# why the large resistor to ground?

#### disantlor

Joined Jun 21, 2006
20
First off I want to say hello and introduce myself as this is my first post on the forum. Though I am fairly new to this I hope to be able to contribute to the community as much as I can.

Anyway, I have been reading up as much as I can, particularly on the subject of guitar tube amplifiers, but I must be missing something fundamental. I can't seem to come up with a reason why there is usually a high value resitor between the input and ground on most things that have an input.

A particular example can be seen in the schematic of the P1 guitar amp project: http://annex.ax84.com/media/ax84_m311.pdf

Having built this amp, I'm trying to now go back and understand what's going on. I read the theory document provided on the site (http://195.178.239.50/ax84/media/ax84_m35.pdf) and understand pretty much all of it except that darn resistor (well and the tone stack to some extent). I'm assuming the answer is pretty simple and therefore I must not understand something fundamental so any wisdom on the subject would be greatly appreciated.

#### beenthere

Joined Apr 20, 2004
15,819
Hi,

The big (large value) resistor to ground has two purposes. The input is coupled in through a capacitor. Without the resistor, the tube grid would pick up charge and gradually bias the tube into conduction. This would not be stable, so the amp would have some obnoxious output resulting from the grid floating. The capacitor serves to eliminate any DC offset in the input that might also do strange things to the amp's output (imagine a 250 mv DC component in the input. The amp will have a gain sitting somewhere between 100 and 1000. Multiply the 250 mv by 1000, and you have a situation where the voice coil will try to exit the enclosure at just under supersonic speeds).

The other function of the resistor is to make a reasonable load for the input signal. There is a sort-of standard for signal inputs that suggests something like 1.25 Vrms developed across something like 47K ohms. The 1.25 Vrms is 0 dB. The value of the resistance is to give the input signal a path to ground that won't load it, and to allow signal components to pass, down to some frequency roll-off set by the lo-pass filter action of the input capacitor and resistor.

The component values will be a compromise between the best low-frequency response and a grid reference resistor that allows the input tube to behave predictably.

#### richbrune

Joined Oct 28, 2005
106
Originally posted by beenthere@Jun 24 2006, 08:37 AM

The big (large value) resistor to ground has two purposes. The input is coupled in through a capacitor.
[post=17888]Quoted post[/post]​
I don't see the capacitor, too me it just looks like the input goes from the jack through R24 into the tube, how does that 10k resistor eliminate a dc offset?

#### BhrMzf

Joined Nov 1, 2004
3
disantlor said:
First off I want to say hello and introduce myself as this is my first post on the forum. Though I am fairly new to this I hope to be able to contribute to the community as much as I can.

Anyway, I have been reading up as much as I can, particularly on the subject of guitar tube amplifiers, but I must be missing something fundamental. I can't seem to come up with a reason why there is usually a high value resitor between the input and ground on most things that have an input.

A particular example can be seen in the schematic of the P1 guitar amp project: http://annex.ax84.com/media/ax84_m311.pdf

Having built this amp, I'm trying to now go back and understand what's going on. I read the theory document provided on the site (http://195.178.239.50/ax84/media/ax84_m35.pdf) and understand pretty much all of it except that darn resistor (well and the tone stack to some extent). I'm assuming the answer is pretty simple and therefore I must not understand something fundamental so any wisdom on the subject would be greatly appreciated.

----------------------------------------

Think about charge accumulation. If the input impedance without this resistor is some 10 MOhm and just a small 5 microC of electric charge accumulates on the input trminals, 5 volts of DC voltage would develop across the inputs of the amplifier! Don't you think that it is too much for input?

#### AllVol

Joined Nov 22, 2005
55
I thnik the resistor is merely a "pulldown" resistor which keeps the input jack at ground potential when nothing is plugged in. This prevents stray input.

AllVol

#### nomurphy

Joined Aug 8, 2005
567
The resistor to ground (1 Meg) prevents the input from floating whenever no input is connected. It also sets the input impedance, which typically is desired to be be fairly high relative to the source impedance.

The 10K is a current limit to the input.

#### nomurphy

Joined Aug 8, 2005
567
The 1 Meg resistor prevents the input from floating when no input is connected, a high impedance is desired relative to the source impedance. The 10K current limits the input.

#### disantlor

Joined Jun 21, 2006
20
I guess my post got lost in the switch to the new forum.

Anyway, what I said was that perhaps the capacitor often found in the guitar pickup/output circuitry is the "missing" capacitor. It's just a guess but I noticed from looking through my book on guitar amp schematics that most of them lack this capacitor so it's unlikely that it's a misprint in the schematic.

#### freetek

Joined Aug 10, 2005
8
The grid of a tube will develop a (space) charge which will cause it to conduct in the absence of signal and produce severe distortion of any input signal and can damage the tube.
The resistor (in addition to terminating the input signal line) will drain off this charge and establish the tube's quiescent operating current.

When a capacitor is present (it isn't in your case), there is no path to drain the accumulated (DC) charge without providing an internal path to ground.

#### windoze killa

Joined Feb 23, 2006
605
richbrune said:
I don't see the capacitor, too me it just looks like the input goes from the jack through R24 into the tube, how does that 10k resistor eliminate a dc offset?

#### pebe

Joined Oct 11, 2004
626
disantlor said:
First off I want to say hello and introduce myself as this is my first post on the forum. Though I am fairly new to this I hope to be able to contribute to the community as much as I can.

