0
\$\begingroup\$

I’m using the ADS1115 with 5V because my measuring device provides 0 to 5V measurements but since the microcontroller is not tolerant to 5V, I need to shift it down to 3.3V.

I plan to use the BSS138 which in a popular configuration uses two 10K resistors to do so (R1 and R2 for one line and R3 and R4 for the other in my case).

The configuration on the right is for one device and the basic single configuration. The datasheet for the ADS1115 calls for pull ups in the range of 1K to 10K in the SDA and the SCL lines.

Lower values more power consumption, faster speed, higher values lower power consumption, lower speed, as far as I understand. Power and/or speed are not a concern.

My questions are

  • can I delete R5 in SDA and R6 in SCL and connect those pins directly to the High Level (HL) side of the shifter since it already has 10K resistors?
    • And better yet delete the 10K resistors (R2 and R4) without a connection to 5VIN?
    • And by doing so, R1 and R3 become the only pull ups?

I’ve read that for 5V a good pull up resistor is a 4.7K and for 3.3V is around 2.?K.

  • If not, can I substitute the two 10K resistors (R2 and R4) with a 5K resistor (4.7K for standard value resistor)?
    • How would you minimize the use of resistors and have a reliable design?

enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ "5VIN" is the same thing as "5V" right??? \$\endgroup\$ Apr 15, 2020 at 10:28
  • \$\begingroup\$ Is there a reason to not simply use a resistive voltage divider on your 0-5V input voltage and do the ADC'ing with 3.3V? Maybe even integrated in the MCU? \$\endgroup\$ Apr 15, 2020 at 10:31
  • \$\begingroup\$ Yes, 5VIN is 5V. Mr. Muller, I have considered the resistive voltage devider solution as well. I just don't know which one is the more appropriate and reliable one. \$\endgroup\$ Apr 15, 2020 at 10:41

1 Answer 1

0
\$\begingroup\$

yes

You can delete those. only one pull-up is needed on each side of the level shifter.

How much pull-up resistance is low enough mostly depends on the capacitance of the traces that carry the signal and the signal speed you need.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.