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I have implemented a step up circuit using an IC. I implemented the component selection from the datasheet. This is its datasheet.

step-up converter schematic from datasheet

Image source: LCSC - AZ34063U datasheet

The step up feeds an H-bridge. The H-bridge offers two pins with an alternating power output that feeds a load.

I received my boards, and I connected the load and put the PCB to the test, to see how hot the chip would get.

After some minutes, I accidentally disconnected one pin of the load that was connected to the H-bridge.

Then I saw smoke coming out, and two seconds later I saw sparks coming out of the diode. The diode was destroyed.

I am not sure if the destruction of the diode is due to the removal of one pin of the load from the H-bridge output, or if that happened at that exact time coincidentally and perhaps other reasons, such as the temperature were the culprit.

This is the diode I used.

The output voltage was 30V, and the load was consuming a 0.25A current.

EDIT: Suggestion made by user Russell McMahon

enter image description here

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    \$\begingroup\$ You need to show the H-bridge and please explain which diode you are referring to. Add links to relevant data sheets please. \$\endgroup\$
    – Andy aka
    Commented Oct 9, 2023 at 10:43
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    \$\begingroup\$ The diode D1 is NOT a protection diode - it is the output rectifier for theboost converter. Under load the circuit will regulate using R1 and R2. Under no load the energy in the inductor in one switching cycle MAY be enough to drive the output voltagfehigher or far higher than designed. The1N5819 is rated at 40 volts. A disconnected load is a reasonable reason for its demise. Placing a zener diode at the output with perhaps a small series resistor will stop no load deaths. 29v < Vz < 40V || (MC34063 - very olde indeed. Still useful :-). I've used several 100,000 of them). \$\endgroup\$
    – Russell McMahon ♦
    Commented Oct 9, 2023 at 12:26
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    \$\begingroup\$ @user1584421 - Hi, I see you have written comments, but didn't use @username notifications - explained here and here. FYI without those: (a) people may not guess who each comment is intended for (so you may not get replies, or may not get them from the expected person); (b) unless comment-writers clicked the "Follow" link, they won't get any notification about your comments and therefore may not see them & may not reply. || I always recommend using @username in comments :) \$\endgroup\$
    – SamGibson ♦
    Commented Oct 9, 2023 at 21:06
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    \$\begingroup\$ @user1584421 A zener plus maybe resistor in series is placedfrom Vout to ground. || My "solution" is not magic but rather that something which provides a sacrificial target in place of the diode IF Vout rises higher than intended, regardless of why. This will work for any power supply - for a long as the zener survives :-). || What it does is absorb any energy above Vz. If Vz is say 33V and robust enough to absorb a single cycle of inductor energy then it will face any energy surge first. IF the IC is regulating correctly there will only be one initial suge, if any. \$\endgroup\$
    – Russell McMahon ♦
    Commented Oct 13, 2023 at 7:38
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    \$\begingroup\$ @user1584421 Yes - zener position is correct - BUT you show its polarity reversed - it should block voltages under about 30V. || It's not possible to be CERTAIN that this is the problem as eg the diode may be being marginally stressed by the startup cycle. || I'd expect this NOT to happen as inductor should store about 150 uJ at about 1.5A current limit - If @ 330 uF C2 should store about 500 uJ at 30V so should not be "overvolted in one cycle. Even 2 cycles (first just below Vreg) should only get it to 300 uJ. || IS C2 = 330 uF? If say 100 uF it would be more likely to happen \$\endgroup\$
    – Russell McMahon ♦
    Commented Oct 14, 2023 at 4:29

2 Answers 2

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Made with microcap v12 FREE

Here is the behavior of this IC.
Note that there is no capacitor at the input which should "break" the input pulse current.

Just note the high current at the start ... current or no current into the load.

enter image description here

Load wired ...

enter image description here

Just change the "slew-rate" of input power ...

enter image description here

And here are the "powers" ...

enter image description here

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  • \$\begingroup\$ Hi Antonio! Thank you for the answer! I just placed some comments on the other answer. Perhaps you find them useful! Thank you for the graphs, but I am having trouble understanding them. Do they reveal the issue for the demise of the diode? \$\endgroup\$
    – user1584421
    Commented Oct 9, 2023 at 20:49
  • \$\begingroup\$ Just "perhaps" change the slew rate of input power (your capacitor C1 at input ...) \$\endgroup\$
    – Antonio51
    Commented Oct 9, 2023 at 21:00
  • \$\begingroup\$ This is the capacitor that I used: lcsc.com/product-detail/… \$\endgroup\$
    – user1584421
    Commented Oct 9, 2023 at 21:08
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    \$\begingroup\$ No. Don't change the capacitor. It is "ok". I simulated the "opening" of the load. Nothing to say (see new simulation). Don't understand why the diode got burned. \$\endgroup\$
    – Antonio51
    Commented Oct 10, 2023 at 11:17
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    \$\begingroup\$ No worries for that. \$\endgroup\$
    – Antonio51
    Commented Oct 10, 2023 at 11:38
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This complements Antonio51's answer.

Summary:

  • A look at the diode's single non repetitive surge current ratings and Antonio51's simulations suggests that diode D1 is being run at at least 150% of its single surge rating and maybe more depending on the startup surge duration. Adding the 1 Ohm series resistor plus capacitor that Antonio51 shows in his final diagram should eliminate this possible failure mode.

The startup current surge is the equivalent of briefly shorting the output. This is due to charging C2, 30 uF.
Fig 15 on page 5 of the datasheet shows the peak allowed non repetitive single cycle current is 25A - which is about 2 x what Antonio51's simulations show.

HOWEVER fig 10 page 3 suggests an allowed 1mS peak of under 8A. The simulation shows somewhat less time BUT 50% more current. This seems very marginal.

Adding Antonio51's extra input R7 of 1 Ohm and C3 of 100 uF will lower Ipeak and reduce this possible cause. (At 8A surge this will drop 8V, so 8A will not happen. As an 8A <= 1 ms surge is (just) acceptable and the actual will be lower again "all should be well".

Note: The output zener that I mention elswhere has no useful role in this fault mode.

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