Dropping heater voltage

[Enzo]

Or it was just a coincidence, may have had nothing to do with the extra half volt.

[kg]

They won't dissipate heat like a resistor, won't use any power.

i know this is kind of an older post, but i have to say that it HAS to use and dissipate the power (voltage drop times current flow) somewhere... otherwise we'd be breaking some thermodynamic rules.

[Enzo]

OK, but it wouldn't be resistive, and I suspect it would run a lot cooler than a resistor doing the same task. If I am missing something, fill me in. I suppose I am overgeneralizing.

[kg]

no, for the most part it wouldn't be resistive. the big benefit there would be that the voltage drop across it would not depend so much on the current being pulled through it. (it would to some extent because of the bulk resistance of the diode, but that's a very small number.)

i've thought about it some more and i've come to the conclusion that the diode HAS to throw off ~0.7v*I watts of power during operation. there's no way around it. the only way to get a "lossless" shift in voltage would be to employ transformers or some other magnetics.

the good thing about using diodes in this case is that because it is an AC circuit it's not continuously conducting so the duty cycle is low. also the low forward voltage drop (0.7v) keeps total dissipation down.

[jrfrond]

While it is good practice to run the heaters as close as possible to their target voltage, I don't think that 7VAC is near enough to cause the heater on the aforementioned 12AX7 to go nuclear. You may have had a defective tube. Line voltage, hence filament voltage, can vary upwards as much as 12% (at least that is the max I've seen), and unregulated heater supplies that are spot-on at 120VAC will vary upwards by the same percentage with no catastrophic failures.

[Shea]

I guess I'm too late to help, but when I had this problem with a small amp I had made, which initially had a neon lamp for a power indicator, I replaced that lamp with an incandescent 6.3v bulb that I connected to the heater supply just like in a Fender amp. That was enough to bring the voltage down to one I was comfortable with.

Shea

[MWJB]

"While it is good practice to run the heaters as close as possible to their target voltage, I don't think that 7VAC is near enough to cause the heater on the aforementioned 12AX7 to go nuclear." Agreed. If that were the case, a lot of owners of vintage Fenders would be in trouble, it's not uncommon for these to be nudging 7VAC. I have temporarily run as high as 7.5VAC & no sudden/spectacular failures.

Most tubes are rated for up to 6.9VAC as far as I can remember?

[Chiamp]

ANd if you have some of those bridges like Marshall heaters use, half of one of those could be used.

Enzo, I have a 25a bridge here , how do you hook up that so it gets parralelled ?

do you cut the filament wire
hook it on the bridge "ac side" to the +/- together ( no connection on other ac side of bridge )
and then just out to the other end filament wire or ?

and is it enough on only one filament wire, or should it be both ?

[Enzo]

Flowing through a diode - or one leg of a bridge - current drops about half a volt. SO if we insert a diode in series with our heaters, they see half a volt less heater voltage. if we use just one diode, we get pulsing DC. if we use two - one pointing each way - then we get AC voltage still.

Since all our tubes are in parallel, we ned to insert this series diode or diodes between the power transformer winding and the tubes.

SO disconnect one transformer wire and connect it instead to your diode. Then the other end of the diode goes to wherever that wire used to be.

A bridge doubled? yes, exasctly as you outline. Look at the diagram of a bridge. What we want to accomplish is to have a diode anode and a diode cathode connected to the transformer wire. That is one diode pointing each direction. Then the far ends of those two diodes are also connected together and go to the circuit. Note the AC corners of your bridge. Each has an anode and a cathode joined together. if we jumper the + corner to the - corner, we thus join the other ends of the two diodes at an AC corner. SO to use a heavy bridge as a voltage drop, short + and - together. The wire from one AC corner to the +/- as your series voltage drop element.

The other half of the bridge is unused, unless you really need two voltage drops.

Think of the diodes as part of the transformer. your diode dropper goes in one leg of the transformer

[Chiamp]

Enzo ,
much appreciated !!