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guitarcapo
03-12-2011, 08:15 AM
What does it do? I've heard it called a "reference point" but it seems weird to me because the signal is so weak there how such a high value resistor does anything. Does changing the value do much?

Blue Strat
03-12-2011, 08:37 AM
It creates an input impedance for your signal. Check out "voltage dividers" online. Your pickups have impedance/resistance which forms a voltage divider with the 1M resistor. The higher the resistor, the more signal/voltage gets dumped at the input of the first stage allowing the amp to create more overall gain (which is amplification, not overdrive as per the common misunderstanding).

guitarcapo
03-12-2011, 08:39 AM
So a higher value resistor make for more gain?

I read THIS somewhere:

The 1 meg resistor is the “grid load” resistor for the first tube stage of the amplifier. All tubes have to have a grid load resistance or they will not operate

Blue Strat
03-12-2011, 08:42 AM
So a higher value resistor make for more gain?


To a point. 1M has proven to be optimal so I wouldn't change it, but feel free to experiment. Lower values will definitely be noticeable...at least on a scope.

Ronsonic
03-12-2011, 09:25 AM
What does it do? I've heard it called a "reference point" but it seems weird to me because the signal is so weak there how such a high value resistor does anything. Does changing the value do much?

This has been discussed here before, and pretty heatedly.

That R references the grid to ground when nothing is plugged in (good housekeeping) and represents most of the load resistance that your pickups or pedals see when plugged in.

The only change in sound you'll get is to load down the pickups if you go too much smaller. Going larger might make the guitar a little brighter depending on cables and a bunch of other factors.

It does not set gain. I'd look elsewhere for tweaking opportunities.

brad347
03-12-2011, 09:29 AM
In audio circuits like guitar amps, sometimes we're concerned with the best voltage transfer, and sometimes we're concerned with the most efficient power transfer.

An example of the latter would be matching impedances between the amp's output and the speaker load. Matched impedances gives the most efficient power transfer. Since we're doing work (actually physically moving the speaker cones to create the sound), power is the big concern.

However, when you have a tiny guitar signal in millivolts that you wish to amplify, we're more concerned with getting as much of that voltage to the first amplifying stage of the amp as is practical--there's not a lot of work done at this stage.

Since fluctuations in voltage (not power) are what's used to represent the audio, voltage transfer becomes more important than power transfer. We do this by carefully setting our impedances, and the 1M resistor is part of what helps that happen.

To understand this better, as Mike K said, look up 'voltage divider.' The circuit is simple, and looks like this:

http://www.beam-wiki.org/w/images/8/83/Voltage-divider.png

In this graphic, V(in) is the voltage input, and V(out) is the resultant output. By varying R1 and R2, we can control how much voltage is transferred to the output.

The higher the value of R1, the lower the voltage on the "out" side. As you lower R1, voltage on the "out" side goes up. Fairly intuitive, and easy.

R2 is between "out" and ground. The lower the value of R2, the more voltage is bled off to ground. A high resistance at R2 makes the path to "out" more attractive.

by varying R1 and R2, we can vary the output voltage... best case scenario is for V(out) to approach (never exceed) V(in).

In this scenario, there's a voltage divider created, as Mike said, between your guitar's impedance and the 1M resistor. In this scenario, the 1M resistor is "R2" in the graphic above. The higher this resistor is compared to R1 (guitar's output impedance), the greater the voltage transfer from the guitar to the amp. Such a drastically mismatched impedance would not be good for power transfer, but then we're not doing "work." We're only concerned with the voltage.

brad347
03-12-2011, 09:30 AM
It does not set gain.

Correct. Being part of a passive voltage divider, it can offer no positive gain... it will also never be able to offer unity gain (though we can get practically 'close enough').

However, it CAN, if the value is too low, cause signal loss.

guitarcapo
03-12-2011, 10:34 AM
Well to get to my reason for asking:

I built a Matchless Spitfire recently. A conversion of a Knight PA with similar tubes and transformers.

I turned it on last night for the first time and the signal (sound at the speaker) was very weak. No distortion...just very little volume. I did a quick check and noticed I forgot this resistor. It was getting late and I had to track down the part. I'll probably track down the part and install it over the weekend but it got me thinking whether a resistor that high in value in a place where the signal is so weak would make that much difference from not being there at all.

