Switching voltage regulator design output voltage issue - Page 1
Forum section: electronics / projects

author: dominicM
author_rank: Regular Contributor
timestamp: August 21, 2014, 03:08:17 am
content: I am new to switching regulators and relatively new to electronics in general. I am having an issue with a regulator I designed.I tested it with a battery input (7.2V) and DC adapter (16V). I get output voltage of just 350mV instead of the expected 5V. There is also a quiet noise when powered. At one point I had forgotten to order a shottkey diode so I tried powering it on without it. I got the expected 5.03V output thought it became unstable at about 7 to 9V range (DC adapter slowly discharging after AC loss). I went and ordered the missing part but now getting the issue describer above.Here is the schematic:And here's the PCB:What could my error be?

author: DanielS
author_rank: Frequent Contributor
timestamp: August 21, 2014, 03:50:58 am
content: By running your circuit without the schottky diode to clamp the inductor's free-wheeling current, you may have busted your PWM. It "worked fine until 7-9V" only because the there was not enough stored energy in the inductor to kill your regulator before then.Put the Schottky in, replace your regulator and assuming your circuit is wired correctly, it should work since you are using the reference design as-is.

author: georges80
author_rank: Frequent Contributor
timestamp: August 21, 2014, 04:07:33 am
content: Yep, the schottky is there to clamp the SW node from going negative. Without it you have most likely destroyed the SW output of the switcher IC.Consider how the voltage across the inductor will change as the VIN is applied to SW and then removed by the switcher at the switching frequency... Since the 'output' of the inductor is feeding a capacitor that initially is at 0V you will get large -ve voltages at the SW node as things power up... poof...cheers,george.

author: dominicM
author_rank: Regular Contributor
timestamp: August 21, 2014, 12:57:54 pm
content: It actually worked fine at all input voltages except at ~7-9V It was fine at 9-16V and 6-7V. Under 6V voltage started to drop as expected. Damaged chip does make sense, though I am not sure why I only got the noise and wrong output voltage after I added the diode... Any insight on this? In any case I will try to replace the chip and see if it works, thanks!

author: Niklas
author_rank: Frequent Contributor
timestamp: August 21, 2014, 01:59:54 pm
content: 350 mV is quite close to the forward voltage drop of a Schottky diode. Check the direction of the diode.The noise can come from the over current protection in the switcher IC that continues to switch and hopes that the fault goes away.

author: dominicM
author_rank: Regular Contributor
timestamp: August 21, 2014, 02:23:00 pm
content: Quote from: Niklas on August 21, 2014, 01:59:54 pmThe end of the diode with white line is connected to ground. Is that right? If I am right the line indicates a cathode (negative). Does it by any chance need to be connected in reverse?

author: DanielS
author_rank: Frequent Contributor
timestamp: August 21, 2014, 03:34:00 pm
content: Quote from: dominicM on August 21, 2014, 02:23:00 pmI looked at your PCB but the picture was too blurry to tell fore sure which orientation your diode was in. Looks like your diode is on backwards.This "free-wheeling" diode is supposed to allow the inductor's current to continue looping through the inductor, capacitor and diode loop so the inductor can transfer its energy to the capacitor while the PWM switch is off. Connected from PWM output to ground, it simply shorts the PWM switch to ground.

author: fcb
author_rank: Super Contributor
timestamp: August 21, 2014, 04:07:48 pm
content: How are you grounding the MP2456 (pin2)?

author: dominicM
author_rank: Regular Contributor
timestamp: August 21, 2014, 05:17:41 pm
content: Quote from: DanielS on August 21, 2014, 03:34:00 pmSo should the end with a white band connect to ground or SW? Now the white band is on the GND side.

author: dominicM
author_rank: Regular Contributor
timestamp: August 21, 2014, 05:18:57 pm
content: Quote from: fcb on August 21, 2014, 04:07:48 pmI was using a battery for testing so I just clipped the two pins to positive/negative of the battery.

author: Niklas
author_rank: Frequent Contributor
timestamp: August 21, 2014, 07:38:15 pm
content: Quote from: dominicM on August 21, 2014, 02:23:00 pmThere you have it! The white band should not be a ground. Desolder and rotate the diode

author: dominicM
author_rank: Regular Contributor
timestamp: August 22, 2014, 02:49:48 pm
content: Quote from: Niklas on August 21, 2014, 07:38:15 pmYup, that was it! Can you give an explanation or a link to one for why it needs to be reversed, seems counter intuitive to me. I think I confused it with a zener diode when reading up on it before...

author: mikerj
author_rank: Super Contributor
timestamp: August 22, 2014, 03:14:19 pm
content: You have the diode symbol correct in your schematic, so I suspect you have simply confused the anode and cathode ends on the symbol.An easy way to remember this is that the cathode end of the symbol looks like a K (remember K for 'K'athode)Cathode ---K|---- Anode

author: dominicM
author_rank: Regular Contributor
timestamp: August 22, 2014, 07:37:37 pm
content: Quote from: mikerj on August 22, 2014, 03:14:19 pmWhat confuses me is why you connect the negative/cathode to the positive and anode/positive to the ground.

