1
\$\begingroup\$

I am trying to set up a circuit to use the LMR16030 buck regulator to step down 60V to 9V. I bought the Powergood version and I'm not sure how to connect the power good pin. As you can see in the datasheet on page 3: http://www.ti.com/lit/ds/symlink/lmr16030.pdf

"PGOOD pin for Power Good version, open drain output for power-good flag, use a 10 kΩ to 100 kΩ pull-up resistor to logic rail or other DC voltage no higher than 7 V."

I'm not sure how to apply this as I do not have a DC voltage at 7V in my circuit.

I based my circuit on the application section of the datasheet page 19 and encountered a few problems with the Powergood pin open. I first forgot to connect the GND and obtained the desired voltage, but oscillating, as my arduino was turning on and off rapidly. When I then noticed my error, I connected the GND pin properly and it instantly shorted with a big spark at the GND pin. Could this be due to the fact that I left the PGOOD pin opened?

Thank you for your help!

\$\endgroup\$
1
\$\begingroup\$

Could this be due to the fact that I left the PGOOD pin opened?

No. That is an optional output. Leaving it unconnected won't cause a problem.

When I then noticed my error, I connected the GND pin properly and it instantly shorted with a big spark at the GND pin.

Until you have enough experience to know when to break it, a good rule of thumb is: Never make changes to a circuit with power still applied.

Even if you are changing to the correct connections, doing so with power applied can cause partial (sometimes hidden) or fatal damage to components. So please, despite the temptation to fix the connection problem quickly when you notice it, power off (and wait for any capacitors to discharge) first! :-)

The spark suggests a large potential difference between the ground at the output (which you connected at the time you saw the spark?) and the ground of your regulator. Although I can't say that this definitely caused damage, it won't have done any good to anything to equalise the potentials so quickly.

Regarding the PGOOD pin:

I'm not sure how to apply this as I do not have a DC voltage at 7V in my circuit.

That topic is answered in the datasheet, section 7.3.9:

Voltage seen by the PGOOD pin should never exceed 7 V. A resistor divider pair can be used to divide the voltage down from a higher potential.

Therefore if you were actually going to use that PGOOD signal, you could use a resistor divider from the 9V output (once your regulator is supplying a stable output!) to generate whatever pull-up voltage you require. However if you are leaving that pin unconnected, it sounds like perhaps you aren't using that signal, in which case you simply don't use it.

I first forgot to connect the GND and obtained the desired voltage, but oscillating, as my arduino was turning on and off rapidly.

Personally, if I'm designing a power supply, I don't use something valuable connected to the regulator's output while I'm testing, in case of unexpected problems which might cause damaging output voltages. Instead of using your Arduino, you may want to consider making a "disposable" load e.g. car light bulbs, or high-wattage resistors. There are also robust electronic loads available.

Finally, I wanted to point out that this IC requires special care regarding the PCB layout (see datasheet section 10) and component choice (inductor, diode and capacitors, see datasheet section 8). If you are trying to make this using some kind of breadboard arrangement, you may not be successful - even if it seems to work, it may not be stable, and/or may emit lots of EMI :-(


Update: As discussed in later comments, TI offer evaluation module LMR16030PEVM for this IC, which provides a "known working" PCB, and was a successful solution in this case.

\$\endgroup\$
  • \$\begingroup\$ Ok thx, the IC properly melted as soon as I turned on the power. And it did twice with two different ICs and arrangements, could such a problem be due to the Powergood pin being opened \$\endgroup\$ – Eliott W Oct 13 '16 at 9:43
  • \$\begingroup\$ @EliottW - Hi - "could such a problem be due to the Powergood pin being opened". I don't believe so. I strongly suspect your problem is elsewhere e.g. component layout or component choice. Assuming LMR16030P, the datasheet explains that PGOOD is an open drain output, so leaving it unconnected should be harmless. If you want to test your concern that having PGOOD unconnected is causing failures, connect PGOOD (pin 6) via 10k - 100k resistor to Gnd (pin 7). When the IC drives the internal open-drain transistor, it will effectively connect Gnd to Gnd - valid & not damaging, but useless :-) \$\endgroup\$ – SamGibson Oct 13 '16 at 10:10
  • \$\begingroup\$ [cont] Just to be clear - making such a connection to PGOOD won't fix your problem. It will just allow you to test your obvious concern that having PGOOD unconnected is causing your problem. If you make only that change (connect PGOOD via 10k - 100k to Gnd, so that it is no longer unconnected) I expect the same IC failure to occur again, since the cause is elsewhere. If you are struggling with this design, I suggest opening a new question, supply your schematic, PCB layout diagram, exact component list & explanation of your choices for inductor, diode & capacitors, and ask for improvements. \$\endgroup\$ – SamGibson Oct 13 '16 at 10:16
  • \$\begingroup\$ Ok thanks, I'll try to rearrange the layout and see, if this fails I guess I'll need a PCB and solderless won't make it \$\endgroup\$ – Eliott W Oct 13 '16 at 10:19
  • 1
    \$\begingroup\$ @EliottW - "is there an easy way I can step its output up to 7-9V ?" - [I will assume 9V desired o/p] Yes it's easy, but don't think of it as "stepping-up" the output i.e. it's not a 2 stage process 60V->5V->9V. Instead you can just modify the eval board to provide 9V instead of 5V. How to do this is explained in section 2.2 "Adjusting the Output Voltage" in the eval board user guide that I linked in my previous comment. Carefully follow those instructions, including removing the PGOOD pull-up resistor R9, or using a resistor divider to generate the pull-up voltage (as discussed before). \$\endgroup\$ – SamGibson Oct 13 '16 at 13:48
1
\$\begingroup\$

First, the issue you are having with the spark has nothing to to with the PGOOD pin being open. The part will work fine without the PGOOD connected to anything.

If you connected the ground pin with the power on the part and saw a spark but now everything is working OK the spark was likely just the ground current making connection. If it's not working you may have damaged something.

Second, assuming you need PGOOD at all, if you don't have a voltage to pull the PGOOD pin up to, you can create one with an LDO or other method. 5V/100K is only 50uA so power dissipation shouldn't be an issue.

\$\endgroup\$
  • \$\begingroup\$ Ok thanks! The pin was properly damaged and the IC melted from the inside, I rearranged the circuit and had the same problem again immediately as I connected the power. \$\endgroup\$ – Eliott W Oct 13 '16 at 9:42

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.