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Need help designing a resistor sequencer.

Discussion in 'General Electronics Discussion' started by elframbo, Aug 20, 2016.

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  1. elframbo

    elframbo

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    Aug 20, 2016
    Hello,

    I'm new to electronics and have started modding guitar effects pedals. I
    need help designing a circuit that will sequence out three resistors. The pedal has a specific resistor that controls gain potential in the circuit.

    I did join the modding community forum as well and someone was nice enough to give me this design.

    http://i174.photobucket.com/albums/w114/quackzed/resistor sequencer_zpsivld4dme.png

    This circuit will be allowing the three resistors from the 4066 chip to switch in and out of the circuit at a controllable rate. I tried to build this on breadboard and used LEDs in place of the resistors R1,2, and 3 that were leading out of the 4066 chip. It did not work and I'm not sure if it was because of using the LEDs or if thee was an issue with the breadboard, a component, or if the design is not a working one.

    This lead me to look at LED sequencers, which have a very simple design, but I'm not sure how to connect those into my pedal's existing circuit. And I'm not sure it would even work properly because instead of LEDs I would use resistors.

    To sum up, the effects pedal has an option to mod out resistor R10 in the pedals existing circuit. I will be removing R10 and would like to replace it with a circuit that will automatically vary the resistance across the point where R10 used to be at a controllable rate.

    I hope this makes sense, as I said I am new to this. Any help is much appreciated.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
    2,820
    Jan 21, 2010
    Please upload your image here. I looked at your image just after you posted the query but because the third party site makes it difficult to return here I could not post my reply (which I have since forgotten, but at least I have found your thread again)
     
  3. elframbo

    elframbo

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    Aug 20, 2016
    AH sorry about that, I have now uploaded the image.

    Thanks!
     

    Attached Files:

  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
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    Jan 21, 2010
    What are the values of R1, R2, and R3? The 4066 has an internal resistance which may be significant if their values are low.

    Do the two circuits (the sequencer and the one where the resistor is being replaced) share a common ground?

    What is the supply voltage for the circuit not shown here? The voltages at the ends of the resistor replacement point must remain within the 0 to 9V range of the power supply of the 4066.
     
  5. AnalogKid

    AnalogKid

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    Jun 10, 2015
    You refer to a resistor R10, but that is not in your image. If that is a part in the original pedal circuit, can you post that schematic?

    Also, there is nothing obviously wrong with the circuit you show, but it will not work with LEDs substituted for R1-R3 for several reasons. Do you want to get into that, or just move on?

    ak
     
    (*steve*) likes this.
  6. elframbo

    elframbo

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    Aug 20, 2016
    Thanks for the replies!
    Steve, the values are R1 = 100k R2 = 220k R3 = 470k. The sequencer will share the same ground and power supply as the pedal (original circuit). The pedal runs on a 9v power supply.

    Analogkid,
    I will attach the schematic for the pedal in this reply. If it is helpful, the pedal is a Daneletro Fab tone. R10 is on the original circuit it is a guitar effects pedal. I do want to know the reasons the leds will not work. But, for now I would like to just move on and I will research that on the side. I want to learn to eventually design my own circuits and effects pedals.

    Also is there a way to make this circuit simpler, or smaller? I ask because I looked into LED sequencers and they do not have the 4066, so is it really needed for this to function as a resistor sequencer? Just trying to shrink it down as much as I can.


    Thanks again for all the help and replies!
     

    Attached Files:

    Last edited: Aug 21, 2016
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
    2,820
    Jan 21, 2010
    The 4066 is a switch (actually 4 of them) designed to allow you to switch things (like resistors in your case) in and out of a circuit.

    A LED sequencer simply provides power to some devices (in this case LEDs).

    Unless your circuit has the resistor connected to a power rail (and the right one for the LED sequencer), a LED sequencer can't be used to switch the resistors in and out of the circuit.

    Your question is a bit like someone asking for a switch and you offering them a lightbulb.
     
    Arouse1973 likes this.
  8. AnalogKid

    AnalogKid

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    Jun 10, 2015
    In round numbers, a typical small red or green LED needs about 2 V or pulsed or steady-state DC across it to light, and the brightness depends on the current through it, usually up to around 20 mA. Nowhere in the signal path of the pedal are there those voltages or currents. R10 is a negative feedback resistor around an opamp, so there is very little current available and not nearly enough DC offset to wake up an LED.

    I haven't seen zillions of effects pedal schematics, but that one seems to be exceptionally complicated. Changing R10 changes the gain and bandwidth of IC1A, which probably has some effect on Q4 but I'm not sure what. The signal through R10 is low level analog audio, so you need a switching device that can do that - an analog switch. LEDs are basically DC devices at medium to high current, and LED controllers are built to handle those signals. While you can control LEDs with a 4066, it is not a good pairing. But you cannot control audio with an LED controller.

    You can replace the 555 and 4017 with a 4060, and use a different analog switch to reduce the chip count from 3 to 2, but that's about it for reductions.

    ak
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
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    Jan 21, 2010
    Another reduction would be to use an 8 pin microcontroller to drive the 4066 and have 4 resistors. You could (with appropriate resistor selection) have 16 values of resistance.
     
  10. elframbo

    elframbo

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    Aug 20, 2016
    Thanks for the information, from what I understand it seems like you are saying that the circuit I was previously given will work. How should I test it on breadboard? Would I connect an ohmmeter across the leads that would connect to the terminals for r10?
     
  11. AnalogKid

    AnalogKid

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    Jun 10, 2015
    If you are talking about testing just the three chips and three switched resistors, then yes, an ohmmeter should show the effects of the mux switches. In the schematic in post #3, connect the meter to the two blue wires.

    ak
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
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    Jan 21, 2010
    You really need to test 3 things.

    Firstly that the 555 is oscillating.

    Secondly that the 4017 is correctly changing state.

    Thirdly, that the resistance is changing.

    I guess there is a fourth, that the change in resistance effects the circuit in the desired manner.

    You started at the last step and are working backward. You could also work the other way. With only 4 steps it doesn't make much difference, but if you start somewhere near the middle you can cut your work in half at each step rather than by a quarter.
     
  13. elframbo

    elframbo

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    Aug 20, 2016
    Hey everyone,

    Just wanted to give you a status update as to where I am so far. I built the circuit on breadboard and my R1 (100k), and R2 (220k) are both 1/4 watt. However the only 470k resistors I have are 2 watt, so R3 is 470k 2watt. I connected a 9V battery for a power supply to test out the circuit and measured resistance as stated earlier. I have a Digital Multi meter and it was set for 20M ohm. The rating on the DMM goes from 200 ohm, 2k, 20k, 200k, 20m. So a setting of 200k gave me an open circuit,

    Having it set at 20M the reading jumped around from 0.60-0.64 it doesn't seem like it is enough of a variance to say the resistors are being switched out. Not sure what I should be looking for with the meter, I suspect the 470k resistor might be causing an issue because the other resistors are 1/4 watt and it is 2 watt.

    I also adjusted the VR1 and tried it at three settings, full on, mid, and full off. The output resistance didn't seem to change much. However, when it was turned fully counter clockwise it did seem to stay at 0.62 with nearly no variance at all.

    Any ideas?

    Thanks!
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
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    Jan 21, 2010
    The resistance may be changing too fast for your multimeter to correctly display. Make sure that your 555 is switching no faster than about once per second.

    For testing, because the multimeter does not have a power supply with a common connection to that of the 4066, connect the common junction of the three resistors to the negative side of your 9V supply.

    Be aware that even with this, you will get different results with the multimeter connected one way vs the other. One will be right and the other will be wrong.
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
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    Jan 21, 2010
    Oh the power rating is of no consequence. The high power resistor is just bigger.
     
  16. elframbo

    elframbo

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    Aug 20, 2016
    Thanks again for the information!

    A quick note, the person that gave me the circuit originally said to switch out R1 (from the 555 timer) for a 10k resistor, and Vr1 with a 100k variable resistor. I overlooked that and did not post that sorry for the omission.

    I also had the 100k in the circuit that I am building and just switched it out for the 10k resistor. Now when I have the variable resistor midway I'm seeing 0.29 - 0.54 on the DMM.

    From that it looks like there was some progress made on the speeds. Now a quick question I want to test this in the pedals circuit. Can I power my build circuit on breadboard with a 9v battery? The pedal will be running on 9v of power as well. Will this cause an issue with two power supplies? Or do I need to tap into the pedals power supply to test this?

    And lastly am I jumping the gun testing it in the pedals circuit?

    Thanks!
     
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
    2,820
    Jan 21, 2010
    There should be no problem running it from the breadboard.

    Just make sure to connect the -ve side of both batteries together.

    If you temporarily connected the junction of the three resistors to ground, disconnect them from ground before connecting them into the pedal.
     
  18. elframbo

    elframbo

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    0
    Aug 20, 2016
    Hey I tested it in the pedal and here are the results so far.

    I first tested with the parts listed including the 100k vr and the 10k resistor leading from the 555 timer to ground. The speed was way to fast. So I tried switching a few things up, changed the 555 pin 7 resistor to power from 10k to 1k, then changed the 100k vr to a 10k pot. Still way to fast and at the high end was just a screech sound/feedback loop.

    Went back to the 100k vr, and changed out the pin 7 to positive power resistor a couple of times I went up to 100k, then up to 470k the highest resistor value I have. This was getting a little closer it was still way to fast, but I could hear the changes.

    I think I need to order more parts to test further thanks for all the help so far. Any other suggestions on the speed?
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,477
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    Jan 21, 2010
    Increase the value of C1. Doubling it will double the time between changes in the switched resistor value.
     
  20. AnalogKid

    AnalogKid

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    Jun 10, 2015
    There are two versions of a 555 astable oscillator circuit, and yours is the classic. Here is an excellent calculator to help you predict its operation.

    http://www.ohmslawcalculator.com/555-astable-calculator

    I don't think you've stated what the effect is that you are trying to achieve, or how fast (or slow) you want the resistors to switch in and out of the circuit. Once you have a guess at that number (something more objective than "way too fast"), the 555 part should be easy.

    For the schematic in post #3, the range of resistor-changes-per-second goes from about 7 to about 50. Increase C1 to 1.0 uF and I think you'll be close.

    ak
     
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