is 240 VDC normal?

[iknowjohnny]

Would you consider 240 VDC a normal voltage for preamp tubes in a JCM style amp? From voltage charts i've seen they run a good 100VDC more than that. So i have long wondered why i only get 240 on mine. The PT is getting 470 after the SS rectifier, then my rail sports a 5H choke and small dropping resistors all the way thru, 8.2k to the V3 and V2 then a final 2.2k at the V1. V1 has 100k/1.5k on plate/cathodes and i get 240V while marshalls even use smaller cathode resistors, at least on V1A. So why am i only getting 240 VDC? Do marshalls use far larger PT voltage? Mine is a mag comp marshall style PT. I like higher voltages in the pre stages and i don't understand why it drops so much.

[Phil_S]

This would easily be explained by having a 22K resistor instead of a 2.2K resistor. The tube math doesn't lie. Lift one end of the 2.2K and put a meter on it. Same for that 8.2K.

Alternatively, on the live circuit, see what the voltage drop is across the 8.2K and the 2.2K resistors (one probe on each side of the resistor). That will tell you where you are experiencing the voltage loss. Since there isn't much difference in the current load at each node (the big draw is at the power tubes), you should see roughly 4x the drop on the 8.2K compared to the 2.2K.

[iknowjohnny]

All parts check fine. Heres how it goes....a 8.2k coming off the choke and supplying the PI at which i get 400 V. (node of course, not plate) then thru another 8.2k and i get 353 V at V2's node. Then thru a 2.2k to V1 and i get 346v at that last node. Then with 100k/1.5k on plate and cathode at both V1A and B, i get around 240 at both those.

What do you think of this idea.....parallel a second 8.2k off the choke and go straight to V1? I know matchless does that, tho they parallel all node. Thats brings up another question.....why DO they do that? same reason as me? I was thinking maybe they have a PT optimized for the PA, but for what they want from the PA it leaves them with too little for the pre, so they parallel the nodes so as not to lose so much voltage. Stupid theory of is that possible? Might be my solution?

[Phil_S]

Put a jumper over the second 8K2 and see what happens.

I was just guessing before and didn't want to show you my numbers before you reported. I guessed that V1, V2, and V3 draw about 10mA, that V1 & V2 draw 6mA, and V1 draws 3mA. I also did not realize you've got two 8K2 resistors in the dropping ladder. Adjusting for that...

If you've got 8k2 at 10mA, that drops 82V. Your actual for the first one is 70V.

The second 8K2 at 6Ma would drop 50V; that would be 470-120 =
350). Your actual is 47V bringing you to 353V. So far so good.

The last, 2k2 at 3mA would drop about 7V; in my guess that brings it to around 343V. You've got 346V. So, this checks out nicely as far as the rail following Ohms Law, as if expectations were that it might diverge?

One problem you've got here is the extra 8K2 sucking 47V out of the supply. I'm looking at a 2204 schematic and that second one isn't there.

The next is the drop across the plate load resistor. At the PI, the 82K and 100K are parallel, making the effective R around 45K. If the tube is drawing 2mA per side, then 45K * 2mA = 90V. What are you seeing at the plates on the PI? Math says it should be around 310V. Current draw varies and that will account for the difference in what you see.

You are going to see a similar effect at V1 and V2. There, you are working with reduced voltage at the next node. If you get a drop somewhere between 50-90V, (depends on current draw of the tube), that makes perfect sense.

So, at V2, your 353v at the node drops to something under 300v, and at V1, your 346V is dropped to 240 (about 90V). This is surely possible. That is why I said to get rid of the second 8K2 node. That will add somewhere around 50 to V1.

If you parallel the 8K2 pair, they are effectively 4K1 and should drop half of 43V (your readings of 400-353), but I don't know why you'd bother. Just run the PI and the two preamp bottles from the same node from one 8K2.

[iknowjohnny]

One node for all 3 ax7's? I've never seen that done, and every time i've ever asked about nodes i've always been told to have a seperate node for each bottle. I just now removed the last node and ran the PI and v2 off one node. No change worth noting in voltage or tone. but i'll try what you said a report back. thanks for your time.

OK, ran all 12ax7's off the 8.2k after the choke and i still get only 257v at V1 a and b and also at the PI. 400 VDC at the node. The PI has JCM 800 values all the way thru.

[Phil_S]

You've got six plate load resistors in parallel after that 8.2K dropping resistor. Thats 5x 100K and 1x 82K, for an effective load of 15K. Again, assuming 10mA for the three 12AX7's, that's 15K * 10mA = 150V drop. 400-150=250. So, it all works according to Ohms Law. You are getting a 143V drop, so that's about 9.5mA and seems right for three 12AX7.

If you want to raise plate voltage, you will need to use smaller dropping resistors in the B+ ladder. If you like what you have at the PI (I don't think you said what), then try a pair of 1K dropping resistors for the 3rd and 4th nodes. I might go as low as 470K. You can experiment by tacking in some resistors in parallel. A pair of 2.2K = 1.1K. 8.2K || 2.2K = 1.7K. See if you can use Ohm's Law, V=I *R to help you approximate the values you need.

To calculate parallel, it is the reciprocal of 1/R1+ 1/R2 (for as many as you use). For two of the same, it is simply half.

Good luck with this.

[iknowjohnny]

Thanks, i'll give lower values a try. I was once told 2.2k was too low and that you should use higher values, so i tried to stay near 10k but had to go lower on V1 to keep it from going way too low. It sounded lousy with less nodes so i think i need them to stay in place, but i will try dropping all the resistors. Thanks.