Is XL4005 Current-Limited? - Page 1
Forum section: electronics / beginners

author: johnyradio
author_rank: Newbie
timestamp: June 27, 2014, 12:02:16 am
content: Hi,i purchased this board with adjustable voltage, but non-adjustable current.http://r.ebay.com/JIwHXtDatasheet:http://www.xlsemi.com/datasheet/XL4005%20datasheet.pdfWith a heatsink on the chip, the diode burned out while charging a 12Ah li-ion at 4.2V.I want to try again with a heatsink on the diode, but my buddy says: "if the board was current-limited, then the diode would not have fried, so it is NOT current-limited."Is he correct?Someone else said: "there is no current limit. the datasheet says it has thermal and current limiting functions-- it doesn't. furthermore, reducing the frequency to 60khz (from 300 according to the datasheet) will only wreak disaster on the circuit because the inductor if it wasn't saturated already, will certainly be now with a 5 fold reduction in frequency. ".Is he correct?Thx!

author: wraper
author_rank: Unknown
timestamp: June 27, 2014, 12:22:01 am
content: So you were charging 12 Ah Li-ion (considering the capacity sounds more like Li-po, whatever not big difference) battery with 5.5A DC/DC converter without proper current limiting and controlling the charge? Nice way how to burn your house.

author: johnyradio
author_rank: Newbie
timestamp: June 27, 2014, 12:24:30 am
content: no, this chip is SUPPOSED to be current-limited. That's what i'm asking about.

author: c4757p
author_rank: Super Contributor
timestamp: June 27, 2014, 12:24:36 am
content: The XL4005 has a current limit, or so it claims, but that doesn't mean the whole board has a useful current limit. Using that diode on that board, given the current limit spec of that chip, and claiming 5A output, is mental. Let's do a little math:Assuming (pulled out of my ass) 12V in, 4.2V out: duty cycle is 4.2V/12V = 0.35.The limit current is actually given as 8Atypicalwith no actual tolerance given.The average diode current will be roughly (8A)(0.35) = 2.8A.The datasheet for the SS54B gives a forward voltage of 0.55V.The power dissipated in the diode will be (2.8A)(0.55V) = 1.54W.The datasheet gives a thermal resistance of 55°C/W, so the diode will rise 85°C above ambient.Assuming 25°C ambient, the diode is at 110°C.This iswaytoo hot, but still under the absolute maximum. Now, to push it higher, all you need to do is:Drop the input voltage. For example, at 10V input, the diode goes about to the absolute maximum of 125°C.Increase ambient temperature. Is it summer where you are?Use a cheap, possibly faulty diode. Not unlikely.Not give a sufficient path for heat to exit the diode, causing the thermal resistance to go above 55°C. Not unlikely.Have the current limit kick in at greater than 8A due to the fact that it doesn't even specify a tolerance. Not unlikely.tl;dr: yes, the chip has a current limit, but whoever designed the board didn't give a shit.Also, what wraper said.

author: johnyradio
author_rank: Newbie
timestamp: June 27, 2014, 01:44:12 am
content: Doesn't this forum say something about being nice to noobsjk. appreciate the detail. Not tldr. You gave me greater understanding of how to interpret the specs.Quote from: c4757p on June 27, 2014, 12:24:36 amso basically, to extend your summary slightly, "didn't give a shit" means they used a diode that's under-rated (thermally speaking) for this board, right?Question, would a sufficient heatsink on the diode suffice to allow me to drive it to 5A or 8A?Quote-Actually i'm giving it 19V. Again, with sufficient heatsink on that diode, can i get away with that? Let's see:-Duty Cycle: 4.2/19 = .22-Diode Current: 8A*.22 = 1.77A-Diode power dissipation: 1.77A * .55V = .97W-Diode temp: 55*.97 = 53.5CHmm, that's actually seem cooler. Is my math wrong? Seems counter-intuitive that a higher supply V would run the diode cooler. Which makes it stranger that the diode blew. But hey, i'm a noob.QuoteOk so you're saying there must be another variable that's pushing the diode to hot. I think i'm almost getting this.QuoteMy V is higher, not lower...QuoteSF does get not particularly hot...QuoteNow i'm thinking about replacing that diode with a hotter-rated diode, same board, because i like the price and size of this board. Now trying to figure out how to find a correct replacement part.... Suggestions welcomed!QuoteYep, gonna throw a heat-sink on there. Your analysis makes me think this (plus a higher input V), same diode, might actually work.QuoteThis other ebay page (same board) says "Load regulation:±0.5%". Did they pull that out of their ass?http://r.ebay.com/QRwl6v

author: c4757p
author_rank: Super Contributor
timestamp: June 27, 2014, 01:57:57 am
content: Again:do not charge the battery this way!!It's not safe. Evenwithgood current limiting, you're not keeping the battery from being overcharged, which with this type of battery is actually somewhat complex. Use a proper charger.QuoteHey, I did your damned math homework for you!Quote from: johnyradio on June 27, 2014, 01:44:12 amIn theory. In practice, two problems: 1) That diode's not meant to take a heat sink, so it probably won't get good thermal transfer to it. 2) Many regulator chips don't perform well in current limiting. Even the "decent" ones can blow themselves up. A current-limited regulator isnota constant-current supply.QuoteEr... yeah, my math was wrong. Quick sanity check: If the supply voltageequalsthe output, the current flows through the FET in the regulator chipallthe time, and through the diodenoneof the time.QuoteWorth a try. MBR2045CT is one of my favorites when heat-sinking is required. It's a TO-220 with a tab that can take aproperone.QuoteLoad regulation has nothing to do with the current limit - it's how much the voltage changes (assuming the current limit isnotactive) when the load varies.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: June 27, 2014, 02:28:33 am
content: Hi,A small error in the maths for calculating the diode dissipation.Duty cycle for the switch in the Buck converter is:D= Vout/VinSince the diode is conducting when the switch is not conductingDuty cycle for the diode is:1-D or 1-Vout/VinSo with 19V in and 4.2VThe diode duty cycle is 1 - 4.2/19 = 1 - 0.22= 0.78If the current limit is 8AThen diode dissipation is 8A x Vfwd x 0.788 x 0.55 x 0.78 = 3.4WThis is about4x more dissipationand therefore a4x temperature rise.Regards,Jay_Diddy_B

