LM3671 step-down converter - does it require additional filtering? - Page 1
Forum section: electronics / projects

author: kacper
author_rank: Newbie
timestamp: December 29, 2016, 05:46:11 pm
content: Hi,I'm designing a AVR XMEGA(ATxmega32A4U) board using a LM3671 step-down converter to convert the 5V input voltage to 3.3V. Does the LM3671 require additional filtering in it's output. I suppose a couple of 0.1uF capacitors wouldn't hurt but then how does one calculate how many to use? I'm mostly worried that high ripple on the supply will screw up the ADC.

author: KhronX
author_rank: Frequent Contributor
timestamp: December 29, 2016, 06:04:12 pm
content: "Additional" = in addition to what, exactly?I assume / hope you've read through the datasheet:http://www.ti.com/lit/ds/symlink/lm3671.pdf(hint: pages 16-17)Also, on the product page on the TI site, in the "Technical Documents" section, there's a link to this article:http://electronicdesign.com/analog/comparing-power-supply-architectures-12-bit-adc

author: Benta
author_rank: Super Contributor
timestamp: December 29, 2016, 06:33:46 pm
content: I wonder why you would want to use a switching regulator to go from 5 V to 3.3 V.An LDO is much less complicated, and we're not talking about a lot of current.

author: kacper
author_rank: Newbie
timestamp: December 29, 2016, 07:15:57 pm
content: Quote from: KhronX on December 29, 2016, 06:04:12 pmYes I've read through the datasheet. The datasheet specifically states "The output filter capacitor smoothes out current flow from the inductor tothe load, helps maintain a steady output voltage during transient load changes and reduces output voltage ripple.". So if the output capacitor reduces the ripple, does it reduce the ripple to low enough amount or do should I put additional bypass capacitors to lower the ripple further?Quote from: Benta on December 29, 2016, 06:33:46 pmThe USB max current is 500mA, a 1.7 voltage drop in a LDO would generate 0.85W. It's a lot do dissipate without a heat sink (at least what I've been able to gather through my research.) Will by design draw so my current, probably not, but it will add some room for drawing more current.

author: Retep
author_rank: Regular Contributor
timestamp: December 29, 2016, 08:25:23 pm
content: Quote from: kacper on December 29, 2016, 07:15:57 pmDefine "low enough ripple", then check section 8.2.1.2.3 (page 18) of the datasheet. In fact I recommend to read the entire chapter 8 of the datasheet carefully.QuoteIt is a bit strange to take the max. supply current of USB as a starting point. I'd recommend you get some ballpark figures first for the current draw of your circuit first, before dismissing an LDO as a possible option.

author: kacper
author_rank: Newbie
timestamp: December 29, 2016, 09:30:45 pm
content: Quote from: Retep on December 29, 2016, 08:25:23 pmI can't. I guess what I'm reaching for is what the maximum amount of ripple would be acceptable to not crash the mcu or affect other components so they malfunction. The lower the ripple the better but how much is acceptable?I've seen 0.1uF and 10uF capacitors been used to filter out noise and ripple but the number of how many bypass capacitors are needed and how to calculate that has eluded me.QuoteI was unsure how to calculate those numbers correctly, so I took the worst case scenario current draw knowing full well that the circuit would probably draw less than that. The maximum current draw for the mcu is 200mA, but how much current is needed to hold a pin high, do pwm and so on. Forgive me if it all sounds stupid, but I would rather not spend the entire day trying to figure out the numbers. Maybe I'm trying to hard and should just go with an estimation, but it's hard to estimate something without experience.

author: kacper
author_rank: Newbie
timestamp: December 29, 2016, 09:39:21 pm
content: Quote from: Benta on December 29, 2016, 06:33:46 pmAre switching regulators that much harder to use so that it doesn't warrant the effort and learning curve? The typical application example of the LM3671 didn't look to hard assuming one can somewhat adhere to the layout guidelines.

author: Benta
author_rank: Super Contributor
timestamp: December 29, 2016, 10:13:22 pm
content: Quote from: kacper on December 29, 2016, 09:39:21 pmHere comes my personal opinion, OK?Switching regulators are a major pain in the arse, and should only be designed by people whoreallyknow what they are doing.Ask your neighbours, who might like to listen to radio or watch TV.Old-fashioned DC/DC converters at 20 kHz make enough trouble as it is. And your 2 MHz converter has the potential formajortrouble.The EU EMC Directive may be regarded as limiting creativity, but it's there to protect other people.Rant over.In your application, a power loss of 0.85 W is nothing. Using an LDO will give you a wonderfully quiet supply for your ADC. There are even low-noise LDOs available.Benta.

author: Retep
author_rank: Regular Contributor
timestamp: December 30, 2016, 03:13:13 pm
content: Quote from: kacper on December 29, 2016, 09:30:45 pmIf you can't define how much ripple is acceptable, how are we supposed to tell how much filtering is needed without knowing your circuit and application? Crashing the MCU is one consideration, though in my experience the 8-bit AVR's are quite tolerant if you take the basic precautions. But you also mentioned the ADC, what do you expect from it and what are you going to use as voltage reference? I bet there are many other factors to consider...but then again we don't know anything about your circuit and application.QuoteThe formulas are in the datasheet (page 18), and the capacitors values (andtypes!) used in their examples are no doubt good enough forsomeapplications. Whether they are good enough foryourapplication is a question only you can answer.QuoteA maximum current draw of 200mA seems to be rather excessive for a 8-bit AVR microcontroller; I'd expect the typical power consumption being at least 10x less than that. 200mA seems to more like the maximum current that the power supply pin can handle rather than the typical current draw of the MCU itself (those are not the same thing!). How much current is needed to hold a pin high depends pretty much on what is connected to that pin. If nothing is connected the state of a output pin will have negligible impact on the power consumption of the MCU. Otherwise the current needed to supply of whatever is connected to the output pin needs to be added to the total power consumption. Of course I don't know what other hardware is in your circuit and how much current that is expected to draw.QuoteGetting some ballpark numbers wouldn't take all day. And to determine what capacitor you need to get an "acceptable" ripple with you will need those numbers anyway. But if your approach to engineering is that you can't be bothered to make some rough estimates, why not also take the gamble that an LDO will do just fine? Based on the information you supplied sofar I'd expect your chances of success with an LDO would be better than with a switching step-down converter.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 30, 2016, 05:00:45 pm
content: If in doubt, prepare to have a filter. For just a single piece you don't really care about the maybe 20 cent's.Usually the switched mode regulator should be OK for the µC but the ADC might like some filter to get full performance.I would also guess the LDO is the way to go. If there are other high power loads (e.g. LEDs, relays) these might run from 5 V anyway.Even with an LDO or an other clean source, you may need a filter for the analog supply of the µC, as the µC itself can cause some noise too.