dropping resistors

[daz]

Can someone explain how and why resistor values in the PSU rail for the preamp node are chosen ? I read that values are not usually chosen to control voltages in the preamp as much as decoupling. I'm really not sure what that means, but i always thought it meant a way of isolating the nodes so they don't affect each other, or interact. Is that correct? And how are values chosen if they are for isolation? The whole concept is unclear to me and i've seen as low as 2.2k and as high as 20 something k.

[R.G.]

The resistors are the series part of a filter. The capacitor(s) to ground from each section between resistors are the shunt part. This forms a simple R-C lowpass filter to all the AC signals on the various parts of the high voltage supply.

It does two things. First, it reduces power supply ripple, and second, it prevents power stage drains on the B+ line from being carried off down into the sensitive preamp stages.

Let's say you have 10V of ripple at (about) 120Hz at the first filter cap, a 2H choke, and a screen decoupler of 10uF, and thereafter 10K and 10uF decouplers for three more dropping sections.

The power stage sees B+ with 10V of ripple, and imposes 2x signal frequency drops on the B+ to the extent that it makes B+ sag. Let's say that's 10V too.

The screens see B+ as filtered through 2H and a 10uF to ground. At 120Hz, the impedance of the inductor is Xl = 2*pi*F*L, or 6.28*120*2 = 1507 ohms inductive and the cap is Xc = 1/(2*pi*F*C) = 1/(6.28*120*10e-6) = 133 ohms capacitive. Simple ohm's law application would say the ripple is reduced to 133/1640 = 0.08 of the 10V, or 0.8V. That's a crude way of estimating the reduction in ripple, but it gets us going in the right direction, especially since it's the resistors you want to know.

So let's say that the L and C reduce the ripple to that 80mV. The next 10K series, 10uF to ground reduces the 120Hz ripple by a factor of 133/(10000+133) or 0.013, to 0.08*0.013 = 1.05 millivolts. Each succeeding section reduces it by 98.7% again, so ripple stops being a factor.

The reduction of signal passed down from the first filter cap is simlar. If the signal there is 164Hz (two half-waves of the 82Hz fundamental of the lowest guitar note) then the reduction into the screens and succeeding stages is similar to the losses for the 120Hz ripple.

With a 10K/10uF decoupling stage between each section of the amp, signal residue on the B+ passed to earlier stages is reduced by that factor of 0.013, or 98.7% at each stage. That means that the stages can't pass signal to one another very well, so you can't get odd little feedback loops.

Of course, you're free to choose other values than 10K and 10uF for other attenuation between stages.

[Tooboob]

You don't want to run your preamp tubes at 400V+. Well, you might want to, but the preamp tubes definitely aren't in favor of it. The "dropping resistors" do drop voltage. The preamp is trying to amplify a comparatively weak signal, usually a few times, and supply noise tends to make it all the way to each stage's output. big caps after the dropping resistors keep the supply stable with changing preamp current, and create a low-pass filter for any noise coming in from the other side of the resistor.

Sometimes more than one triode will share a node. if they are amplifying the same signal, with one inverting the signal and feeding the other triode's input, the current fluctuations tend to cancel, allowing the resistor and capacitor to generate a voltage that's more stable.

Many adventures are possible, but probably ill-advised. Lookie here: http://www.carvinservice.com/crg/sch...0Schematic.pdf

If you look under the box with the board number in the middle of the schematic, you'll see that the screen voltage is connected to 22uF through 47K. This voltage drives V1, which forms separate triodes used as the input receivers for the clean and lead channel. They share 22K in series, which splits to separate 100K plate resistors with no capacitor. Both triodes are amplifying the same signal, but one cathode has a large bypass capacitor. Their plate voltages affect each other through the resistors. I've always wondered if this was intentional, if they forgot the cap and found it sounded good, initially had the cap and removed it, or just had a brain fart.

