Best way to drop filament voltage

[dorrisant]

I'm using a 15vac 2a supply. It has no center tap.
Can a virtual center tap be used with some zeners to drop it?
I don't have much room for a step down tx. I could remote mount a different tx though.

It is a 5e3 build with octal preamp tubes and a 5Y3 on a 5v tap along with the pilot.
2x 6v6 = 1a
6sn7 = .6a
6sl7 = .3a
Total = 1.9a

The tx is good for 2a @ 15v.

I will go do some searching, thanks in advance!

Tony

[tubeswell]

You could try 2 x parallel DC heater supplies (one for the 6V6s, and one for the 6SL7s). That way you could use voltage dividers after full wave bridge rectifiers to make sure the heater supply is 12.6V per circuit.

[davent]

[s]You could regulate your supply to 12.6vdc and put the two 6s*7's in series and put the two 6v6's in series, long as the current capacity is ok for your transformer.[/s] My experience with 6sl7's is they appreciate vdc for the heaters.

edit: My brain failed to note that there are two different preamp tubes with different current demands so that messes that series string up. Can the current demands of the two tubes be balanced?

dave

[/code]

[Leo_Gnardo]

I could remote mount a different tx though.

I'm not much likin' the alternate solutions. Too much fiddle faddle. And zeners would likely put noise in your signal - boo to that! Save headaches and get a 2 or 3 amp 6.3V transformer.

[Phil_S]

Here's what I think I'd try short of getting a 6.3V transformer.

Get a 12.6V pilot and remove it from the rectifier circuit. It will help with loading the circuit, even if just a little bit. Auto parts stores have racks of 12.6V bulbs.

Put the 2x 6V6 in series. Put the 6SN7 + 6SJ7 tubes in series. Parallel the pairs.

Math says, put a 1.25 Ohm 5W sandblock resistor between the transformer and the first filament connection on one leg. This should drop 2.4V/1.9A (actually 1.26r). I think I'd start with a 1 Ohm. The sandblocks are usually 5% tolerance, so pray for one that reads a little high. I wouldn't use one that is over 1.25r and that is why I said use 1R. Put the resistor against the chassis if you can to allow it to sink some of the heat, or make sure it is in free air.

If you can get it around 13V, that's good enough.

If you can build the regulated DC supply, that's great, but this is much simpler and is worth trying first.

[tubeswell]

15VAC rectified in FW bridge is 21.2VDC. You want to drop that to about 13VDC, so that's say 61.28% of 21.2V - so your voltage divider wants to be geared to that. You ideally want the resistors to pass as much current as possible through the divider so that you aren't starving the filaments, so you should shoot for lower-value, higher-wattage resistors. e.g. 68R (R2) and 39R (R1) on upper leg, will give 63.5% drop i.e. 13.4V, which should be about right. Use a largish capacitance cap to decouple the R2 to ground. The voltage drop across the 39R will be 7.7V, which amounts to 0.2A lost through the divider. The dissipation needed is (0.2 x 7.7 = 1.5W so) 3W minimum for the 39R resistor, and 5W for the 68R resistor.

[img:914:388]http://farm4.staticflickr.com/3759/9856 ... afc1_o.png[/img]

If you want to lose (say) 0.4A of current (to help the heater filaments pull down the supply voltage a bit further), just halve the resistances in the divider, try 22R (R1) and 33R (R2) , which will deliver 12.7V - just about perfect. But increase the resistors' power ratings to at least 10W for the 33R and 6W for the 22R.

[martin manning]

My first thought is to take Leo's advice and just get a 2A, 6.3VAC filament transformer.

If you want to use the 15VAC and drop a few volts to 12.6, that would go like this:

Two pairs of 6.3V in series/parallel is a good idea, but you can't just stack the 6SN7 and the 6SL7. They have different current requirements so they won't split the voltage equally. You could fix that by putting a shunt across the 0.3A 6SL7 to bring it up to match the 6SN7 at 0.6A. That would require a 6.3/0.3=10.5R resistor, at 0.95W, so make it a 2W.

With the two 6V6's stacked and paralleled, you have 0.6 + 0.45 = 0.95A draw. To drop 2.4VAC (15-12.6) at 0.95A requires a 2.5R resistor at 2.25W, so make it a 5W. This doesn't sound too bad, just two resistors of relatively small physical size.

I wouldn’t recommend going to a DC supply for the entire filament string. There will be some large capacitors and a large cold-start surge current to deal with, requiring a very robust diode bridge.

