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

  • \$\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


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.

  • \$\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

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