Now look at V2. V2B has it's own RC supply circuit, but the resistance is 180K, and the capacitance is only 0.047uF. The plate resistance is a meager 22K. Eyeballing it, you'd think it was a dropping resistor and filter cap, but the values are very strange. You look harder, and you might think that at low frequencies, it will look like a 200K plate resistance, and it will drop to closer to 22K at the high end. In fact, at low audio frequencies, the DC component of the voltage at the 0.047uF cap drops something like 100V, the tube clips, and the top of the clip droops. This was a change to a circuit they'd used for years, and they hand-wired it in as an improvement, calling it "Hot Rod Mod 1B" for the remaining 5 years of product life. They switched from 6L6s to EL34s at about the same time. It's in the artwork for the Series IV version, re-born after over 10 years without production. The world hasn't exactly tripped over itself trying to copy it, though it has its fans.

So the dropping resistors and their associated capacitors provide voltage reduction and filtering, unless the design is atypical. I think Mesas gets a bit creative with RCs here too. A capacitor or RC in parallel with a plate resistor can have interesting frequency dependent effects if the cap is sized right. Gain of high frequencies is reduced, but the clipping is still abrupt. If you're actually trying to create distortion, there are lots of ways to do it.

[daz]

Thanks for that RG, tho i was pretty much lost after "The". LOL!

[daz]

Ok, well i'm lost on much of this but it did seem to confirm that they are for voltage and decoupling, which again i *think* means to isolate the nodes. But i still am not sure whether they choose the value purely for voltage or what.

[Gingertube]

The pictures help ...
The Valve Wizard
Cheers,
Ian

[eschertron]

Resistors do drop voltage based on their current consumption. For each 12AX7 triode section, 1ma consumed is a decent thumbrule. So if the first tube (2 triode stages) is on a node, a first iteration guess is there's a 2ma draw on that filter node, dropping voltage across the filter's resistor. If I have a 10k dropper, then 10k X 2ma = 20 volts. The next node up will have THAT load plus the load of its valves. So the values of the resistors can be guessed at that way.

Now for the filter decoupling caps, I read "somewhere" on the internet (I'm not good at noting my references, D'oh!) that for the preamp filter sections, a time constant (R value X C value) should be greater than 0.16 seconds - the number is actually 1/(2pi) - and the author refers to this as a 'stiff' filter node. Once the R value has been chosen based on a target voltage, then the C value can be calculated by the formula C = 1/(2pi*R). So for the 10k resistor, a cap that is = 1/(2pi*10k) or about 16uF is sufficiently large for decoupling.

That's how I understand it, at least.

[R.G.]

Resistors do drop voltage based on their current consumption. For each 12AX7 triode section, 1ma consumed is a decent thumbrule. So if the first tube (2 triode stages) is on a node, a first iteration guess is there's a 2ma draw on that filter node, dropping voltage across the filter's resistor. If I have a 10k dropper, then 10k X 2ma = 20 volts. The next node up will have THAT load plus the load of its valves. So the values of the resistors can be guessed at that way.

Yep. About 1ma per 12AX7 plate powered by a node is a good starting guess. At 1ma, each 1K drops a volt. At one tube (2ma) each 1K drops 2V. A 10K drops 20V.

Now for the filter decoupling caps, I read "somewhere" on the internet (I'm not good at noting my references, D'oh!) that for the preamp filter sections, a time constant (R value X C value) should be greater than 0.16 seconds - the number is actually 1/(2pi) - and the author refers to this as a 'stiff' filter node.

RC = 0.16second is the same as saying F = 1/(2*pi*R*C) = 1/(6.28*0.16) = 1Hz, which is a way of saying that the R and C make a filter that has a "corner" at 1Hz. A 10K/10uF as I pulled out of my... er, out of the air for my example is R*C = 0.1, or a filter rolloff of 1.59Hz.

The reality of it is that there is a fair amount of slop in the calculations, not least that high voltage electros have wide tolerances. So yeah, RC=0.16 is not a bad choice. Often you are forced to use a cap, so you can then pick a resistor to match the RC target, choose the nearest standard resistor, and probably be OK if you get close.