0
\$\begingroup\$

I have been running some tests on the AP63356Q switching regulator to view the performance at different loads. I plan to use the regulator to step the voltage down from either 12 or 24 V to 5 V. After conducting multiple tests, I have discovered that mean Vout decreases with higher current while the Vout ripple increases.

I've plotted these points in the screenshot below: Switching regulator test results

This is the schematic of my circuit setup: switching regulator schematic

The inductor being used in this circuit is the NRS8040T6R8NJGJ. I use different resistors on a breadboard to get the results for different loads.

I get that adding an extra capacitor or two may limit the peak-to-peak voltage but any ideas on how to maintain a steady 5 V at higher loads? Would using the external compensation solve this issue? Any advice on how to choose a suitable inductor for the job (I have noticed that the inductor's temperature increases as well)?

Here are additional images of the layout: 3D PCB PCB design

\$\endgroup\$
6
  • 2
    \$\begingroup\$ Can you show the layout? It's very important for DC-DCs \$\endgroup\$
    – bobflux
    Jul 13 at 20:00
  • 1
    \$\begingroup\$ How and where are you measuring the output voltage? Is it right across the output or after a long run of wire? \$\endgroup\$
    – Big6
    Jul 13 at 20:01
  • \$\begingroup\$ Did you follow the PCB layout guidelines in the datasheet you linked to? \$\endgroup\$ Jul 13 at 20:02
  • \$\begingroup\$ I've added some screenshots of the layout. @Big6 Yes, I am measuring the output voltage after the wire using an oscilloscope. I tried following the PCB layout guidelines as best as I could but this board design had to be modified in order for it to be mountable onto another board. \$\endgroup\$ Jul 13 at 20:17
  • \$\begingroup\$ At least the 5V output comes directly from the coil, and the caps are on the other side of coil. The output would contain ripple as it does not go through the caps. Feedback seems to come from caps though. \$\endgroup\$
    – Justme
    Jul 13 at 20:23

1 Answer 1

1
\$\begingroup\$

Since you are measuring voltage at the load resistor on a breadboard, there is some resistance between the regulator output and your measurement point. If this resistance is about 200 mOhm it would explain the measured voltage drop with increasing current. You can confirm this by measuring the output voltage directly at the output capacitors.

The output ripple increasing is less obvious to me. In general the ripple is higher than you would expect. This could in part be a measurement problem picking up a lot of noise with the oscilloscope probe. A picture of the measurement setup and a screenshot of the waveform could clarify this.

Suboptimal layout could also contribute to noise and ripple on the output. Since you are space constrained you might not be able to fix all of this, but here is a list of most important things I would try to do about the layout:

  • Input cap should be placed directly at the input and GND pin
  • SW node should be as short as possible and without vias
  • Bootstrap capacitor should be connected between BST and SW IC pins as short as possible

Regarding the inductor choice, the inductor given should be OK. It should not get very hot. Given the datasheet values I would expect a temperature rise of less than 10°C with your maximum load current.

\$\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.