Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: Rp3703 on February 19, 2018, 09:00:31 pm
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So I built a version of Steve Hoffman’s two channel Plexi Preamp a while back and paired it with a Marshall 9000. It sounded good but not great. So then I changed the preamp circuit to an 800. Again, sounded good but not great. I put it to the side for a few years and now, having no other use for this monster of a power amp, I’d like to revisit the project but this time, use the 9000 to power the Plexi Pre. In doing some research on this, I took some B+ measurements and realized they have the PI is running on only 130V. I’m guessing this was done to accommodate the low power mode where the 9000 uses a 12AX7 for the power amp but I’m also wondering if this is why I was not so pleased with the sound of the amp before. Anyhow,(schematic attached) there is 338V before the R14 220K resistor and 466V before R21 35K resistor. I would guess B+ to the EL34’s would be what it needs to be, so if I were to eliminate all connections to the 12AX7 power tube and use the standard 20K resistor at R21 and replace R14 with a jumper, would the PI be running at the standard B+ voltage? If so, then I could continue on with a 10K resistor to the next tube and another 10K resistor to the tube after that? My plan is to swap C1 and C2 with .022uf caps and run a jumper to where C4 and C6 connect to R7 and R8. I will pull at least one leg of everything in between to take it out of circuit. I need to swap C14 for a .022uf cap, replace R2 and R5 with 1M, replace R6 with 10K, eliminate R25 and replace R 26 with a jumper. After that my plan is to run B+, ground, the input of C14 and heaters to the Plexi Pre. The bias circuit in the 9000 looks way different than the standard Marshall bias circuit but I don't see any reason to mess with it other than removing it from the 12AX7.
As far as the Plexi Pre goes, I plan to go with 70's Super Lead components including an internally jumpered input. So here is a question I have. When you jumper an authentic Plexi, does signal go through the 1M resistor before being fed through the 68k resistor and then into the tube? Anyhow, that’s probably enough for now.
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Yes, I realize my original post was a bit long winded. So I'll try to reduce it to one question at a time. Looking at the schematic below, when you jumper a four input Plexi, does signal travel through the 68K resistor, then split, with half the signal going to half of V1 and the other half going through the other 68K resistor, out through the input jack, then back in through the other channel input jack and through another 68K resistor and into the other half of V1?
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You have it jumped wrong. Guitar plugs into the hi input of one channel. The jumper cable plugs into the lo jack of that same channel and then into the hi input of the other channel. There is no signal division. You essentially put full signal into one channel and the same full signal into the other channel.
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So the input with the 1M resistor to ground is the hi input. How is the signal not split though? A guitar plugged into the Hi input sends signal into half of V1 and at the same time sends that same signal out through the Lo input of the same channel, through the jumper cable, into the Hi input of channel 2 and into the other half of V1. What am I missing?
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Well, it does split and go two ways. But it does not split in half. All the signal goes both ways. You will have essentially the same signal level at the grid of each tube.
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So the jumpered portion of the signal goes through a 68K resistor on the way out and another on the way back in, correct? Does the 1M to ground reduce signal? Is that why it's on the Hi input?
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Would this be the equivalent of a jumped Plexi done internally?
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So the jumpered portion of the signal goes through a 68K resistor on the way out and another on the way back in, correct? Does the 1M to ground reduce signal? Is that why it's on the Hi input?
The 1M ensures that the tube grid has a dc path to ground (sometimes called grid leak).
The full guitar signal goes through the lower HI jack and passes thru a 68K to the grid on the right in my pic. There is no voltage divider so the full signal reaches the grid.
Then that full guitar signal goes through a 68K on the LO jack, up through the jumper, and into the upper HI jack to the 1M resistor. That 68K and 1M form a voltage divider. Approx 7% of the signal is lost in the 68K but 93% appears at the top of the upper 1M. That 93% signal level reaches the left tube grid with no further reduction because the upper 68K resistors do not form a voltage divider.
So, yes, there is a small signal reduction, hardly worth mentioning. That's why I used the word "essentially" in my earlier posts.
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Would this be the equivalent of a jumped Plexi done internally?
Close enough. But just use two 68Ks and only one 1M directly on the input jack. I've done this on all my plexi clones.