Designing circuit for research proj.

Discussion in 'General Electronics Chat' started by giantimi, Jul 27, 2009.

  1. giantimi

    Thread Starter Member

    Jul 15, 2009
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    So the circuit that I am designing with a current input and an output that is a binary signal who frequency is proportional to the magnitude of the input current. The input range is 0.2 to 2 microamperes. Attached is the circuit. ( Ignore the bottom part of the diagram, just focus on the portion labeled current signal processing block)

    [​IMG]

    The way this circuits works is the current mirror charges the capacitor, until the voltage to the schmitt trigger goes below the lower threshold. When that happens the PMOS is "turned on" and the capacitor discharges until the high threshold is reached and the PMOS is "turned off".

    The problem I am having is when I run a transient simulation ( in Cadence ), the input voltage to the Schmitt Trigger stays constant. I tried different ratios of Width to the Length for the current switch and the PMOS Switch and nothing works and different capacitor values. What am I missing?
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    What are you using as a schmitt trigger?
    What is the input resistance of this schmitt trigger?
    Perhaps a buffer between capacitor and schmitt trigger would help.

    Greetings,
    Bertus
     
  3. giantimi

    Thread Starter Member

    Jul 15, 2009
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    here is my schmitt trigger design.
    RC2 was changed from 4.7 k ohms to 10 k ohms.

    [​IMG]
     
  4. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    With the low current you have (0.2 - 2 μA) you must use a capacitor of good quality (low leakage)
    and a schmitt trigger with a very high input resistance (perhaps even fet input).

    Greetings,
    Bertus
     
  5. giantimi

    Thread Starter Member

    Jul 15, 2009
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    so inserting a source follower with the input from the pmos switch to the schmitt trigger would be a good idea?
     
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You could try.
    You do not want the current to get lost in the schmitt trigger.

    Greetings,
    Bertus
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    Might even be a good idea to have a FET input op amp buffer driving the schmitt trigger, to really decrease the load impedance.
     
  8. giantimi

    Thread Starter Member

    Jul 15, 2009
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    i'm still having trouble despite your suggestions... Here is a photo of the circuit in Cadence.

    In the photo, i have marked the dc operating points and voltages. These don't seem right at all, such as with current mirror. Vds is below V_overdrive. I have tried different solutions and nothing seems to be working.

    [​IMG]
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Your Schmitt trigger is not inverting, as shown in your block diagram. You need another inverter in the feedback loop.
     
  10. giantimi

    Thread Starter Member

    Jul 15, 2009
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    yea, inserting the source follower did help as you said. However, the voltage gain is not at all 1. The highest I could get was about 0.6 gain. Which is terrible. Are there any other simpler buffer topologies other than making an op-amp?
     
  11. Ron H

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    A current sink in place of the source follower pulldown resistor will raise the gain closer to unity.
    Can your circuit tolerate the source follower Vgs? You already have to deal with Vbe in your Schmitt trigger. Things could get a little tight with a 1.25V supply.
     
  12. giantimi

    Thread Starter Member

    Jul 15, 2009
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    I think my circuit can tolerate the source follower Vgs? Anytime I changed the W/L ratio of the source follower I noticed that the Vgs for the current mirror would change. However, it did not affect the current leaving the current mirror. It was always the same. So I think my circuit can tolerate it.
     
  13. Ron H

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    Did you see this post that I made earlier? Any comments?
     
  14. giantimi

    Thread Starter Member

    Jul 15, 2009
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    Yea, I put another inverter in there. I think now that I did that, if the buffer works as intended, then the entire circuit should operate properly. Thank you for that suggestion.
     
  15. giantimi

    Thread Starter Member

    Jul 15, 2009
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    Does anyone know of any other ways to make a good buffer with such a low voltage supply?
     
  16. Ron H

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    How about this? NCS2200 is available from ON Semiconductor. I simulated it using an op amp macro model because ON Semi doesn't seem to have a spice model available, and it worked.
     
  17. Ron H

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    I doubt that you will find a commercially available MOSFET with Vgs(th) low enough to work in your current mirror, due to the low supply voltage, unless you use something like ALD110800 from Advanced Linear Devices.
     
  18. giantimi

    Thread Starter Member

    Jul 15, 2009
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    Yea i figured. I am designing the entire circuit in Cadence with biCMOS6hp process. I am sure you familiar with that. So vth tends to be about 0.6 V. I think I might have to design a simple op amp from scratch. It just needs to have a high gain, that's all. Thank you for your suggestions and your patience in dealing with me.
     
  19. Ron H

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    I'm not familiar with the process, but now I understand why you are mixing MOSFETs and bipolar transistors. I had erroneously concluded that you were using discrete parts.
    I still don't understand how you can get your bipolar Schmitt trigger to have a lower threshold less than Vbe. It seems to me that Q1 will turn off when the input is less than about 0.6V.
     
  20. giantimi

    Thread Starter Member

    Jul 15, 2009
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    Yea, I should have explained that I'm doing things in Cadence with the biCMOS6hp process. But anyway, it's ok for the lower threshold to be about 0.6 V, so long as the high threshold is around 1V. When I did a simulation of just the Schmitt Trigger that turned out to be the case, so it works well enough. All that really matters to me is the OFF voltage of the schmitt trigger. That needs to be low enough for the inverters, considering the power supply is so low.
     
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