But then I read statements like the one I cut and pasted from Dan Torres site:

"The 1 meg resistor is the “grid load” resistor for the first tube stage of the amplifier. All tubes have to have a grid load resistance or they will not operate"

....and I'm thinking that might be it?

brad347
03-12-2011, 10:54 AM
add that resistor. The input signal has to know that ground "exists." It does so through that resistor... it's your ground reference.

guitarcapo
03-12-2011, 11:00 AM
Hard to replace it when it was never there. ha ha

brad347
03-12-2011, 11:06 AM
gotcha. edited to fix.

1guitarslinger
03-12-2011, 11:47 AM
Some amps do not have that 1 meg resistor. The 5F10 Harvard and the old GA5 come to mind.

http://www.ampwares.com/schematics/harvard_5f10.pdf

http://www.gibson.com/Files/schematics/GA-5.pdf

The signal realizes ground through the guitar's circuit.

My guess is that the OP's amp has another issue.

brad347
03-12-2011, 12:14 PM
good catch. Thanks for setting me straight.

1guitarslinger
03-12-2011, 12:36 PM
good catch. Thanks for setting me straight.

No problem.

I believe that resistor is only absolutely necessary when certain types of microphones and transducers are used. But with good, old-fashioned electric guitars, the guitar's electronics act as the grid load resistor.

1guitarslinger
03-12-2011, 01:50 PM
The voltage divider analogy is not correct. The guitar's electronics are in parallel with that grid resistor, not in series with it.

Drawn as a schematic, the guitar would be shown as an AC signal source with one side going to the tube's grid and the other side to ground. For DC purposes, the signal generator symbol could be replaced with a resistor.

brad347
03-12-2011, 03:04 PM
Guess it never hurts to make a fool of yourself once in awhile if it means you learn something from it. Thanks! Sorry for spreading misinformation... I hate that. I won't delete my posts though, just so others can learn from my erroneous thinking.

Structo
03-12-2011, 03:24 PM
Grid Leak resistor

1guitarslinger
03-12-2011, 04:00 PM
Guess it never hurts to make a fool of yourself once in awhile if it means you learn something from it. Thanks! Sorry for spreading misinformation... I hate that. I won't delete my posts though, just so others can learn from my erroneous thinking.

Nobody made a fool of themselves, and we all make mistakes in public. Fact of life.

I think that "group think" is easy to get going on message boards. I also think that we all have difficulty visualizing the relationship between the guitar and the amp electrically, because we view them as such separate devices physically. It's as if we think that the guitar cable "isolates" the two from each other rather than connecting them to each other making them one electrically.

Prattacaster
03-12-2011, 06:18 PM
Like structo said it's the grid leak resistor, sometimes(like a lot of fender low/high inputs) it also doubles as the "R2" of a voltage divider.
Back to what structo simply stated, grid leak resistor doesnt mean much to a person if they are wondering what the resistor actually does in the circuit. Just knowing what its called doesnt offer much of an explaination. So ill try to simplify my take on what this resistor does.
People say this resistor has an effect on the "Load" that the tube grid has. Load was always a funny word to me so I tried to think why it was called "load" and it really does make sense. Think of this as a buffer between your signal and ground and also the tube's grid and ground. Its easy to see that the less resistance to ground we have the more likely the signal will be able to leak to ground. At the same time the grid will be closer to ground if we make this R a lesser value. For maximum signal we want the preamp to be as far away from ground as practical. 1 meg seems to be a great compromise because as for resistors go, the higher the value, the more noise we could introduce into the circuit. But at the same time it keeps us far from the ground reference point. Being close to ground would weigh down or "load" the grid since it would be closer to ground. Sure, even without a resistor we get some signal though but far less than if we place some resistance between ground and our signal

1guitarslinger
03-12-2011, 07:00 PM
Like structo said it's the grid leak resistor, sometimes(like a lot of fender low/high inputs) it also doubles as the "R2" of a voltage divider.
Back to what structo simply stated, grid leak resistor doesnt mean much to a person if they are wondering what the resistor actually does in the circuit. Just knowing what its called doesnt offer much of an explaination. So ill try to simplify my take on what this resistor does.
People say this resistor has an effect on the "Load" that the tube grid has. Load was always a funny word to me so I tried to think why it was called "load" and it really does make sense. Think of this as a buffer between your signal and ground and also the tube's grid and ground. Its easy to see that the less resistance to ground we have the more likely the signal will be able to leak to ground. At the same time the grid will be closer to ground if we make this R a lesser value. For maximum signal we want the preamp to be as far away from ground as practical. 1 meg seems to be a great compromise because as for resistors go, the higher the value, the more noise we could introduce into the circuit. But at the same time it keeps us far from the ground reference point. Being close to ground would weigh down or "load" the grid since it would be closer to ground. Sure, even without a resistor we get some signal though but far less than if we place some resistance between ground and our signal

I have to respectfully disagree with the majority of your post. See my previous posts, and read up on what those 68k resistors do. And though that 1 meg resistor does leak a teeny-tiny amount of current and does develop a voltage across it, we are not dealing with grid-leak bias so grid-leak is not its function.

