Dropping B+

[Rp3703]

I am building Doug Hoffman’s dual channel guitar preamp but I used a different PT than the one he did. I just checked my B+ and I have 300V running to the first 12AX7 and 320 running to the second. Doug’s layout drawing shows 269V and 284V. In each channel, B+ is already running through a 10K resistor before it runs to the second 12AX7, could I just raise the value of this resistor to get it where it should be and if so, what value would be recommended here? Should I drop it further back in the power supply?

[kagliostro]

You certainly can easily drop B+ to the preamp tubes (increasing the resistor values - do it experimentally)


different thing is to drop the voltage for power tubes, if you decide you must do it, consider a VVR, an amplified Zener, a big zener or a buck transformer (this one on HT AC, not at the input of PT or your filament voltage will drop with B+)


K

[eleventeen]

Of course you can do this. You calculate the required increase in resistor value by measuring volts across the first dropping resistor. It's 10K.

With your particular transformer, let's say you find that resistor drops 46 volts. (I am not finding the schematic at this moment, but we know, generically, that 12AX7 triodes in preamp use only draw about 1 mil each....so this is a kind of reality check against the mathematical results you get or should get)

From ohm's law: E = IR, 46 = 10000 * I? (current) ....divide both sides by 10000 and you find that the current thru that resistor is 4.6 mils. You want to knock off another 35 or so volts for a total of 81 volts. (46+35) So, you now have your ohm's law equation 81 = R? * .0046, divide both sides by .0046, and you get 17,608 ohms. An awkward value, to be sure, perhaps a 7500 in series with your 10K, if you can find one. Or just change to a 15K or an 18K. I'd go 15K, wanting to shade a bit towards the higher voltage side.

Once you know how much current is pulled through that dropping resistor, you just solve ohms law to find the R amount you need. To state it differently, you *must* determine the pass-through current to figure what R value you need to change to.

Real down and dirty style, you could also count the 12AX7 stages, multiply by 1, get 4 mils. (Pure guess, again, I don't have the schemo in front of me, I am assuming 2 tubes = 4 sections) Want to drop 35 (more) volts? Need to add 35/.004 = 8750 ohms in series with what you already have.


HOW do we know that a typical preamp stage draws 1 mil? Because we go look at a Fender amp and we see all those 12AX7 cathodes sitting on top of 1500 ohm resistors, and we see 1.2 or 1.6 or 1.8 volts at the cathodes. We basically drop 1.5 volts across a 1.5K ohm resistor, that's 1 mil. That's the best way to determine if a particular stage is working for an amp on your bench: When we see that 1.5 or so volts across that 1500 cathode resistor. When we gang two cathodes together and share a resistor, we go to 820 ohms, basically, 1/2 of 1500 ohms.

[PRR]

> 320  ...   284V

12.7% different.

20% different is "Nothing!" in tube work.

Play hard and enjoy.

[Rp3703]

According to the schematic, there is a 10k resistor between B+ for the first and second tube that shows a voltage drop of 15V(284-269). Can that be used to determine the current draw for B+?

[goldtop1]

This really works well and is inexpensive, have installed this on two amps that I have built with larger transformers.I actually used the diagram on the left.

[Rp3703]

Thanks Goldtop for the suggestion but I'm not sure that this would work since my transformer is not center tapped. Plus, I have already built the preamp and I would much rather fix this by simply replacing one resistor with another. That way, I do not have to install any more eyelets on the board.

[sluckey]

You have the cathode voltages given for every tube. Calculate the current thru each tube by dividing the cathode voltage by the cathode resistor value. Do this for each triode. Then add all the numbers together. This is the total load current that unit is drawing from the power supply. I get 8.72mA. Check my math.

Your B+ voltage is 16v higher than Doug's B+. So, you need a resistor that will drop 16v when load current is 8.72mA. 16V/.00872mA = 1835Ω.  Half watt is big enough for testing but I'd use a 1W or 2w or 3w once I knew exactly what value I would end up with. I would start with 1.8K or 2K or 2.2K and recheck voltages. Go up/down to fine tune to Doug's exact value.

I would unsolder the positive leg of the bridge and stand it up straight. Then solder the resistor between the flying positive lead of the bridge and where it was previously connected.

[jjasilli]

According to the schematic, there is a 10k resistor between B+ for the first and second tube that shows a voltage drop of 15V(284-269). Can that be used to determine the current draw for B+?

1.  The FIRST preamp stage (or tube) is to the left for ea. channel.  You're naming them backwards.  Tube count is done from the signal input side, not from the power supply side.

2.  I agree with PRR -- that your voltages are close enough to nominal spec. 

3.  Better way to measure voltage drop: measure the voltage in your amp where it says "to power supply".  Does that match Doug's schematic?  (This voltage is not shown on the portion of the schematic you posted).  Anyway, then measure the voltage at the next node down from the power supply.  This is the SECOND tube, not the first.  Subtract and get the voltage drop across that 10K resistor.  (Alternatively, if you have a modern DMM with a floating ground, you can measure the voltage drop directly across the resistor.)  Knowing the voltage drop & the value of the resistor, use Ohm's Law to calculate the current draw. 

Now select the voltage drop you want in order to hit your target voltage.  Plug that voltage drop number and current draw into Ohm's Law to calculate the new value to use for a dropping resistor.  Rinse & repeat for the first stage (but this time use the voltage at the second tube's node and drop from there).