Analogue Switches - Output Behaviour

Discussion in 'General Electronics Chat' started by jbriaris, Apr 11, 2013.

  1. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    Hi All

    I'm a relative newbie to all things electronics and have a question regarding the output from a Dual SPDT analogue switch.

    I have implemented the ADG884 switch in LTspice and get the expected voltage levels (1.25V and 1.75V) at the outputs D1 and D2 when toggling the switch with a PULSE waveform at IN1 and IN2 (see circ1.png attached).

    However, when I connect D1 and D2 to give the single output Dout (see circ2.png attached), the votages are greatly reduced, i.e., ~0.62V and 0.88V.

    I'm trying to understand why this is happening. Is the voltage feeding back into the switch causing this behaviour? Should I buffer the outputs before joining them? I'm sure there is an obvious reason, and would greatly appreciate someone pointing me in the right direction.:)

    Thanx!

    I'm sure there is an obvious reason for this, e.g. the voltage feeding back into the switch causing this behaviour? and I would very much appreciate a pointer in the right direction.
     
  2. #12

    Expert

    Nov 30, 2010
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    Diodes use up about .6 volts when you pass current through them. The best you can do to minimize this in one quick move is to use Shottky diodes. They waste less voltage than other diodes.
     
  3. Ron H

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    Apr 14, 2005
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    If you want to switch between two inputs, just use one spdt switch, e.g., S1. Connect V1 to S1a and V2 to S1b. Don't use S2. You only need one control input. If you want dead time, use your circuit, but leave S1a and S1b floating, instead of connecting them to ground.
    The way you are doing it now, each switch output connects to either a voltage source or to ground.
    #12 apparently thought that your D's referred to diodes.
     
  4. #12

    Expert

    Nov 30, 2010
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    True. Completely fooled me. He's talking about D1 and D2 and losing .62 volts but there isn't a diode in either of the drawings, so I guessed that "D" meant a diode in the drawings.
     
  5. crutschow

    Expert

    Mar 14, 2008
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    When both switches are on and the outputs connected together than one output is shorted by the other output back to the other source. The switches are bidirectional so it's like connecting the two sources together.

    If you want to sum the two signals together than you would need to use a summing amplifier.
     
  6. Ron H

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    The control signals are complementary, with some dead time between them. I think the sources are alternately connected to ground through two switches, making voltage dividers, where the divider resistors are the ON resistances of the switches.
     
  7. crutschow

    Expert

    Mar 14, 2008
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    You are correct. I didn't carefully examine the circuit. :p

    If he removes the connection to ground from S1B and S2B then the voltage reduction should not occur.
     
  8. Ron H

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    Doesn't hurt to reiterate what I said in post #3.:)
     
  9. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    Thanks all for the replies! Great to see such an active forum in action, and very helpful. :)

    Yes, Ron H is correct. What I'm trying to achieve is switching between two reference voltages with some dead time (0V) inbetween, hence the requirement for the dual SPDT. I realise now (quite obvious after stepping back to see the woods from the trees :p) that as currently set up I'm shorting the circuits back to the source (as stated by crutschow).

    I set S1B and S2B to float, and now what I get at the connected output is the desired 1.25V and 1.75V levels at the switch on times (see attached circ3.png - I've plotted all voltage signals to hopefully provide more insight). However, during the 'dead' time I now get spurious voltages at ~2.2V and ~2.6V, when this ought to be 0V. I'm assuming this might be due to capacitance in the switches?:confused:

    I reconnected S1B and S2B to ground with a series resistance of 100 Ohms, and get closer to the ideal (see circ4.png), but they're still slightly off. I guess this is me trying to balance the potential divider circuit as mentioned by Ron H. Is this the best way to do this? Or am I missing something?

    Note: The single output Vout will ultimately be used to provide a reference voltage to the non-inverting input of an opamp.

    Thanks again all! :)
     
  10. Ron H

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    Apr 14, 2005
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    Why do you have so much dead time? What voltage do you want during the dead time? The voltage will be undefined during that time. Also, a single section of the ADG884 is already break-before-make, with about 16nS dead time.
    And why are you using such beefy switches? The input of a noninverting amplifier is very high impedance, so you don't need low on resistance switches. You could use 74HC(T)4066 or CD4066, which are lots cheaper.

    Tell us what the switched reference voltages will be used for. There may be an even simpler solution.

    Please answer ALL the questions.
     
  11. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    >> Why do you have so much dead time?

    I'm using the switches to provide a reference voltage to an opamp that forms the start of a constant current source circuit. The constant current source circuit will drive two LEDs in back-to-back configfuration, hence the two different voltage levels. However, I also want the LEDs to be off for an equivalent peiod of time, and hence the desired 0V dead time, i.e., zero current through the constant current source.

    >> What voltage do you want during the dead time?

    0 volts, as explained above.

    >> Break-before-make with 16ns dead time

    Can live with those timescales.

    >> Why are you using such beefy switches?

    Totally agree with you - I'm not drawing current through the switch, and thus could use ones with higher ON resistance. The choice was made simply for modelling purposes in SPICE, i.e., I could easily find the SPICE model for ADG884 :p. Would certainly use ones like you suggest for actual implementation.

    Really appreciating this guidance - I'm learning quickly! Thanks.:)
     
  12. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    BTW

    Just found the LTSpice yahoo group with a spice model for the CD4066B. Will also have a play with this! :D
     
  13. Ron H

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    What will your actual ON and OFF times be? Your original post shows on and dead times ≈1mS. Are those realistic, or does it need to be slow enough to be visible?
    In either case, you can simply use two SPST analog switches (there a 4 in a package) with a single resistor from the op amp input to ground. The resistor gives you zero volts during the dead time.
    Tie unused switch control inputs to ground.
     
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  14. jbriaris

    Thread Starter New Member

    Apr 11, 2013
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    Perfect!:D

    Using SPST switches with output reference to ground via resistor gives exactly what I need. Thanks Ron H.
     
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