At first glance, the arrangement may seem a voltage divider, and theoretically it is, but that's not the purpose. The purpose of those 68Ks are to isolate both inputs from each other when both inputs are used at the same time. The function is similar to what the 270Ks after the volume controls are doing in this 5E5A schematic. http://www.ampwares.com/schematics/pro_5e5a.pdf

When only the #1 input is used, the resistors are in parallel, making the value 34K. When both are inputs are used, each is 68k. Whether 34k or 68k, the value is so much lower than 1 meg that it is insignificant in the voltage divider scenario. If the #2 input is used by itself, the #1 input's isolation resistor is a 68k path to ground, which loads down the signal substantially, as if your guitar's volume is rolled back.

With amps that have only one input, those isolation resistors are not necessary, and not used unless the builder does not understand their purpose.

Not to sound harsh, but the number of amplifiers that exist that do not have that grid resistor or isolation (summing resistors? mixer resistors? I forget) are proof that much of your post is incorrect.

Want to test this for yourself? Take your favorite amp, lift that 1 meg resistor and see what does or does not change. Then short across the 68k on the #1 input, plug into the #1 input, and see what does or does not change. It's very doubtful that you will notice any change in the tone, gain, noise, etc.

1guitarslinger
03-12-2011, 09:09 PM
Amplifiers without the grid (damper resistor? I've forgotten)

http://www.ampwares.com/schematics/harvard_5f10.pdf

http://www.gibson.com/Files/schematics/GA-5.pdf

And an amp without the 68k (mixer resistor? Again, I've forgotten)

http://www.prowessamplifiers.com/schematics/misc/Trainwreck_Express.html

It is hoped that the this faulty voltage divider mis-info can be dropped and that all can move on.

If in doubt, research this fully for yourselves...

guitarcapo
03-12-2011, 09:36 PM
Well I fixed the problem and it wasn't the resistor. It was a "cold solder" at V2.

It's up an running and sounds really good but:

1. I discovered that the 1Meg pots I used were linear taper (oops)
2. I didn't have 68K input resistors so I used some 25K ones. The gain is OUTRAGEOUS...a little too much actually....I'll probably get the right value ones in there to get a less metal sound going. Maybe that's just how a Matchless Spitfire is.
I might try lower a gain V1 tube.

3. There's a bit of hum. I'm thinking maybe increasing the input resistors will help but a lot of it might be lead dress issues that will be hard to fix because
of the space and how it was retro fit in. I might play around with shielded cable in places. Increasing filtering didn't help. The box probably would do better closed up for shielding too.

Still a fun amp.

pdf64
03-13-2011, 04:49 AM
The 68k resistors used on the amp inputs serve as mixers and grid stoppers. As above, they aren't part of a potential divider arrangement with the 1M, but rather are between the 1M and the tube grid (not between the guitar and the 1M).
Grid stoppers work with the tube's internal capacitence to limit the bandwidth / roll off frequencies above ~18kHz.
It's best practice to fit a grid stopper even with single input amps, in order to roll off radio frequency interference that finds its way into the chassis and also to make the circuit resistant to oscillation at high gains. So while some 50s designs left them out (resistors were expensive then!), most amps since then have them.
I'm happy to call the 1M a grid leak resistor, even though grid leak bias is not being used, as it's the value of the resistor that determines whether the grid leak effect becomes significant in the tube's bias scheme; and it has to be called something to distinguish it from the other resistors in the grid circuit, eg grid stopper, mixer. AKA grid ground reference resistor?
For further detail of how the 1M resistor, guitar lead etc affect pickup tone, see
http://buildyourguitar.com/resources/lemme/
Pete.

1guitarslinger
03-13-2011, 09:20 AM
Ken Fischer referred to that grid resistor as a grid return resistor.

Here is something that Fischer wrote that might help a little bit with understanding its purpose.

