of the zener. Your solution quickly become complicated. Even if you remove one or two tubes, you have to be careful on the power dissipation. Not to mention the dissipation of the dropping of the resistor.

Above all, I don't believe it's over power that burn the regulator. I think it's something else after I read the schematic. Regulators do not burn on over drawn, they won't be in business if the regulator burn on over drawn a few seconds. This is such a basic requirement for a regulator to survive on over current!!! If you collapse the input voltage and the filter cap at the output still keeping the output at 12V, you really risk burning the regulator.

[olddawg]

Yeah.. I see your point. Use several zeners in parallel or a discrete transistor regulator with a Zener.

[Alan0354]

Yeah.. I see your point. Use several zeners in parallel or a discrete transistor regulator with a Zener.

Now you are talking, a zener and a pass transistor. Still have to have a reverse diode from emitter back to the base to protect the transistor if the input voltage collapse faster than the output voltage. This is actually a serious problem that people don't talk about. The old 78xx would not survive if the input drop lower than the output voltage.

I have to check the temperature coef of zener for parallel a few of them. If the coef is +ve( ie the zener voltage increase as it heat up), then yes, that will work beautifully as the one that initially have lower zener voltage will conduct more current and heat up, then when it heat up, the zener voltage increase and the current decrease as the others will take over.

But if the coef is -ve, then you'll have a run away problem.

[Merlinb]

To be honest, I don't think it's the over current that burn the regulator. Now that I saw the schematic, regulators have current limit and it just doesn't burn like you describe.

What are you talking about??? There OP has said nothing about burning. It simply won't start up into a low-resistace load.

[g1]

How about Mick's solution from post #12? Isn't this just a problem with inrush current that needs to be limited?
Seems a very simple solution, or is there more to the issue that I am missing?

[exclamationmark]

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). I don't think the current limiting should be a problem, since the datasheet states it's a linear type rather than foldback (so it should essentially act like it's own soft limiter). Since a pass transistor worked fine in the past, maybe the regulator is dissipating too much heat for it's package size and going into shutdown? Assuming you're sitting in the range of 16-17 volts before the regulator, you're going to be dissipating easily 20+ watts into the regulator with 6 tubes plugged in. 20 watts is a lot of heat, but since it's only a transient I don't exactly know if this would be enough to cause the die temperature to skyrocket that quickly.

At this point if you really don't want a re-design, I'd literally stick in every 12v regulator I could in place of this one and see if it works - the main criterion would be non-foldback current limiting and high junction temperatures. I think there are some 150 degree c regulators floating out there somewhere (or maybe I'm just thinking of normal transistors...) Ideally, you would want some sort of soft-start circuit that would ramp up the voltage slowly, but that would obviously require a little bit of modification (well except for the thermistor mentioned before).

[olddawg]

Another simple solution: Use your same regulator. Put a diode on the ground pin to elevate it .7 v. Put a forward biased diode on the output of the regulator as a current limiter. The voltage to the heaters should be very close to the same as without the diodes.

[Mick Bailey]

. 20 watts is a lot of heat, but since it's only a transient I don't exactly know if this would be enough to cause the die temperature to skyrocket that quickly.

When I've encountered this problem the package remains stone-cold. I wonder if some devices have short-on-startup detection circuitry in addition to thermal limiting?

[Alan0354]

Not to hijack this thread, but the topic is related.

Can I put the filaments of two power tubes in series and use 12V transformer to power the filament? Mainly I am referring to 6L6 and 6V6.

[Alan0354]

What are you talking about??? There OP has said nothing about burning. It simply won't start up into a low-resistace load.

My bad, I miss read f*ing as fry!!!


Well then, monitor the initial current drawn before doing anything, find out how much initial current drawn first before worrying about finding a solution. Still, it's all speculation at this point.

[Wombaticus]

It's just a filament supply? 1.2 amps? What are you concerned with. I've seen supplies like that in stereo equipment for 35 years. I thought filament supplies were not that critical. What "crudeness" are you worried about?

If so, they are lousy filament supplies! There will be atrocious ripple rejection, a ton of current required in the shunt, and high dissipation from any actually useful amount of series resistance that is added before the Zener. Take my word for this, it's not a viable option. Such approaches are really only worthy of serious consideration in low current situations.

An amplified Zener or Mick's solution might be indeed be a possibility, and I'll consider this further if necessary, but again, my goal is to avoid a redesign and to make use of an existing PCB that was designed around the Micrel part that won't light up.

And yes, Merlin is absolutely correct that the part is NOT burning out, but is rather just refusing to start into a heavy load (like many modern LDO regulators, it is designed so as to not require a diode between input and output).

exclaimationmark is perhaps right that it is thermal shutdown rather than current limiting that is preventing it from starting... although I'm not sure that it makes a practical difference.

This brings me back to my original question "does anybody know of a 12V regulator (preferably three pin) that can, no questions asked, light up a string of six 150mA @12.6V preamp tubes *from cold* without locking up?". At this point, if the answer is "no" we can wrap up this thread, as I think I now have a pretty good handle on what my options are... 1) Try a bunch and see if one works; 2) Graft in an amplified Zener instead of the LDO regulator; 3) Try Merlin's current boosting resistor hack; 4) Redesign to incorporate some soft startup circuitry a la Mick, or a power transistor current booster as I've used in the past.

[olddawg]

If so, they are lousy filament supplies! There will be atrocious ripple rejection, a ton of current required in the shunt, and high dissipation from any actually useful amount of series resistance that is added before the Zener. Take my word for this, it's not a viable option. Such approaches are really only worthy of serious consideration in low current situations.

An amplified Zener or Mick's solution might be indeed be a possibility, and I'll consider this further if necessary, but again, my goal is to avoid a redesign and to make use of an existing PCB that was designed around the Micrel part that won't light up.

And yes, Merlin is absolutely correct that the part is NOT burning out, but is rather just refusing to start into a heavy load (like many modern LDO regulators, it is designed so as to not require a diode between input and output).

exclaimationmark is perhaps right that it is thermal shutdown rather than current limiting that is preventing it from starting... although I'm not sure that it makes a practical difference.

This brings me back to my original question "does anybody know of a 12V regulator (preferably three pin) that can, no questions asked, light up a string of six 150mA @12.6V preamp tubes *from cold* without locking up?". At this point, if the answer is "no" we can wrap up this thread, as I think I now have a pretty good handle on what my options are... 1) Try a bunch and see if one works; 2) Graft in an amplified Zener instead of the LDO regulator; 3) Try Merlin's current boosting resistor hack; 4) Redesign to incorporate some soft startup circuitry a la Mick, or a power transistor current booster as I've used in the past.

Try the 2 diode trick as I suggested earlier. You won't have to mod the board at all, the diodes can be soldered directly to the regulator pins, and will cost you 10 minutes and 10 cents. If the problem is low load resistance at start up and instantaneous over current, it will fix that.

[Wombaticus]

Another simple solution: Use your same regulator. Put a diode on the ground pin to elevate it .7 v. Put a forward biased diode on the output of the regulator as a current limiter. The voltage to the heaters should be very close to the same as without the diodes.

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.