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Analog Switch question

Discussion in 'Electronic Basics' started by MRW, Jun 27, 2007.

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  1. MRW

    MRW Guest


    I have an analog switch question. For this example, I would like to
    use Texas Instrument's TS5A3159A SPDT Analog Switch.

    For the digital input (triggers the switch ON or OFF), it has the
    following electrical characteristics:

    Input Logic High - 2.4 to 5.5 V
    Input Logic Low - 0 to 0.8 V

    I have the following questions:

    1. What happens to the switch at 0.8 < Voltage at digital input < 2.4?
    Does the switch either turns ON or OFF?

    2. What happens to the switch if I leave the digital input open?

    3. What other ways can I use to control the digital input
    electronically? As of now, I can only think of using a comparator.

    4. Would it be okay to leave the NC or NO terminal floating until a
    signal source is connected to it?

  2. Possibly. Or possibly leaks a bit of signal through a high
    resistance, and sucks power supply current. It is an ill
    defined operating point, so generally you try to get through
    it pretty quickly.
    The input leakage current spec is a couple nanoamperes at
    room temperature, and in either direction, which implies a
    very high input impedance. If left floating, the voltage on
    the input is indeterminate and may float to anywhere between
    the supply rails, and if any static electricity is nearby,
    it may get driven a bit beyond that. This is not a good
    idea. If it is possible that the input may be disconnected
    from its logic signal, even briefly, with the power on, I
    would connect a pull up or pull down resistor that is a low
    enough resistance to make sure the input voltage stays on
    one side or the other of the .8 t o2.4 volts no man's land.
    For instance, if you want it to default to the low voltage
    state (less than .8 volts with the absolute maximum leakage
    current of 100 nA trying to pull it positive) I would
    connect a .8V/100nA=8 meg resistor from input to ground.
    Actually, I would probably use a 1 meg resistor, since that
    would add an insignificant load to the logic driver.
    Logic gates and inverters can also be used.
  3. MRW

    MRW Guest

    Awesome, thanks again for your help, John! :)
  4. MRW

    MRW Guest

    I actually have another question regarding analog switches. I have
    this image:

    I'm trying to convert the voltage reference of a source in another
    reference for the analog switch. If the source is referenced to
    ground, then I would like to be able to reference it to Vcc/2 instead.

    In my diagram, does it accomplish the task?

    Also, for the capacitor values, I am assuming that I have to pick the
    value based on:

    - minimum frequency to pass thru the low pass filter
    - frequency relationship is also dependent on the parallel combination
    of the voltage divider resistors
    - capacitor voltage rating is based on the voltage difference between
    VCC/2 and reference voltage of the source

    Are these assumptions correct?

    Also, for audio bypass capacitors, what other characteristics should I

  5. It is one general way to do it.
    Yes, C1 and the parallel resistance of the two resistors
    connected between it and fixed voltages (assuming the analog
    switch sends signal to a high impedance load) have an RC
    time constant. At frequency 1/(2*pi*R*C) the response of
    this low pass filter will be 3 db down. More for lower
    That would be the minimum. More doesn't hurt anything.

    But there are other considerations.

    How much supply current will these dividers consume? How
    regulated is that Vcc supply?

    This divider scheme will add some of whatever that noise is
    to the incoming signals.

    What is the loading effect (attenuation) of this divider
    load on the source impedances feeding it?

    How high can these resistor values go, before leakage
    current through the input capacitors and the switch shift
    the DC operating point enough to cause some trouble?
    Otherwise, higher resistance has only good effects.

    In many cases, it is better to use an opamp to make a low
    pass filtered Vcc/2 supply, and tie all the nodes that need
    this reference voltage to that output, with single resistors.
    They don't catch fire? It is a fairly undemanding
    application. Perhaps you might define what *you* mean by
    the term, "audio bypass" so we are all talking about the
    same thing.
  6. MRW

    MRW Guest

    This would just be a simple Ohm's law of the voltage divider right?
    i = Vcc / (R_div1 + R_div2) ?

    So there is a balance between picking a higher resistance and limiting
    the voltage divider supply current draw, right?

    I am assuming this would be the parallel combination of the voltage
    divider resistors in series with the source impedance. In order to
    minimize the loading effect, the parallel combination must be a higher
    resistance (i.e. source impedance and parallel combination in voltage
    divider configuration). Is this a correct assumption?

    I'm not to familiar with leakage current, yet. But is this the same
    concept of adding a high resistance resistor to the input nodes of an
    opamp? The input bias current of the opamp would "drop" a higher
    voltage if the resistor value is high... (? .. I think).

    Something like this one? I figured if I put a capacitor at the output
    to ground, then that would filter out the higher frequency components.
    I picked the capacitor value based on the parallel combination of the
    three output resistors. Did I make the right assumptions for this

    I was just referring to the capacitors at the input terminals (Mic in
    and CD in).

    Thank you very much, John!
  7. MRW

    MRW Guest

    Forgot the link:
  8. Yes, times the total number of such dividers in the system.
    Both the current draw and the ability of supply noise and
    ripple to get into the signals through the divider get
    better as the divider resistors go up in resistance.
    That's right. The parallel impedance of your divider forms
    a second signal divider with the source impedance as the
    input side. For instance, if your two sources each have a
    1k ohm impedance, then your two 1k resistors in parallel
    form a divider with that source impedance that divided the
    signal voltage down to .33 of its unloaded amplitude.
    You want the bias divider to look like a high impedance,
    compared to the signal source impedance. Having the bias
    divider have a parallel impedance at least 10 times the
    source impedance wastes less than 1/10th of your signal
    Right. if you want to provide an accurate bias voltage to
    an opamp input, you need the bias network to have a low
    impedance compared to the opamp bias current, so that the
    bias system dominates the voltage.

    In this case, the leakages are the input capacitor, anything
    that leaks from the supply into the switch channel (the data
    sheet should spec that) and any bias current injected into
    this node, by opamps or whatever that are downstream of the
    switch, when it is switched on.

    You may be able to use 100k or 1 meg resistors for your bias
    network and still control the average bias voltage of the
    signal passing through. You have to total up all the leakage
    sources and also come up with an acceptable DC error voltage
    you can tolerate, to calculate the maximum resistor values
    you can use.
    Yes, something like that. Though I think I would use 100k
    or 10 k resistors for R1 and 2, and add a capacitor across
    R2 to form a low pass filter for power supply noise.
  9. Oops. Get rid of the cap from the opamp output to ground.
    It doesn't filter much, since the opamp output impedance
    approaches zero, but it does tend to destabilize the opamp.

    If they are large enough to hold essentially a constant
    voltage during the audio signal swing, then linearity is not
    important. And since there is a definite DC voltage across
    them, electrolytics (aluminum or tantalum) could be used.
    But if the divider impedance is high enough, film caps are
    probably the best way to have low leakage through them.
  10. MRW

    MRW Guest

    Thank you again, John! :)
  11. neon


    Oct 21, 2006
    this is a logic switch takes vcc to turn it off and 1 to .8v to on. if the input is 2.4 or whatever it will just pass it thruogh or open output
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