Anyway, I have been reading up as much as I can, particularly on the subject of guitar tube amplifiers, but I must be missing something fundamental. I can't seem to come up with a reason why there is usually a high value resitor between the input and ground on most things that have an input.

A particular example can be seen in the schematic of the P1 guitar amp project: http://annex.ax84.com/media/ax84_m311.pdf

Having built this amp, I'm trying to now go back and understand what's going on. I read the theory document provided on the site (http://195.178.239.50/ax84/media/ax84_m35.pdf) and understand pretty much all of it except that darn resistor (well and the tone stack to some extent). I'm assuming the answer is pretty simple and therefore I must not understand something fundamental so any wisdom on the subject would be greatly appreciated.

As beenthere said, the 1M resistor is to drain away the static charge that would otherwise accumulate on the grid from electrons thrown off by the cathode. It was called a 'grid leak'. The 10K resistor was often fitted and was known as a 'grid stopper'. It formed a lowpass filter with the input capacitance of the grid and its purpose was to prevent parasitic oscillations at RF.

The normal input coupling capacitor used with a 1M resistor was typically 100nF. I had never before seen a circuit without the capacitor fitted internally. A typo perhaps?

#### disantlor

Joined Jun 21, 2006
20
thanks for all the replies!

let me see if I can try to explain how I understand this to make sure I have a grasp on whats actually going on.

charge can build up on the grid (stray electrons in the tube hit the grid?) and if not dealt with, would mess up the bias of the grid which is obviously bad. The resistor provides a path to ground for these stray electrons.

To avoid grounding out the input, the resistance is set sufficiently high so that only a very small amount of the input current follows the path (current divider) to ground.

one question I have is, if there were no path to ground, why would the electrons be attracted to the grid; would it not effectively "see" infinite resistance and avoid it (or is that notion just an idealization of what really happens).

#### mozikluv

Joined Jan 22, 2004
1,437
pebe said:

As beenthere said, the 1M resistor is to drain away the static charge that would otherwise accumulate on the grid from electrons thrown off by the cathode. It was called a 'grid leak'. The 10K resistor was often fitted and was known as a 'grid stopper'. It formed a lowpass filter with the input capacitance of the grid and its purpose was to prevent parasitic oscillations at RF.

The normal input coupling capacitor used with a 1M resistor was typically 100nF. I had never before seen a circuit without the capacitor fitted internally. A typo perhaps?
hi pebe,

i don't think it's a typo error. i have seen several known brand schems using the 12AX7 as their input stage without the coupling cap. and i agree that a coupling cap should be inplace. as to why there is one or there is none i can't comment on that being tube amps is not my cup of tea.

moz

#### disantlor

Joined Jun 21, 2006
20
disantlor said:
let me see if I can try to explain how I understand this to make sure I have a grasp on whats actually going on.

charge can build up on the grid (stray electrons in the tube hit the grid?) and if not dealt with, would mess up the bias of the grid which is obviously bad. The resistor provides a path to ground for these stray electrons.

To avoid grounding out the input, the resistance is set sufficiently high so that only a very small amount of the input current follows the path (current divider) to ground.

one question I have is, if there were no path to ground, why would the electrons be attracted to the grid; would it not effectively "see" infinite resistance and avoid it (or is that notion just an idealization of what really happens).
first off I wanted to make sure my logic is correct in the above quote. Second I've been reading Art of Electronics and I came across a similar problem, but with Op Amps this time. The text reads as follows:

"Once again we have a dc amplifier. If the signal source is ac-coupled, you must provide a return to ground for the (very small) input current, as in Figure 4.6:"

(figure 4.6)

Could someone maybe explain whats happening on a particle/microscopic level so I can understand how the electrons are flowing/ why this resistor is necessary.

Thanks again!

#### pebe

Joined Oct 11, 2004
626
disantlor said:
first off I wanted to make sure my logic is correct in the above quote. Second I've been reading Art of Electronics and I came across a similar problem, but with Op Amps this time. The text reads as follows:

"Once again we have a dc amplifier. If the signal source is ac-coupled, you must provide a return to ground for the (very small) input current, as in Figure 4.6:"

http://filebox.vt.edu/users/mdisanto/public/ex4.6%20pg179.gif (figure 4.6)

Could someone maybe explain whats happening on a particle/microscopic level so I can understand how the electrons are flowing/ why this resistor is necessary.

Thanks again!
I missed the last sentence of your previous posting that asked how the grid became charged. The reason is this. The cathode is heated to a high temperature and that causes electrons to be thrown off. They are attracted by the positive charge on the anode and this gives rise to the anode current - the amount of this being controlled by the potential on the grid. But the grid is more negative than the cathode and so repels some of the electrons to give a cloud of them in the space between cathode and grid (if I remember correctly it's known as 'space charge'). Some of these electrons will land on the grid and would make it negatively charged if it were not for the resistor fitted - hence its name of grid leak.

The op-amp is a different case because they use semiconductors. Your drawing is of an FET input type. The resistor of the input gate is extremely high and the gate is effectively 'floating'. Without the resistor the gate potential would be affected by static voltages, leaks across the PC board, and leakage through the coupling capacitor. So it is necessary to provide the resistive path to ensure correct biasing.

In the case of a bipolar inputs, The resistor would be to the +ve rail because NPN transistors are usually used in the input stages. In that case the resistor supplies current for biasing the base.