THE GIBSON GA-5 This is a Gibson most like a Fender Tweed Champ. It uses one 12AX7, one 6V6, and one 5Y3. The GA-5 has two inputs, using one as a normal gain and the other as a low gain by use of a 47K shunt resistor. There isn't the usual grid return resistor on the input stage. Your guitar's pickups and controls act as this resistor. This means the shorting connection on the high gain jack must close when nothing is plugged in or the tube can "run away." Also, when using a mike for harmonica you should add a grid return resistor to the circuit. 5.6 meg ohms works fine. The amp has just one control besides the on-off switch. This is volume. The amp uses an 8" speaker but unlike the Fender Champ, this one is 8 ohms.

jay42
03-13-2011, 11:28 AM
5.6M...showing his past. Most of the Ampeg V series amps used 5.6M to ground and no grid stopper...mine used to pick up radar sweeps. The common PCB (V-2, V-4, V-4B, VT-22, VT-40) has pads for both a shunt cap and a grid stop, iirc. Easy mods.

phsyconoodler
03-13-2011, 12:07 PM
One thing you have to realize is that electricity follows the path of least resistance.Lower the value of the 1 meg resistor buy more than half and some of the AC signal does not get to the tube,just like the low input on a hi/lo input arrangement on a fender amp.
Take the amps that don't have the 1 meg resistor and add one and the amp is louder.
If that isn't a voltage divider then it's still confused in it's description.I have left them out on a few amps by mistake at first and not one has caused the tube to 'run away' like it's 'supposed' to do.
Now if you lift the ground of the jack it can by introducing RF to the grid of the tube.Try plugging your guitar cable into the input with no guitar and turn the volume up high and leave the room.Sometimes the amp will run away and start squealing and howling.Not so when the cable isn't in and the grid is grounded.No 1 meg resistor in sight.
So in my experience the tube does not need to see a 1meg to ground grid resistance at all to keep it stable.
And that comes right back to what it's real purpose is.
I am not well versed in the theory of it,but I do know it acts a lot like a voltage divider to me.

guitarcapo
03-13-2011, 12:26 PM
The amp was sounding way too high gain so I took some voltage readings and the V1 tube is seeing a plate voltage of 264 instead of 150. The B+ voltages seem fine and the resistor seems to be reading right. Is it possibly a power rating issue of the plate resistor?

1guitarslinger
03-13-2011, 12:42 PM
One thing you have to realize is that electricity follows the path of least resistance.Lower the value of the 1 meg resistor buy more than half and some of the AC signal does not get to the tube,just like the low input on a hi/lo input arrangement on a fender amp.
Take the amps that don't have the 1 meg resistor and add one and the amp is louder.
If that isn't a voltage divider then it's still confused in it's description.I have left them out on a few amps by mistake at first and not one has caused the tube to 'run away' like it's 'supposed' to do.
Now if you lift the ground of the jack it can by introducing RF to the grid of the tube.Try plugging your guitar cable into the input with no guitar and turn the volume up high and leave the room.Sometimes the amp will run away and start squealing and howling.Not so when the cable isn't in and the grid is grounded.No 1 meg resistor in sight.
So in my experience the tube does not need to see a 1meg to ground grid resistance at all to keep it stable.
And that comes right back to what it's real purpose is.
I am not well versed in the theory of it,but I do know it acts a lot like a voltage divider to me.

Again, much of what you have written is simply incorrect.

It's clear that after all of the info posted, you still do not understand the purpose of the grid return and mixer resistors. All I can suggest at this point is to hit the books.

EDIT - My suggestion to everyone is to not take anything that I, or anyone else says as "the gospel." Research things for yourself. Experiment with things for yourself.

1guitarslinger
03-13-2011, 12:43 PM
The amp was sounding way too high gain so I took some voltage readings and the V1 tube is seeing a plate voltage of 264 instead of 150. The B+ voltages seem fine and the resistor seems to be reading right. Is it possibly a power rating issue of the plate resistor?

What is the actual B+ of your amp compared to what is expected?

Blue Strat
03-13-2011, 12:44 PM
Correct. Being part of a passive voltage divider, it can offer no positive gain...

That wasn't my point. The point was that, taken as a black box INCLUDING that resistor, the resistor value WILL affect the gain of the entire black box if it's beyond an acceptable value (too low, loading down the input source).

