This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

parasitic cap caused leaking?

Hi,

If we have an IC with CMOS output, and cut of its power supply by means such as a switch disconnecting its VCC, then what happens at output pins? Would parasitic cap's voltage keep one of the MOSFET on?

 

Calino

  • I've never heard of C_para being a problem in practice; I'd guess it's discharged somehow. (There might be a megaohms resistor somewhere.)

    The parasitic body diode of the P-channel MOSFET (from the output to VCC) could be a problem when VCC=GND. CMOS devices with 5V-tolerant outputs are constructed so that this is not a problem.
  • Hi Calino,

    I brought this up to the Logic Applications team and we have some thoughts on the effect of the capacitor that is part of the push pull amplifier. It is likely that the capacitor will try to discharge through the mosfets or logic device which although have a large resistance but are not perfect. Mosfet gates tend to be on the order of 10 Megaohms. We are worried that the capacitor will cause the input to the push-pull amplifier to become an RC curve which will reduce the bandwidth your device can operate at. Also depending on the logic device you could have problems forcing current through it in High Z mode.

    Pulling from college courses generally there are diodes and resistors on the input to fix bias at the output and allow somewhere for the parasitic capacitance can discharge from, however I can't speak for industry design.

    Out of curiosity what logic device are you using to run the amplifier?

    Thanks,
    Daniel
  • Hi Calino,

    I just noticed that you were talking about inside of our part and not a push pull amplifier that you had hooked up to one of our logic parts and that changes things significantly.

    The two things I would say are that I need to know which part this is to give you full confirmation. The second being that this part you are using likely has ESD Diodes in it that the capacitance will discharge through.

    Thanks,
    Daniel
  • Daniel,

    What about the cases for:
    i: 1.8V/3.3V TI processor GPIO, like for MSP430, Keystone, C2000, etc.?
    ii: TI signal muxes.



    Calino

  • Hi Carlino,

    Q1: The 10Mohm resistances that is generally seen on MOSFET devices is because their gate must have a finite resistance. I used 10 Mohms because generally that is the order of magnitude that MOSFETS will have as their resistance however that is not the rule. You can assume non ideal devices have a resistance, just not necessarily 10Mohm.

    Q2: Below I have an attached a diagram that shows the ESD diodes and their interactions with parasitic capacitances and the paths through the diodes is where the capacitors will drain. I am not allowed to comment on any of the devices that you have listed since you should ask about those in their respective forum.

    i) Microcontrollers forum

    ii) ON Semi's own forums

    iii) The Switch/Mux forum

    iv) The amplifiers forum

    I do believe however that the below diagram will be the case, i do not guarantee this however for the parts you listed.

    Thanks,

    Daniel

  • Daniel,

    Is the path on the left or right diodes?


    Calino

  • Hi Calino,

    Q1: The path is highlighted below. The Vcc in this circuit is 0V in this case because the Vcc has been switched off by whatever you are doing. 

    Q2: I suggest looking at HSPICE and LTSPICE. If you are at a university sometimes they have a commercial license you can get by talking to some of the professors.

    Q3: From what I have heard most people use HSPICE however that doesn't happen in my area.

  • Hi Calino,

    If you are completely disconnecting the supply from anything at all the current will decide to go through the ESD diodes to ground because of the diodes leakage current. Since diodes are not ideal devices they will let some current through backwards it will just be much slower than if the cap could discharge through Vcc.

    Thanks,
    Daniel