0
\$\begingroup\$

After a day of struggling with trying to simulate an LM2678 based buck converter (see: this post) another user suggested using TIs online power designer web bench tool to simulate. Selecting a design on the site gave me just about the exact same circuit I had designed by hand for a 5v buck converter using the design equations found in the LM2678 datasheet. I'm struggling to make sense of the simulation results from TIs web-bench, in fact they seem entirely nonsensical to me:

TI circuit

Transient Load Analysis: Transient Load Analysis

Transient Steady-State Analysis:

Transient SS analysis

Could someone explain to me what this means, if anything? I'm expecting to see a constant 5v out of the steady state analysis voltage over the load but maybe I'm just not understanding what the results are telling me?

\$\endgroup\$
  • \$\begingroup\$ If you really want to get frustrated you can export the design from Webench to the TINA-TI tool and be able to probe whatever nodes you want to figure out what it's actually simulating. \$\endgroup\$ – The Photon Aug 29 '18 at 20:47
  • \$\begingroup\$ @ThePhoton is the TINA-TI tool as horrible a user experience as the Power Designer? \$\endgroup\$ – Blargian Aug 29 '18 at 21:18
  • \$\begingroup\$ It's slightly better. But still in the horrible zone... \$\endgroup\$ – Dejvid_no1 Aug 29 '18 at 21:29
  • \$\begingroup\$ The ADJ simulation seems bogus. But it actually works if you try the design for the fixed 5V verison. \$\endgroup\$ – Dejvid_no1 Aug 29 '18 at 21:34
  • \$\begingroup\$ @Dejvid_no1 Thanks for the heads up. That adds to my suspicions that the ADJ spice file just does not work at all. \$\endgroup\$ – Blargian Aug 29 '18 at 22:24
0
\$\begingroup\$

The waveform that you're seeing labeled transient response shows how the output voltage fluctuates due to a changing demand in current downstream from the converter. This can tell you a few things such as: stability, overshoot / undershoot, response time, and more.

When the red waveform (load current) goes from 3A towards 0A, the energy that was stored in the inductor L1 doesn't change instantaneously, and therefore must be absorbed into the output capacitor Cout. This results in a voltage increase in the voltage output. The opposite change in load current causes the voltage at the output to drop since the power converter cannot respond as quickly as the load is being demanded. The energy is temporarily supplied by Cout and causes a dip.

Usually you want to know how the output responds under rapid load changes of different frequencies so you can determine if your output voltage is stable and within the max/min voltage of the downstream parts that power is supplied to.

This is a very brief explanation, but hopefully gives you some insight.

\$\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.