author: johnyradio
author_rank: Newbie
timestamp: June 27, 2014, 03:25:29 am
content: I'm slightly confused. Which is the thing you're saying" don't do it!", and which is the thing you're saying "worth a try"? Sorta sounds like you're referring to the same thing in both cases.

author: c4757p
author_rank: Super Contributor
timestamp: June 27, 2014, 04:34:29 am
content: Thank you Jay_Diddy_B, you found my stupid.Quote from: johnyradio on June 27, 2014, 03:25:29 amDon't do it: Charge a battery this way.Worth a try: Improving the buck converter. For other uses.

author: johnyradio
author_rank: Newbie
timestamp: June 27, 2014, 04:08:36 pm
content: Quote from: Jay_Diddy_B on June 27, 2014, 02:28:33 amso, the diode's duty cycle is the inversion of the regulator's duty cycle. Makes sense.So therefor, with a higher input V, the regulator runs cooler and the diode runs hotter, right? Ah-ha! Thx!Ok, so you calc the power through the diode at 3.5W. I'm looking at the diode c4757p suggested, the MBR2045CT. It says:Peak Repetitive Reverse Voltage 45VPeak Repetitive Forward Current 20A45V * 20A = 900 Watts!http://www.onsemi.com/pub_link/Collateral/MBR2045CT-D.PDFThat can't be right. What am i doing wrong?Quote from: c4757p on June 27, 2014, 01:57:57 amOk, soconstant-currentmeans the current stays steady with changing load, correct? Or no, with changing supply V. Er, both? So there must be an input minimum current to get the thing going. I don't see Likewise, what if there's no load?Current-limitedmeans the output current can vary a lot with changing load (or supply?), but will never go higher than the limit, right?just got an email from XLSEMI saying the 4015 is the improved successor to the 4005-http://www.xlsemi.com/datasheet/XL4005%20datasheet.pdfhm, the 4015 runs at lower freq, with a lower short-circuit freq. Trying to understand that. This article says lower freq is more efficient, but min Vo is higher.http://www.eetimes.com/document.asp?doc_id=1272335this one is based on the 4015, and is listed as a "lithium charger". c4757p, do you advise against using this one as a li-ion charger? Is it going to suffer the same diode problem? It appears to have the same or similar diode!http://r.ebay.com/pHg5Ksthx!!

author: wraper
author_rank: Unknown
timestamp: June 27, 2014, 08:02:26 pm
content: Quote from: johnyradio on June 27, 2014, 04:08:36 pmThis is just yet another DC/DC conwerter, chinese guy just pot lithium charger in header while it obviously is not.Cheapest what you can get to properly charge lithium batteries while being relatively safe:http://www.hobbyking.com/hobbyking/store/__7028__Turnigy_Accucel_6_50W_6A_Balancer_Charger_w_Accessories.htmlAlso you should not charge lithium batteries unattended, especially if they do not have pressure fuse which breaks terminal connection on pressure buildup. There are not so small amount of RC hobbyists who burned down their houses while charging lithium batteries unattended. Like this:

author: johnyradio
author_rank: Newbie
timestamp: June 27, 2014, 08:17:06 pm
content: Quote from: wraper on June 27, 2014, 08:02:26 pm

author: c4757p
author_rank: Super Contributor
timestamp: June 27, 2014, 08:23:22 pm
content: Quote from: johnyradio on June 27, 2014, 04:08:36 pmUse theforwardvalue for voltage. It's at the top of page 3, about 0.5V. The "Peak Repetitive Reverse Voltage" (often called PIV, Peak Inverse Voltage) is the highest voltage that the diode willblock, when applied backwards.QuoteConstant-current: If you attempt to draw more thanxamps, the regulator will limit you at exactlyxamps for as long as you draw current. This is done by dropping the voltage, and is how bench / lab power supplies typically work.Current-limited: A more vague term; the regulator will try to prevent you from drawing excess current in some undefined way. This is typically only meant to protect the regulator. It could be cycle-to-cycle (cuts the switching cycle short when it hits a certain limit), an entirely separate, linear control loop (in which case it'll be similar to constant-current), foldback (where the limit is lower at smaller output voltages, so that you could draw your full 5A at full voltage but only a few mA into a short), hiccuping (where the supply shuts down briefly and then tries again), etc. In general, don't rely on these features unless you really know what you're doing.QuoteYes, like wraper said. If you're going to charge the battery, do it right with a proper charger that has some smarts in it. The Turnigy looks good.

author: wraper
author_rank: Unknown
timestamp: June 27, 2014, 10:27:02 pm
content: Quote from: johnyradio on June 27, 2014, 08:17:06 pmMobile phone batteries have internal protection circuit, tight charging control in the phone and over-temperature control. So you can't compare them with free standing battery without any protection and with much higher capacity. BTW I have personally seen burned mobile phone battery, a little bit thou but phone plastics also were melted a bit. Pressure fuse usually disconnects "+" terminal if internal pressure becomes high, thus preventing any charging/discharging, circuit becomes open. Therefore battery is no more unusable. Such protection always have laptop battery elements (genuine ones).