power string wiring

[andrew]

Hi all, I am looking at a Matchless Clubman schematic and it looks like each resistor in the power string originates from one point (screens) rather than the usual style power string. I saw a Vox clone done this way also. Why is it sometimes done this way? Thanks

[andrew]

I forgot the schematic: http://www.prowessamplifiers.com/schema ... ubman.html

[Phil_S]

Each node on the B+ supply goes to typical connections. The highest voltage goes to the CT on the OT (power tube plate supply), next goes to the screens on the power tubes. The rest go to the plate load resistors in the preamp section. Maybe I don't understand your question?

[andrew]

Hi Phil, I see this in other Matchless power amps also. In a Fender or Marshall the resistors are in series each with its cap. On the Matchless one end of each resistor is attached to the screen node and the other end goes to its cap. I hope this is clear- I lack the technical terminology. Thanks for the reply.

[andrew]

Here is another http://www.ceriatone.com/images/layoutP ... iatone.jpg

[jhaas]

Yeah, I've wondered the same thing.

It looks like they run the plates and screens supply in series, with a choke between, and then run parallel supply nodes to the preamp.

I could see where this method would make tweaking preamp voltages easier, as changing one dropping resistor won't effect the voltages of other nodes, as much as they would if they were wired in series. But the preamp nodes don't benefit from the filtering of the nodes that come before it when they are wired in parallel. Maybe this was desirable.

[Zippy]

If you draw big current from the power amp, the Matchless power supply will not experience as great a voltage drop in the preamp as that in a series-wired power supply (ie, less sag effect on the preamp plates).

[andrew]

Thanks for clearing that up guys.

[andrew]

One more question please. I have 450 on B+1 with kt66s PP with a choke between B+1 and B+2. If I use the above method for the power string what wattage would you suggest for the resistors? Thanks!

[Structo]

I like using 3w metal oxide flame proof resistors for my dropping strings.

[M Fowler]

I also use 3w metal oxide flame proof resistors.

Mark

[andrew]

Thanks for the advice. I had been sourcing 25 watt wirewounds $$$. Three watts are fine with me.

[tubeswell]

You really only need a power rating about double what the power dissipation is to be safe. In most 2 x EL84 or 2 x 6V6 amps 1W will work fine for the first resistor in the line.

Remember that voltage drop across the resistor x the resistance = the current through that resistor. And voltage drop x current = power dissipated.

Taking the BFPR for example, 20V drop across the 1k = 20mA of current, and multiplied by the 20V drop is 0.4W dissipation.

[rp]

I always wondered how resistant to failure you want to make resistors. At some point aren't you better off smoking a resistor and maybe loosing a power tube rather than have whatever is causing the high current draw meet up with 25w resistors and go on for way too long and take out a tranny or have your amp melt through the core of the earth? I guess that's what fuses are for and I always fuse the B+, so I'm not sure if I'm logical here. If they made small 1000V filter caps I'd overrate there, but I keep my resistors moderate, 2-3W. I posted a few weeks back about leaving my draining jumper in and powering up and getting a smoke signal, had I had 25W ps resistors in there good bye PT! What's right here? Is it incorrect to think of resistors as fuses?

[PCollen]

You really only need a power rating about double what the power dissipation is to be safe. In most 2 x EL84 or 2 x 6V6 amps 1W will work fine for the first resistor in the line.

Remember that voltage drop across the resistor x the resistance = the current through that resistor. And voltage drop x current = power dissipated.

Taking the BFPR for example, 20V drop across the 1k = 20mA of current, and multiplied by the 20V drop is 0.4W dissipation.

P = I x E
= I x (I x R) = (I X I) x R
= (E / R) x E = (E x E) / R > (20 x 20) / 1000 = 0.4