1
\$\begingroup\$

we are trying to use a battery charging buck converter design in reverse direction to be used as a boost converter to drive a LED panel of 17.4 volts / 1.32 A While the LTSpice simulation seems to deliver right voltage and current; but we are not able to eliminate the output ripples completely and could see huge inrush current (probably due to the capacitors) on Mosfet drain output.

Minimum capacitor value calculated from this document https://www.ti.com/lit/an/slva372d/slva372d.pdf?ts=1695520854803&ref_url=https%253A%252F%252Fwww.google.com%252F enter image description here

D (Duty cycle) = 10.56
I (max) = 1.4 A
f = 25 KHz
ΔV (acceptable output ripple voltage) = 50 mV

Questions,

  1. How we can reduce the ripple spikes?
  2. Will the inrush current damage the MOSFET?

The circuit from LTSpice enter image description here

High side driver enter image description here

Low side driver enter image description here

Capacitors selection enter image description here

Output waveform enter image description here

Ripple enter image description here

U1 Drain Inrush current enter image description here

\$\endgroup\$
3
  • \$\begingroup\$ @daydreamer2100 - Hi, Stack Exchange doesn't work well when questions materially change, after an answer is given (see the term "chameleon question" at Meta.SE). We've been alerted that the question changed, meaning you got answers both before & after significant changes. It's a bit of a mess :( We have been asked to make a decision. My compromise is allowing rev 4 of the question, but not later. Your "update" (which was introduced in rev 5) seems mainly to be confirmation of other answers & should either be comment(s) on other answer(s) {...} \$\endgroup\$
    – SamGibson ♦
    Sep 24 at 16:01
  • \$\begingroup\$ (continued) (I can explain how to link to an image in a comment, if needed) or you can write a self-answer to explain the result of simulating other answers. Note: Since that self-answer would just be confirming other people's answers, it would be inappropriate for you to "áccept" it in that case. Instead, please consider whether an answer which told you the cause of the problem (i.e. shoot-through) should be áccepted. || Now knowing that the original problem was shoot-through, if you have a further question (e.g. MOV/TVS etc.), then ask a new question & link back to this one for context. \$\endgroup\$
    – SamGibson ♦
    Sep 24 at 16:02
  • \$\begingroup\$ (I have removed an answer-in-a comment and a follow-up question from the comments, since (a) answers should not be written in comments and (b) it discourages other people who don't realise the problems caused by answering in comments, from also answering the follow-up question in comments.) \$\endgroup\$
    – SamGibson ♦
    Sep 24 at 16:07

2 Answers 2

Reset to default
0
\$\begingroup\$

How we can reduce the ripple spikes?

If you are designing your own boost converter this is something you have to be aware of and add circuitry to control to initial overshoot.

Basically, as soon as you power your circuit from the incoming supply it's like applying a step change to a highly resonant LC low-pass filter. Correction; it's not like it is. And, you get an overshoot that can be up to twice the incoming supply voltage without even activating the lower MOSFET. If you activate the lower MOSFET at that same time, you will get more than twice the input voltage as a surge on the output due to resonance.

To avoid this either: -

  • implement a slow-start circuit that can waste this surge energy as heat,
  • have a load dump circuit on the output that acts like a surge suppressor/Zener,
  • pre-charge the output capacitor to the incoming supply via a resistor or
  • use an integrated solution from the usual vendors.

Will the inrush current damage the MOSFET?

You have to choose MOSFETs that can handle this current as with any boost converter design. However, the high peaks that look to be around 80 amps are a problem and, you need to check that U2 is turned off before U1 turns on. Leave yourself a time gap of about 100 ns. It does look like shoot-through and this can be a serious problem if not resolved.

\$\endgroup\$
3
  • \$\begingroup\$ Spot-on, it was a overshoot and after adding some delay the current is now withing acceptable range of MOSFET tolerance, IRF540N can withstand upto 33A. on the other issue of the spikes I am thinking of a surge suppressor TVS diode, have added the spec on my original answer \$\endgroup\$
    – daydreamer2100
    Sep 24 at 12:35
  • \$\begingroup\$ @daydreamer2100 if we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. \$\endgroup\$
    – Andy aka
    Sep 24 at 14:27
  • 1
    \$\begingroup\$ I've also rolled your question back to how it was originally phrased so as not to contradict my answer given. Please only update a question in three scenarios (1) no answers currently given or (2), minor inconsequential changes that don't make answers look like they didn't read your question properly or (3), with the agreement of the persons who have already left answers (and I don't agree because you are just covering ground that my answer explained) hence, the roll-back. \$\endgroup\$
    – Andy aka
    Sep 24 at 14:33
0
\$\begingroup\$

You implemented shoot-through. :)

Notice your driver circuits are inverting:

enter image description here

enter image description here

enter image description here

When HIN_PWM or LIN_PWM is low, the respective gate is high (on). V1 is low from 28.9 to 40µs, and V2 from 0 to 29.5 and 39.2-... µs.

You may've also noticed the output voltage is not as expected, because the duty cycle is 1/(what you thought it was) due to the same inversion.

As for the output voltage transient, that is normal behavior of an open-loop [lack of] control. I assume this is the power stage for a controller like UC3843 and you won't actually run with fixed PWM, so this behavior is N/A.

\$\endgroup\$

Your Answer

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

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