Boost Converter Current Limit Question - Page 1
Forum section: electronics / projects

author: archnemesis
author_rank: Contributor
timestamp: September 11, 2019, 05:42:12 pm
content: I have designed and build a simple boost converter 12V to 25.2V, the schematic for which is attached below. The circuit is part of a larger battery pack project, and forms the integrated charger. The converter works well (is stable and ripple within design specs) and the current limiting circuit seems to work just fine mostly, but I ran across one detail that is giving me some confusion.I would have expected the voltage drop across R58 to be at least 0.6-0.7V (a standard diode drop) before Q13 would turn on and start pulling feedback voltage up, but I only measure about 350mV drop in steady-state while the pack is charging in CC mode, resulting in half the current I designed for (1A charge current). So I doubled up R58 to give me a 0.3R resistor and now I get the full amp I designed for. I even swapped out Q13 for a fresh one and it's the same behavior, which leads me to believe I am mistaken about something.Thanks!

author: MasterT
author_rank: Frequent Contributor
timestamp: September 11, 2019, 06:00:32 pm
content: Things to consider:1. Leakage current Ic.2. Missing protection resistor in series with base of the BC807, likely it's already dead.3. Current is not pure DC, some spikes goes up to 0.8V, but average RMS stays below 0.4V.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 11, 2019, 06:35:51 pm
content: You've got a mighty big resistor in series with Q13, implying it's operating at low currents. Vbe goes as 60mV/dec, so if Vbe is ~0.6V at 1mA, it'll be ~0.48V at 10uA.Possibly the remaining 200mV can be accounted for with noise? With only 15uF on there and no LC filter (and without layout, I can only guess at strays), there may be a fair bit of noise on R58, which if the load is a low impedance (depends how long the battery cable is, or anything else attached?), can easily have that much drop at radio frequencies. In that case the transistor briefly turns on very hard, discharging C22.I'm not completely sure what C22 was intended to do (or if it does it as well as it was intended to). It can't act as a Miller capacitance, because the base source impedance is so low. It will slow down the current feedback path, but only on the throttle-up slope. So as compensation, it's lopsided, not at all linear. The fix would seem to be simple: increase the base source impedance by placing a resistor (say 10k) between R58 and Q13-B. You can also introduce additional RC or LC filtering here, to keep switching noise away from Q13 without having to filter the whole output.Everything else looks alright:- The controller is old, but it's a peak-current-mode type which is good.- Turn-on (R52) is pretty slow but that may be okay; this will incur higher switching losses in CCM, though.- Q11 is reasonably dimensioned. It's newish, good Rds(on), low enough Qg(tot).- D12 isn't labeled, but I guess it's schottky which isn't really going to matter, there's not much difference between types.- There's slope compensation, allowing mild CCM operation and somewhat reducing the voltage drop required on R51.- Compensation is (R+C) || C, lots of freedom to adjust values.- There's a power LED on both sides, although the output side is returned to ground so it will always be lit (to some degree); this might be better returned to +12V, so it's off when the output is off. (Not that there's an enable function on this circuit, so it'll be on whenever the thing is powered.)- If you can afford a current sense amp, you can save a few milliwatts on R58, instead of using the BJT. Pretty inconsequential.- If you can afford a split ground, you can place R58 in the ground-return path instead, and use, probably the same thing ultimately, but it'd be a collector pulling down on COMP, rather than up on FB.Also, completely alternative to sensing output current -- why not use the intrinsic current limiting of the controller? It won't be constant, but if it's more for protection or limiting consumption, that's still fine. (For a battery charger, I'm not aware of any chemistry that's so anal about how fast it's charged, as long as it's in the right ballpark. That is to say, your current limit could be +/-30%, even 50%, and as long as nothing else burns up, the battery probably won't care.)Of course, a boost can't limit current for voltages below Vin anyway, so you can't expect 100% constant current regardless; if that is an advantage (also, possible short-circuit or reversed-output protection), consider a SEPIC converter instead. This works same as boost or flyback, with L1 replaced by a dual-winding inductor, and the opposite side being tied with ground, instead of to +V. (You can also put a coupling cap across the "hot" ends of the inductor, since they have the same AC voltage, and this reduces ripple current in the inductor.) This is probably a cost adder, but coupled inductors are very common in these values at least.Tim

author: archnemesis
author_rank: Contributor
timestamp: September 11, 2019, 07:03:15 pm
content: Quote from: T3sl4co1l on September 11, 2019, 06:35:51 pmThis is a good point, I think the capacitor I chose has too high ESR and is contributing a bit more noise than there should be. My plan for the next revision is to increase capacitance a bit and lower ESR by paralleling capacitors.Quote from: T3sl4co1l on September 11, 2019, 06:35:51 pmYes this is the part of the circuit I was most unsure about, that it didn't oscillate wildly once I built it really surprised me. I was under the impression that the cap should slow down the current limiting action so as not to introduce instability. I will add that resistor as suggested and perhaps experiment with filtering.Quote from: T3sl4co1l on September 11, 2019, 06:35:51 pmThanks for this feedback, very reassuring and encouraging. I'll have to digest the rest of your advice as I continue on. Thanks!

author: jmw
author_rank: Frequent Contributor
timestamp: September 11, 2019, 07:19:29 pm
content: What's the purpose of the circuit around Q13? To change the voltage output when the current exceeds 1 A? 25.2 V into R57 / (R57 + R56) is 2.5 V, so if Q13 is on above 1 A of output current, the voltage divider is R57 / (R57 + R56 || R59), which puts V+ at 17.8 V to deliver 2.5 V to FB. Or am I misunderstanding this part of it?

author: archnemesis
author_rank: Contributor
timestamp: September 11, 2019, 07:43:31 pm
content: Correct, the load is a 6-cell Lithium-Ion battery with a max charge current of 1A.Dave has a good video covering how and why: