Grid resistors X2

[caps]

Guys,

The resistor on the grid....

ok, it in conjunction with the tube capacitance gives you your input impedance. You have that resistor connected to input load and straight into the grid,

Sometimes though along the way to the grid, a second resistor is connected in series connected to ground.

Im thinking the second one to ground, references the grid to 0V.

Sometimes though, either one or the other is missing. Im trying to find out how they interact together.

 

[pstamler]

First, let's get terms straight. A resistor (or anything else) from the source to the grid is deemed "series". A resistor (or anything else) from the grid to ground (or, for that matter, from source to ground) is deemed "shunt". So you're talking about a series resistor, a shunt resistor and shunt capacitance from the grid to ground. (Actually it's from the grid to the cathode, but if you bypass the cathode resistor with a capacitor then the grid capacitance is effectively in shunt to ground.)

The total shunt impedance at any frequency will be the total effective grid capacitative reactance in parallel with the shunt resistor (if the latter is present). Don't forget to use the root-mean-square method of adding those parallel impedances, since you're dealing with a reactance. And don't forget the effective grid capacitative reactance includes that of the Miller-effect capacitance from grid to plate.

The impedance the grid sees will be the total shunt impedance (previous paragraph) in parallel with the total series impedance (which is the series resistor plus the output impedance of the source, whatever that is).

The impedance the source sees will be the total shunt impedance plus the series resistor.

Peace,
Paul

 

[caps]

Thanks Paul, that makes perfect sense.

Often there will be no shunt resistance in this situation. Wondering why that is?

Often the common "1MEG" resistance you see connected in shunt. I presumed this was to take care of any possible grid current??

Sometimes, the only resistor in the grid circuit is a shunt resistance. This is how examples in the book ("Beginners Guid to tube design" by Bruce Rozenblitz ), are set out.

For such a well written resonably in depth book, Im surprised it dosent mention anything about series and shunt resistance in grid circuits. Its always a simple shunt resistance and away you go and calculate Z.

 

[Rob Flinn]

In the UK the series resistor is sometimes known as a "grid stopper" resistor. The purpose is often to prevent HF oscillation. The body of the grid stopper should be soldered so it is as close to the grid pin on the valve base as possible.

Often the common "1MEG" resistance you see connected in shunt. I presumed this was to take care of any possible grid current??

If the valve is biased correctly there should not be any grid current. The grid should be negative with respect to the cathode & is therefore like a reverse biased diode i.e no current flow. The 1M resitor holds the grid at ground when there is no signal (quiescent state), & the the input signal is developed across it in parallel with the reactances present in the valve.

 

[CJ]

there might be something in the first two chapters ....

http://vacuumbrain.com/The_Lab/TA/RDH4/rdh4.html

 

[PRR]

> If the valve is biased correctly there should not be any grid current.

There's always grid current.

In normal audio systems, it is so very small that it causes "negligible" voltage in a 100K or 1Meg resistor. (And yes, there is a null-point where grid current is dead-zero, but it isn't stable.)

> along the way to the grid, a second resistor is connected in series connected to ground.

It would be a hell of a lot easier to figure out what you are asking if you would point to a picture. "on the grid", "second one", "input load".... huh?

If I get the picture: there is often a 1Meg (or similar very high value) from grid to ground. This drains the small leakage so was can say the grid DC voltage is essentially ground. This shunts the signal source, but most sources are not seriously loaded by 1 Meg.

There is sometimes another resistor, 100 Ohms to 47K, going from the source to the grid. In most audio situations, the grid itself is "infinite" impedance. 10K+infinity= infinity, so input resistance and gain is not affected. In normal use, it does nothing and may be ignored. In overload, often the grid tries to go positive. When that happens, grid resistance drops from infinity to a very low value, about 1K. That can upset the driving circuit. A series resistor keeps the input impedance from sinking very low.

Sometimes (much too often) you find low-value resistors in series with the grid. At audio, they have no effect. But the tube does not know that its job is "audio". Some tubes have gain up to many MegaHertz. At these frequencies, a little piece of wire is a resonant circuit, and the tube wants to take-off and oscillate. A small resistor right at the grid pin adds enough loss to kill the oscillation.

 

[CJ]

"The grid resistor on the preamp stages typically ranges from 0 to 68K, although very large values, such as 470K, are sometimes used in high-gain preamps to shape the frequency response and prevent "blocking" distortion in the preamp section under heavy overdrive conditions. The Miller capacitance of a typical 12AX7 is around 151pF, so the upper frequency response -3dB cutoff point of a stage using a 68K grid resistor is around 15.5kHz. The frequency response drops to around 2.2kHz if a 470k grid resistor is used. This "free" response rolloff can be used to tame the "buzziness" of high-gain preamp stages without having to add additional rolloff capacitors. Perhaps the most important grid resistor is the one that goes to the grid of the very first stage, right after the input jack. This resistor is the one that prevents oscillations and pickup of radio stations and other noise due to long or poorly-shielded cables. It is not usually a good idea to eliminate this resistor. Ideally, it should be soldered directly to the grid pins of the socket, with very short leads."

No, that was not me talkin.

Another use is for multiple input situations, like a Fender amp, when your "buddy" wants to jam. Plug him into Jack #2, and plead stupidity when he asks why he is not as loud or brite as you are. :razz:

 

[caps]

[quote author="PRR"]> If the valve is biased correctly there should not be any grid current.

There's always grid current.

[/quote]

I thought there might be , hence the question. One reason for possible grid current I seem to recall, was that electrons colliding with any gas molecules in the vacuum can result in a positive Ion. Which then would be attracted to the grid.

 

[Rob Flinn]

PRR

> If the valve is biased correctly there should not be any grid current.

There's always grid current.

I stand corrected. From which electrodes does the grid current flow from & to ?

 

[PRR]

> From which electrodes does the grid current flow from & to ?

http://vacuumbrain.com/The_Lab/TA/RDH4/rdh4.html

Chapter 2, section 2.2 paragraph iii

Yes, there are zero-current points, but one is at huge negative grid voltage (and not really zero), the other is a steep crossover that you can't catch without snazzy tricks. So in general, "there's always grid current", though it may be very small.

 

[caps]

tubes 202 is a fantastic lot of info. Best Ive found on the net yet, thanls to previous link from PRR. :thumb: