help/advice with transistor switch circuit

Discussion in 'General Electronics Chat' started by minkey01, Sep 20, 2016.

  1. minkey01

    Thread Starter Member

    Jul 23, 2014
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    Hi! Could someone look at this new circuit I have? It was built from this original circuit as reference. See attached 2 pics.

    The jack takes a synth gate which is just an off/on DC signal. The timing is about what a key held down on a piano or synthesizer would be. The voltage amount just depends on which instrument model you would be using. Typically it is 5V, but other common gates are from 2V to 10V. I would like the circuit to be protected from -30V to +30V though, because these could be accidentally plugged in with synthesizers. The output of this circuit is to go into a Teensy 3.1 micro-controller just as a off/on signal.

    My specific question is about my new jack. It is not a switched jack like the original. The new circuit is losing that ground connection from the left side of the R12 resistor. What would happen to my new circuit with nothing plugged into the jack and the left hand of the R1 resistor just floating?

    Thanks for your time!

    new gates circuit.jpg original gate in.jpg
     
  2. Ylli

    New Member

    Nov 13, 2015
    26
    7
    Floating or shorted, there is not going to be any forward bias on the transistor base. Without the grounding jack, the circuit *may* be noise a bit more noise sensitive, but it would take very poor layout and a lot of noise to turn the transistor on. You should be good to go.
     
  3. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,518
    1,247
    You are correct, floating bases are asking for trouble. Safest thing is to decrease R1 to 33 K and add a 100 K from the base to GND.

    For good schematic practice, add a dot to all connecting net lines. In your "new circuit" drawing it looks like the center point of the diodes is not connected to anything. Better yet, stagger the diodes so there never is a 4-way cross.

    ak
     
  4. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    I don´t think that´s true, because leakage of the diodes and the transistor BC junction will have effect and can bring the transistor much closer to turning on.
    I would change R1 to 4k7, add another resistor R2=100k from the right side of R1 to ground, and add another resistor R3=100k between the diodes and the transistor. This will make sure that there is something to keep stray signals away (albeit not as good as the original did), have only a bit different threshold than the original, and keep the base impedance roughly the same.
     
  5. minkey01

    Thread Starter Member

    Jul 23, 2014
    171
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    Awesome! Thanks guys for the quick and great advice. Fixing the floating base now.

    Enjoy the rest of your day!
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    At +30 V in, thats around 6 mA of input current, maybe more than the source wants to supply.

    ak
     
  7. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
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    Well that is true, we don´t know what those devices are and what can the withstand. However, when you insert the jack into the original it will be shorted until you push it in far enough to disconnect the grounding contact, and that I think would be far harsher even if it were just for a brief moment.
     
  8. minkey01

    Thread Starter Member

    Jul 23, 2014
    171
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    I couldn't find any info on the typical amount of amps the source GATE would like to supply. Only some info about impedance should be :

    Signal Input Impedance : 100K
    Signal Output Impedance : 1K

    GATE OFF : 0V GATE ON: 5V
    THRESHOLD : 1.5V

    I did find this other circuit with a web search which is a gate buffer. See attached. Do you think it is better than mine? Do you think this one is protected from -30V or +30V? Leakage on this one?

    Thanks!!

    https://synthnerd.wordpress.com/2016/03/17/synth-diy-gate-buffer/

    gate buffer circuit.jpg
     
  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    It is non-inverting, which may or may not be an advantage. Also, it has a relatively high output impedance when pulling the next stage low. This won't work for some devices, such as older TTL and LSTTL gate inputs.

    ak
     
  10. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    See
    2SD2704K.png
     
  11. minkey01

    Thread Starter Member

    Jul 23, 2014
    171
    0
    last question ;) : back to my first post (new circuit), do you know what the maximum voltage protection is with the 511-BAT42 diodes? that is at the jack input.

    thanks!!
     
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