Ditching the regulator: ways to eat up current in a tube circuit?

[brad]

I've got a 7812 voltage regulator supplying three tubes from an unregulated 12v 1000mA wall wart.  Without a load, the wall wart puts out 19v, but whilst supplying the circuit, it drops to about 15.8v before the regulator.  I'd really like to ditch the regulator by loading the supply even more to bring the voltage down to around 12.6v...and with three tubes running with a regulated 12v on the heaters, the current draw would be already be 450mA right?  Now all I have to do is kill another 500mA! 

Ideally, I'd prefer to avoid semiconductors...so...I'm thinking I could just use a resistor in series with the positive rail, but is there a way to work out what value would be needed without having to power up the tubes and risk frying the filaments?

Hopefully I'll get a schematic drawn in the next few days that I can post, because I'd love to get some opinions on this circuit.

[R.G.]

What you outline is possible, but there are some gotchas.

First, even if it worked perfectly, you'd always be running the wall wart at full-bore maximum. That won't add any longevity to it.

Beyond that, the only passive ways to do this wastes the rest of the power available as heat. The WW puts out 19V open circuit, drops to 15.8V under 450ma load. That means that there is an effective resistor inside the wall wart of (19V-15.8V)/0.45 = 7.1 ohms. That resistance is there no matter what the load is doing. So if you load the thing down to 1000ma, then the voltage drop is 1A*7.11ohms = 7.11V, and the output shoudl be 11.89V, near enough to 12, and verification of our "estimate" of the internal resistance. That means that the wall wart itself is wasting 7W internally at full load, the external 550ma load is wasting 6.6W, and only 5.4W of this is going into your heaters. That's pretty inefficient. You'd need a ten watt resistor to eat that 6.6W of extra load safely.

If you just put a resistor in the positive rail, you need (19V-12V)/0.45A = 15.56 ohms, of which 7.111 is already there in the wall wart. So you'd only need to have an additional 8.44 ohms in series to drop it down to 12V for the filaments. Great huh? one resistor gets you there.

Sure it does - unless you happen to only have one of the three tubes in their socket, or have one  die. The missing or dead one then doesn't eat their fair share of the filament load, so the passive resistor pullup lets the remaining tube(s) have much more voltage on their filament, and suddenly you have tubes burning out because you removed a different tube.

You say you'd prefer to avoid semiconductors. Why is that? All of the protective means that are practical are going to involve a semiconductor.

There are lots of them. You could use a zener regulator. You'd need a 10W zener, but those exist. You could use a half watt zener and a 10W power transistor to do an "amplified zener". You could use the 7812, which is probably the simplest solution. You could use an LM317 regulator. You could use three LM317's, one for each tube, as a current regulator.

But unless you get a much better-regulating transformer, or one with a lot higher voltage, passive resistive "regulators" will be tough to get done properly without having poor life on the tubes.

[brad]

Thanks for the comprehensive answer RG!   

I guess I'll just stick with the 7812 then.  The only reason why I wanted to find an alternative was because I was trying to make the circuit as basic as possible.