Tubeswell, your calc's are way off since you haven't accounted for the current draw of the filaments. Collectively they are going to look like a 13R resistor at their correct operating voltages.

[tubeswell]

6SN7: 6.3V/0.6A = 10.5R (plus 10R series resistor) = 20.5R
6SL7: 6.3V/0.3A = 21R
6V6: 6.3V/.45A = 14R (28R for both)

Oops (thanks Martin)- upon reflection, I make that 16.7R for the heater resistance - if you put 10R in series with the 6SN7 (between the 6SL7 and the 6SN7), and put these in parallel with two 6V6 in series.

So re-looking at this, use a 10k for R2 and with the 16.7R in series, this makes 16.69R for the bottom leg. 12.6V/16.69 = 0.75A. With the 8.6V drop needed over R1, that would make that resistor value 11.4R. (You might be able to find closer-value cement resistors in your stash, but 12R is close enough). 16.7R(12R+16.7R)= 58% of 21.2 = 12.2V which is getting pretty close.

Put a 50W 13V zener in parallel with R2 to clamp the voltage at startup. http://www.newark.com/solid-state/1n331 ... dp/10P4810

At least I think this is how it could stack up. I stand to be corrected as usual.

[printer2]

Use a 12sn7 and 12sl7, you can get them much cheaper than 6sn7 and 12sl7's. Run the 6V6's in series and use the center for a centertap to ground. Or better yet, get some 12V6's instead of 6V6's, I got a whole whack of basically NOS ones for an average cost of $5 each. Not these but,

http://www.ebay.com/itm/12V6GT-Vacuum-T ... 1058032713

Put a low value resistor in line on either side of the heaters to eat up the extra 2.3V.

[dorrisant]

Thanks for all of the replies! I read and reread all of them until I got a decent grasp.

I did some searching and found a few things on the web. Before I got any responses I tried what I found in Merlin's power supply book, p198. While he was writing about dropping just a couple of volts this was quite a bit more. 14.94vac under full load, (yes I did run it this way just long enough to get a stable reading) down to 6.3vac. That would be an 8.64 drop. I used a virtual center tap so according to the author, only one resistor value on one of the filament taps would be needed for the proper drop.
R total = V drop / I heater
R total = 8.64/1.9 = 4.5R
I figured I should at least try it. I had some 3.9R 10W, so I used two parallel 3.9Rs in series with another set of 3.9Rs in order to keep the dissipation per component down... Maybe I was wrong there.
It works at somewhere around 6.68vac... but those 3.9Rs are getting hot, around 300°. I was reading the temperature with a Fluke infra-red but it was too close to the power and rectifier tubes to be sure. I need to recheck with the digital cooking thermometer for a more direct contact reading. Although this was ok for testing I'm thinking I will just take the advice to just order a filament tx.
On the plus side, I did find that the amp sounds great and to my surprise very hum-free. I had to strain to hear any hum at all with both channels bridged and everything dimed. According to some of the stuff I read, this was not supposed to happen with the octals. Nice!

Thanks again for all of the advice!!
Tony

[Phil_S]

R total = 8.64/1.9 = 4.5R
I figured I should at least try it. I had some 3.9R 10W, so I used two parallel 3.9Rs in series with another set of 3.9Rs in order to keep the dissipation per component down... Maybe I was wrong there.
It works at somewhere around 6.68vac... but those 3.9Rs are getting hot...

Nice solution except for the heat problem. I'd like to hear Martin and Tubewell's take on it. They are way more knowledgeable than I am.

If you have 9v * 2A = 18W. Your pairs of 10W are 20W equivalent, but that is probably too close for comfort. See if you can get something around 4.5R @ 25W rating. They will be large (f you have space), but they will handle the heat much better. I don't think I like 300 degrees (assumed 300F, not 300C; if 300C, that is way high.) Standard values for 25W cement resistors should be 3.9, 4, 4.7, and 5 and they should be about $1 each, maybe more if purchased locally. Postage will be killer from a place like Mouser.

[martin manning]

If I understand correctly Tony has four 10W 3R9 in series parallel for 3R9 total, dropping about 15-6.8=8.2VAC. That's 17.2W total, or 4.3W each. Should be fine, but that's a lot of heat to be dumping under the hood.

[Phil_S]

I was only thinking about how to manage the heat a little better. Maybe a modest heat sink would be a better strategy. These can be reclaimed from all sorts of solid state appliances that have been relegated to the junk pile.