I stand by my statement.

pdf64
03-13-2011, 12:54 PM
The 1M resistor is there to ensure that the grid always has a dc path to ground. 99.9% of things that might be plugged in have their own dc path to ground, as it's good design practice. However some old fx pedals don't (see an old rangemaster schematic
http://www.webphix.com/schematic%20heaven/www.schematicheaven.com/effects/ggg_dallas_rangemaster_pnp.pdf
with no 1M resistor, the grid would generate a positive voltage and screw the bias up.
The Valve Wizard describes this process in http://www.freewebs.com/valvewizard1/Common_Gain_Stage.pdf
'During normal operation the grid becomes hot (since it is close to the
cathode) and will emit a few electrons, and so will become positively charged unless
a leakage path is provided to replenish the charge lost. If the grid is allowed to
charge positively, anode current will increase, which increases the valve’s
temperature, making the problem worse. Eventually the bias would drift out of
control and, particularly in the case of power valves, this could easily cause red
plating, thermal runnaway and destruction of the valve. The grid-leak resistor
provides the necessary leakage path for electrons between grid and cathode, and
holds the grid at a fixed quiescent voltage (in this case zero volts). In modern
terminology it might be called a ‘pull-down’ resistor.'

The amp was sounding way too high gain so I took some voltage readings and the V1 tube is seeing a plate voltage of 264 instead of 150. The B+ voltages seem fine and the resistor seems to be reading right. Is it possibly a power rating issue of the plate resistor?
Maybe - what is their power rating? What's the cathode and grid voltage?
Pete.

guitarcapo
03-13-2011, 01:11 PM
Cathode voltage is 1.39 volts which seems about right. The schematic calls for 1.26. I can't get a voltage off of pin 2,7 which I think is the grid(s).
The 220K resistor is 1/2 watt...too low to handle things?

stancotey
03-13-2011, 01:12 PM
Also, a "passive voltage divider" can certainly change gain. It is quite easy to get a gain of 0.5 from a voltage divider circuit. In electronics, gain does not always mean increase.

guitarcapo
03-13-2011, 01:35 PM
Plate voltage at power tube=337 and 340 (about right, schematic says 348)
cathode of PI= 59.5 (schematic says 62)
Cathode of power tubes =11v (about what the schematic says)

plate voltage at the PI is 230 (227 and 233 actually..schematic says 250)

sixstringslut
03-13-2011, 01:48 PM
Has anyone noticed that resistor curcuit is open when cable pugged in? It is only there to reduce noise when cable removed. Some very well made boutique amps don't have a 1m resitor.

Or did I totaly misunderstand the OP?

nateguitar
03-13-2011, 04:20 PM
From Designing Tube Preamps, it is a grid-leak or pull down resistor. It is there to provide a leakage path for electrons when the grid becomes hot (and emits electrons) from proximity to the cathode. This is to prevent + grid build up and is more important to fixed bias amps than cathode biased.

GT100
03-13-2011, 07:49 PM
The 1M resistor is there to ensure that the grid always has a dc path to ground. 99.9% of things that might be plugged in have their own dc path to ground, as it's good design practice. However some old fx pedals don't (see an old rangemaster schematic
http://www.webphix.com/schematic%20heaven/www.schematicheaven.com/effects/ggg_dallas_rangemaster_pnp.pdf
with no 1M resistor, the grid would generate a positive voltage and screw the bias up.
The Valve Wizard describes this process in http://www.freewebs.com/valvewizard1/Common_Gain_Stage.pdf
'During normal operation the grid becomes hot (since it is close to the
cathode) and will emit a few electrons, and so will become positively charged unless
a leakage path is provided to replenish the charge lost. If the grid is allowed to
charge positively, anode current will increase, which increases the valve’s
temperature, making the problem worse. Eventually the bias would drift out of
control and, particularly in the case of power valves, this could easily cause red
plating, thermal runnaway and destruction of the valve. The grid-leak resistor
provides the necessary leakage path for electrons between grid and cathode, and
holds the grid at a fixed quiescent voltage (in this case zero volts). In modern
terminology it might be called a ‘pull-down’ resistor.'


Maybe - what is their power rating? What's the cathode and grid voltage?
Pete.


Here is a practical demo of this effect:

I bought one of the very first Rev C Guytrons way back in 1999.
In 2000 Guy sent me parts to do the X-mod to address the short cummings I felt the amp had.
While the (original) X-mod addressed my concerns it also induced a few new problems. One of which was "scratchy" pots. This was due to a bit of DC on the grids of the tubes.
The fix was to simply put 1Meg resistors in key positions at various tube grid to ground positions. This addressed the noise when you dialed certain (gain) pots and had no (negative or positive) effect on tone...

Lloyd