author: nctnico
author_rank: Super Contributor
timestamp: August 22, 2014, 08:02:21 pm
content: A diode doesn't have a positive or negative. Just a direction in which it lets current through (from anode to cathode).Google for 'step down switcher' and you'll find some interesting theory.

author: dominicM
author_rank: Regular Contributor
timestamp: August 22, 2014, 09:34:19 pm
content: Quote from: nctnico on August 22, 2014, 08:02:21 pmIt has an anode and a cathode, does it not? An LED (type of diode) also has anode and a cathode. All I want to know is why a schottky diode connects in reverse compared to many/most other components like an LED. Is there a general rule or is schottky diode an exception?

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: August 22, 2014, 10:13:03 pm
content: A schottky diode conducts in the same direction an LED does. The usage here is different, though. You don't want it conducting all the time or you've essentially shorted out your load! (which is why you were measuring ~0.3V at the output.)Your next task is to understand why there's a diode there at all

author: rob77
author_rank: Super Contributor
timestamp: August 22, 2014, 10:26:18 pm
content: Quote from: dominicM on August 22, 2014, 09:34:19 pmapparently your understanding is quite limited... there is no positive or negative for any diode ! simply get used to itany diode is a one-way valve - it let the current go from anode to cathode (conventional current direction model) and blocks the current in the opposite way - that's valid for ANY diode even for the LEDs you're mentioning - but those leds have a visible side effect - they emit light while letting the current gonow to the direction of the shottky in your schematic - probably you don't even know... but the switching regulator is called switching because of some switching (on-off-on-off-on) is done inside... when the switch is ON - trhe current goes throught the switch and inductor to the load. when the switch is off - you need to somehow catch the energy stored in the inductor - and that's the job of that shottky diode. and actually it's sometimes called a "catch diode".when you connect the cathode to ground and anode to the switch - then you simply let the current go from the switch to the ground and it doesn't work as you already realized.when you connect the anothde to ground and cathode to the switch, it will exactly do what it's supposed to do - catch the energy stored in the inductor when the switch is off.

author: dominicM
author_rank: Regular Contributor
timestamp: August 22, 2014, 10:56:24 pm
content: Quote from: Galaxyrise on August 22, 2014, 10:13:03 pm"The usage here is different" - this is answer my question. I a have a vague understanding of what it does but will need to come back to it to understand it better.

author: dominicM
author_rank: Regular Contributor
timestamp: August 22, 2014, 11:28:45 pm
content: QuoteI have stated a am new to this in the original question, not sure why you feel the need to say explicitly what is already implied.QuoteIt was said just a few post before, this is why I specifically said cathode/anode instead of positive/negative.QuoteThis only makes me think that in practice it does have what you might call positive/negative even though theoretically it does not. There is anode to which you connect + voltage and a cathode to which you connect -. It's one thing to correct me so I know next time, but you must have known what I meant? Why not correct my miss understandings AND answer my question too? It's kind of like if foreigner asked me a question in bad english but I knew what they meant I would just answer the question instead of saying their english is bad, because it would be rude and unhelpful to just do the latter.QuoteSo you think I learned how to use cad software, drew the schematic, laid out the PCB, ordered all the components, went through the process of making my own PCB and I don't even know the core method of operation of the thing I am building? This is just condescending and improbable to the point where it makes you look silly for saying it...

author: DanielS
author_rank: Frequent Contributor
timestamp: August 23, 2014, 12:34:52 am
content: Quote from: dominicM on August 22, 2014, 07:37:37 pmWhat is the particularity of an inductor in an electrical circuit? Inductors oppose to changes in current.What happens to the inductor's current when your switch turns off? It tries to keep going and by doing that, it inverts polarity across the diode. That's why the diode is "in reverse:" to clamp that and let the inductor dump its energy in the output capacitor.

author: dominicM
author_rank: Regular Contributor
timestamp: August 23, 2014, 02:14:48 am
content: Quote from: DanielS on August 23, 2014, 12:34:52 amThat clears it up even more, thanks!

author: mikerj
author_rank: Super Contributor
timestamp: August 27, 2014, 07:58:32 am
content: Quote from: dominicM on August 22, 2014, 07:37:37 pmJust think about it in terms of current flow. When the switching transistor in the MP2456 turns off, there needs to be a path for current to flow from right hand side of the inductor, through the load, back to ground and then back to the other end of the inductor. In this case the left side of the inductor will be at a voltage which is lower than ground by one diode drop so the anode will be at a higher voltage than the cathode.

author: rob77
author_rank: Super Contributor
timestamp: August 27, 2014, 05:29:03 pm
content: Quote from: dominicM on August 22, 2014, 11:28:45 pmotherwise you wouldn't ask such a question about diodes, would you ?