Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: 12AX7 on January 05, 2012, 12:33:55 am
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Lets say you come off the coupler from the last 12AX7 stage and from there a grid stopper to the next stage with a resistor from the cap also going to ground. On the other hand you can take that resistor to ground and remove it from the coupling cap and put it along with the grid stopper right to the grid. Now it becomes a voltage divider. But theres still that same resistor from the last stage to the grid. I realize the VD drops voltage gain by 50%, but the other the other scenario can be made to match the same amount of gain drop by adjusting that ground resistor.
What i'm trying to understand here in as laymans terms as possible is, what is the difference here? The VD has one of the resistors from last stage to the grid of the next the same as if it weren't part of a VD. So does it still have the same miller capacitance effect and any other effects that a grid stopper has like taming possible blocking or oscillations and all that? And what other differences can i expect between those 2 scenarios. assuming i adjusted the values in each to match thier gain properties? I DO realize they both attenuate highs unless a bypass cap is used.
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Hi 12ax7,
The following URL may help you
http://www.freewebs.com/valvewizard/gridstopper.html (http://www.freewebs.com/valvewizard/gridstopper.html)
Regards!!
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I've read that before but It doesn't address my question. The closest it gets is saying the divider attenuates the signal, but as i said i put a resistor to ground from before the grid stopper if i'm not putting it after to create a VD, and that resistor can also adjust gain with it's value. So the gain aspect is covered. I just want to know what the differences I'd get doing it each way. To clarify i'm taking about 2 resistors creating a VD between a coupling cap from the previous stage to the grid of the next verses the same 2 resistors but lifting the grid end of the one thats grounded and solder that to the coupling cap. I've tried it both ways and i like the way the highs drop with the VD better, but the tone also changes in a way i can't put my finger on and i'm trying to understand whats happening there, and in layman's terms. The VW site leaves be dumbfounded for the most part because when it comes to electronic theory i AM dumb.
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Definitions:
Grid stopper: a series resistor from signal source to the grid of a tube. The "output" of the resistor must be the grid.
Grid reference resistor: a resistor from grid to ground.
It sounds like you're asking why a voltage divider (coupling cap -> series resistor -> grid reference resistor, with output between series and grid reference resistors) reduces signal level, but a grid stopper and grid reference resistor don't. After all, the latter scenario looks just like a mirror image of the former.
That's because the grid is an open circuit; there's no path for current to go anywhere under normal conditions.
Look up Ohm's Law and Kirchoff's Rule for series circuits. Resistors don't just "divide voltage".
For division to occur, and current must flow through series resistors. The current results in voltage drops across each resistance, in proportion to their resistances. The typical voltage divider has an a.c. voltage across it (from coupling cap to ground), which forces a current through the resistors, which then define a voltage drop across each resistance. If you take your output from the junction of the two resistors, then the output a.c. voltage is smaller than the voltage applied to the input of the voltage divider.
When a coupling cap output is applied to a grid reference resistor to ground, and then on to a series grid stopper, the only path to ground is through the grid reference resistor. The "output" of that resistor is the same as its input, and there is no voltage division. The grid stopper does not divide the voltage which is then passed to the grid, because there's no path from grid to ground, no current, and therefore no voltage drop.
You could take an alternate view, and assume there "must" be some kind of current flowing through the grid stopper and into the grid. If no current flows, the apparent resistance of the grid is infinite. If voltage is divided by the ratios of the resistances, and the grid stopper is much less than infinity (and it is), then all the applied voltage is left across the grid. In other words, no voltage drop.
The basics are very important. Ohm's Law, Kirchoff's Laws, the behavior of series, parallel and series-parallel circuits are key building blocks you can't escape if you really want to know how this stuff works. Once you know these items (and how they work with a.c., capacitors and inductors), then most of the function of tube circuits becomes obvious, or very easy to learn quickly.