That 1 Meg resistor at an amp's input jack...

Discussion in 'Amps/Cabs Tech Corner: Amplifier, Cab & Speakers T' started by guitarcapo, Mar 12, 2011.

  1. guitarcapo

    guitarcapo Member

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    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?
     
  2. Blue Strat

    Blue Strat Member

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    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).
     
  3. guitarcapo

    guitarcapo Member

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    So a higher value resistor make for more gain?

    I read THIS somewhere:

     
  4. Blue Strat

    Blue Strat Member

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    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.
     
  5. Ronsonic

    Ronsonic Member

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    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.
     
  6. brad347

    brad347 Member

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    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:

    [​IMG]

    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.
     
  7. brad347

    brad347 Member

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    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.
     
  8. guitarcapo

    guitarcapo Member

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    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?
     
    Last edited: Mar 12, 2011
  9. brad347

    brad347 Member

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    add that resistor. The input signal has to know that ground "exists." It does so through that resistor... it's your ground reference.
     
    Last edited: Mar 12, 2011
  10. guitarcapo

    guitarcapo Member

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    Hard to replace it when it was never there. ha ha
     
  11. brad347

    brad347 Member

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    gotcha. edited to fix.
     
  12. 1guitarslinger

    1guitarslinger Member

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  13. brad347

    brad347 Member

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    good catch. Thanks for setting me straight.
     
  14. 1guitarslinger

    1guitarslinger Member

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    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.
     
  15. 1guitarslinger

    1guitarslinger Member

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    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.
     
    Last edited: Mar 12, 2011
  16. brad347

    brad347 Member

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    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.
     
  17. Structo

    Structo Member

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    Grid Leak resistor
     
  18. 1guitarslinger

    1guitarslinger Member

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    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.
     
  19. Prattacaster

    Prattacaster Member

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    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
     
  20. 1guitarslinger

    1guitarslinger Member

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    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.
     
    Last edited: Mar 13, 2011

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