• State TI Thinks Resolved
  • Locked Locked
  • Replies 1 reply
  • Answers 1 answer
  • Subscribers 18 subscribers
  • Views 1289 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.

CD4010B-Q1: The float VDD voltage is 4.35v?

Frank Xiao
Genius 12055 points
Part Number: CD4010B-Q1
Other Parts Discussed in Thread: CD4010B

Hi team,

My customer is using the  CD4010B chip now. The VCC voltage is 5v. When we remove the D147  , and not use the 12v_ON1 supply, the VDD voltage is 4.35v. Could you help me check why the VDD have voltage? Thank you.

BR

Frank

  • 0
    TI__Guru* 96591 points
    Hi Frank,

    V_DD should always be equal to or greater than V_CC in this part due to its internal construction. If you take a look at the schematic shown in Figure 1 on the datasheet, you can see that the body of the output pFET is connected directly to V_DD. That means that the body diode for that FET is also connected between V_DD and V_CC. If V_DD >= V_CC, this is no problem (ie the diode will not conduct in reverse bias). When V_CC > V_DD, you would expect to see ~1 diode drop fromm V_CC showing up on V_DD.