1
\$\begingroup\$

I designed the following PCB from the schematic below to make a breakout for an I2C counter IC, datasheet is here.

However my RST pin doesn't go up and the IC stays in reset state. I'm feeding 3.3V to VCC. A multimeter is showing 0.29V on the RST pin. I probed also with the multimeter and the 10k resistance is there from VCC to RST PIN.

I added another normal 10k resistor from my breadboard to the RST pin and the pin went up and it's working after that. Any idea why my board 10k resistor is not working?

S-35770 I2C Binary Counter SCH S-35770 I2C Binary Counter PCB

\$\endgroup\$
12
  • \$\begingroup\$ Are you sure that you placed the IC correctly, as in, is pin 1 of the IC on pin 1 of the board? \$\endgroup\$ – LukeHappyValley Dec 15 '20 at 10:47
  • 1
    \$\begingroup\$ You put the other pull-up resistor parallel to the one on the board? Maybe 10k is too weak to pull it up and 5k (10k || 10k) does the trick. \$\endgroup\$ – LukeHappyValley Dec 15 '20 at 10:53
  • 1
    \$\begingroup\$ ESD damage is my guess - change the chip (pay attention to ESD). \$\endgroup\$ – Andy aka Dec 15 '20 at 12:23
  • 1
    \$\begingroup\$ Seems a little bizarre to me. The datasheet says that _RST is a CMOS input with leakage current specified less than a microamp. \$\endgroup\$ – Adam Lawrence Dec 16 '20 at 14:14
  • 1
    \$\begingroup\$ @Trunet - Hi, In response to the concern from Andy aka about possible ESD damage, you said: "I was grounded the whole time". FYI it is easily possible to cause ESD damage, even if you were grounded (e.g. using a wriststrap), depending on other factors including tools, equipment, clothing, environment & procedures. At this time of year in the northern hemispehere, reduced humidity also increases risks of ESD strikes. Due to insufficient info, I am not saying that ESD damage has definitely occurred. However reporting that you were grounded does not prove that ESD damage didn't occur. \$\endgroup\$ – SamGibson Dec 16 '20 at 15:00
1
\$\begingroup\$

I changed the resistor to 4.7k and it's working now. Not sure why is that, but this solved it pulling RST up.

\$\endgroup\$
0
\$\begingroup\$

You have not met the minimum voltage state for RST pin to be low.

See : https://somanytech.com/voltage-drop-across-resistor/ , particularly this part The voltage across series circuit- Practical examples:

Its been a while since i have done some EE experience

\$\endgroup\$
3
  • \$\begingroup\$ I didn't find the current needed on the RST pin in the datasheet to calculate the resistor. \$\endgroup\$ – Trunet Dec 15 '20 at 11:17
  • \$\begingroup\$ any two equal resistors will give you half the voltage across them in series.But then again i could be wrong , little rusty here. \$\endgroup\$ – binaryOps20 Dec 15 '20 at 11:19
  • \$\begingroup\$ I'm aware of that. I will give it a try and replace the 10k with a 1k one and see what happens. \$\endgroup\$ – Trunet Dec 15 '20 at 11:27
0
\$\begingroup\$

Maybe there's a 10k-ish pull-down resistor internally or the internal protection diode got broken and acts like a resistor. This make me think "The higher the external pull-up resistance, the lower the RST pin voltage (Basic voltage divider)." And it worked after decreasing the PU resistance. Sounds reasonable, doesn't it?

I know, the datasheet does not show anything about that. But the behavior seems like there's an internal PD resistor or the protection diode failed.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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