• State
  • Locked Locked
  • Replies 2 replies
  • Subscribers 46 subscribers
  • Views 350 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

ADS805: input current

Len Epstein
Prodigy 40 points
Part Number: ADS805

I have the device configured for internal reference with a 0-5V input range. (SEL to ground; IN_N to Vref).  I am using AC coupling with IN_P to 0.45V using a 2K to ground and 20K to +5V.  When I power up the device IN_P goes to 0.45V as expected.  When the clock starts IN_P rises to 0.6V.  To me, that implies a sampling current of (0.6 - 0.45)/2K = 75uA.  That is larger than I expected.  Is this normal?

  • 0
    TI__Guru* 91947 points

    Len,

    It appears you are using this device as shown in Figure 7 of the data sheet. If so, why are you shifting the common mode to 0.45V? The input should have a common mode voltage centered about 0-5V which would be 2.5V.

    Regards,

    Jim 

  • 0 in reply to jim s
    Prodigy 40 points

    Greetings Jim and thank you for looking at this.

    In my application I am looking at pulses.  The signal line is usually low, but has pulses that can go as high as +5V.  I don't have a problem with 75uA coming out of the input pin, as long as I'm not damaging the device and the current is predictable (say 50uA to 100uA).  If I am damaging the device or the input current can vary significantly in this configuration then I must change something.  (More clarification: I suspect the input current will depend on the input voltage.  If IN_P is 2.5V I suspect input current will be much smaller than 75uA.  My concern of predictability is for my "between pulse" or normal situation when the voltage at IN_P is 0.45, or rather 0.6V.)

    Thanks,

    Len