3
\$\begingroup\$

The problem I have with my ADuM4160 circuit is that I'm getting 5 V from VDD2 pin while getting roughly 3.3 V on pin VDD1. Have I blown the chip?

It doesn't make sense for the VDD2 to be the same voltage as what I am feeding to the VBUS pin, right? Could it be the bypass capacitors I am using? The datasheet says they should be low-ESR, but I'm unsure where to get them from. Most capacitors that I have found are SMD. I'm using cheap eBay-sourced ceramic capacitors.

The author of the circuit told me to check my supply for oscillations, but I'm unsure how. I do have a digital 40 MHz scope.

The only difference in my circuit is that the downlink is powered by my do-it-yourself LM338 power supply. Since my device has its own supply I don't need the switch-mode regulator.

BTW, the VDD1 and VDD2 are the outputs of internal regulators.

I'm testing this on a breadboard:

Enter image description here

Here's an image of what I see on the VBUS2 pin which is coming from my LM338 PSU. I assume that this is normal and nothing to worry about, right? I don't see anything like the following when I turn up the volts per division. I can really see the oscillations when I turn the time base higher. I see something similar on the VDD2 pin as well.

Enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Check you have no short between Vbus2 (or Vbus1 for that matter) and Vdd2 (with power off, and use multimeter on ohms range) \$\endgroup\$
    – Oli Glaser
    Sep 13, 2012 at 16:23
  • \$\begingroup\$ I dont have a dead short between vbus2 and vdd2 but I do have 25K worth of resistence. Vbus 1 and vbus2 is completely open circuit. \$\endgroup\$
    – Ageis
    Sep 13, 2012 at 22:52
  • \$\begingroup\$ Okay, that sounds reasonable. When you scope the lines, use the probe on x10 setting with the bandlimit off. Also, just out of curiosity, try putting a 10k resistor from Vdd2 to ground2. \$\endgroup\$
    – Oli Glaser
    Sep 14, 2012 at 2:15
  • \$\begingroup\$ Thanks for everyones help I think I really have killed it. I soldered on new chip and it now give 3.39v \$\endgroup\$
    – Ageis
    Sep 14, 2012 at 19:59

1 Answer 1

1
\$\begingroup\$

The datasheet says that the \$V_{DD1}\$ and \$V_{DD2}\$ behaviour is the same: if \$V_{BUSx}\$ is connected to 5V, the signal at \$V_{DDx}\$ should be the 3.3V output of the LDO. If \$V_{BUSx}\$ is 3.3V, directly connect \$V_{BUSx}\$ and \$V_{DDx}\$ together.

Ceramic capacitors are by nature low-ESR, and using a linear regulator to feed the \$V_{BUSx}\$ signals should be fine.

You should scope all the \$V_{BUSx}\$ and \$V_{DDx}\$ signals with a short-ground probe, AC-coupled with bandwidth limiting turned on to see if there are any disturbances in the bus voltages (ballpark: anything in excess of 1% of the DC level may be trouble). Check multiple time scales (10s of ms, 10s of ns, etc.).

Some ceramic decoupling capacitors (100nF or so) close to the IC on the \$V_{BUSx}\$ voltage pins wouldn't hurt.

\$\endgroup\$
2
  • \$\begingroup\$ what is a ground probe never heard of such a probe. I just have regular probes that came with my scope. Where can I buy one if I should have one? \$\endgroup\$
    – Ageis
    Sep 13, 2012 at 17:45
  • \$\begingroup\$ He just means that the ground wires from the probes you have should be as short as possible. \$\endgroup\$
    – Dave Tweed
    Sep 13, 2012 at 19:36

Your Answer

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

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