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I'm simulating an LM2596 DC-DC voltage regulator to give me a 5-volt output given a 5 ohm load (1 amp). The circuit is shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

In my simulation, after throwing the switch after the LM2596 has reached a stable voltage, the load voltage after the switch spiked to around 6.6 V before stabilizing to around 4.968 volts. I'm trying to mitigate that spike by adding a capacitor and a zener diode before the switch in order to maintain the load voltage at approximately 5 volts. Without D3 and only C5, it spikes to 5.5 V before settling to 4.968 V, and with D3 and C5 together, it gets reduced down to 5.04 V and settles to 4.965 V in a matter of milliseconds. Is this a good method at 'stabilizing' the load voltage when using a regulator and a switch? What are some drawbacks of using a zener diode in this method?

EDIT: The SPICE model I'm using in my simulations is an adjustable regulator which I set to 5 V using a voltage divider, but the IC I plan to use is fixed at a 5 volt output. That's what I'm showcasing in my schematic, a fixed-voltage regulator.

EDIT: I'm basing my circuit off of TI's recommended schematic, a fixed output voltage regulator. The image is below, but I used TI's webbench tool to get the values shown in my schematic above.

Recommended Schematic

EDIT: Below is a simulation of the output DC waveform at the load. Using a 15 V input, I simulated four different variations of the LM2596. The first plot shows the recommended datasheet's parts while the other three show Webbench's part suggestions with and without a post-ripple filter as well as a small 10 nF capacitor on the load at the last plot.

enter image description here

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  • \$\begingroup\$ ??? before settling to 5.968 V ??? \$\endgroup\$
    – Andy aka
    Apr 20, 2020 at 18:13
  • \$\begingroup\$ @Andyaka Sorry, my bad. Fixed it. Typo. \$\endgroup\$
    – BestQualityVacuum
    Apr 20, 2020 at 18:24
  • \$\begingroup\$ Can you add waveforms to see what is happening? \$\endgroup\$
    – Ricardo Nunes
    Apr 20, 2020 at 21:06

2 Answers 2

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Is this a good method at 'stabilizing' the load voltage when using a regulator and a switch?

All your problems come from the extra LC you have added namely L2 and C4+C5. Together they convert a low pass filter into hyper resonant tuned circuit that resonates at 3.85 kHz and has a Q factor of nearly 70. Any step change in the load will cause significant ringing and over-voltage problems on your output.

It seems you are going down the road of setting a spider to catch a fly then sending in the bird to catch the spider. You need to step back and work out why you have introduced L2.

BTW - don't try using this within the feedback loop of the switching chip because you'll just make an oscillator.

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  • \$\begingroup\$ Thanks for your reply. L2 and C4 are part of a optional post-ripple filter circuit suggested by TI. I introduced it to reduce the ripple at the output, but seeing as how you said it might be part of the problem, I'm running some sims to see if TI's suggested circuit is better than what I am running. \$\endgroup\$
    – BestQualityVacuum
    Apr 20, 2020 at 18:51
  • \$\begingroup\$ I ran some more tests with and without the optional ripple filter that I mentioned, and it seems the the addition of the filter smooths out the voltage at the output load, reducing the ripple voltage at the output. Without that filter, with nominal voltage being 5 volts, the output voltage oscillates between 4.23 and 5.28 V. With the filter, it goes down to \$\endgroup\$
    – BestQualityVacuum
    Apr 20, 2020 at 19:43
  • \$\begingroup\$ Sorry, pressed wrong button.With the filter, the output voltage spikes to 6.53 V, but after 1.3 ms, it goes down to between 4.97 and 4.96 volts. Perhaps it's my SPICE model, but is the fixed LM2596 known for ripple voltage at the output? Also, what were you going to say at the end there? \$\endgroup\$
    – BestQualityVacuum
    Apr 20, 2020 at 19:49
  • \$\begingroup\$ How odd. Maybe it was a rogue sentence I’d started that got lost along the way. \$\endgroup\$
    – Andy aka
    Apr 20, 2020 at 20:10
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    \$\begingroup\$ Yes, an addition of a 10 pF capacitor across the feedback and the output (in parallel with the feedback resistor in my simulation that's not 1K) aided in my simulation by speeding things up. Thank you for that, though ideally, I'll be using the 5V-fixed version, so I would just directly connect the output pin to the feedback pin like on the datasheet. I went with it as it seemed to suit my needs as a power supply, and it's affordable. \$\endgroup\$
    – BestQualityVacuum
    Apr 23, 2020 at 14:54
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You definitely need larger capacitors at the input. Start with 100uF and see how much it helps, then 220uF, then 470uF.
Your C3 also seems small.
Finally, you should take the feedback from the output (at C4), not before it.

EDIT: I have re-read a part of your question and realized I had missed the part about the switch(ing) causing the problem. DOH!
Try adding some small capacitor (like 100nF to 10uF, see what range of values works) at the output AFTER the switch, NEXT to the load. I am suspecting switch/contact bounce and lead inductance.

Also, you shouldn't stress too much about simulations, as they often show a problem that isn't there in a real-world circuit, or they miss a problem that IS there in the real world.

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  • \$\begingroup\$ Thanks for your reply. I was basing my circuit off of TI's recommended layout, whose values were chosen thanks to TI's webbench tool. I think I see what you mean about taking the output at C4, and I'll run some sims to see how that changes things. \$\endgroup\$
    – BestQualityVacuum
    Apr 20, 2020 at 18:55
  • \$\begingroup\$ I ran some sims based on what you said, and taking the feedback at C4 caused the output voltage to oscillate from 4.996 V to 5.09V. Taking the output at C3 produced a smoother output (with almost no oscillation) at around 4.97 V. Adjusting the value of C3 given my current values seems to cause distortion at the output even when I increase it to a larger value. \$\endgroup\$
    – BestQualityVacuum
    Apr 20, 2020 at 20:04
  • \$\begingroup\$ Just now I have re-read a part of your question and realized I had missed the part about the switch causing the problem. DOH! Try adding some small capacitor (like 100nF to 10uF, see what range of values works) at the output AFTER the switch, NEXT to the load. I am suspecting switch/contact bounce and lead inductance. \$\endgroup\$
    – Edin Fifić
    Apr 20, 2020 at 20:11
  • \$\begingroup\$ I posted my simulation circuit, showing comparisons between the original recommended values in the datasheet and what was calculated using TI's webbench tool. Results when adding a 10 nF capacitor at the output is shown at the last plot. It seemed to have a negligible effect when viewed with the parts before. \$\endgroup\$
    – BestQualityVacuum
    Apr 20, 2020 at 22:21
  • \$\begingroup\$ 10nF is very small for a 1A circuit, but it does have a noticeable effect. Try 100nF, 1uF and 10uF. It is also important what type/chemistry of the capacitor you're using at higher values. Electrolytics are slow, while ceramics can ring, so you could also combine a 100nF-1uF ceramic with a 10uF electrolytic. Finally, try increasing the C4 or Cout to 470uF and parallel it with a 0.1uF-1uF ceramic. You will often see up to 3 or even 4 different value capacitors paralleled to prevent ringing and because they have different pulse and high-frequency responses. \$\endgroup\$
    – Edin Fifić
    Apr 21, 2020 at 5:19

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