DC-DC woes - Page 1
Forum section: electronics / beginners

author: UHF
author_rank: Contributor
timestamp: August 16, 2023, 03:22:07 am
content: Hello, I've got a circuit powering USB via a 5V to 5V DC-DC convertor (from another USB) for isolation and an external 7-12V DC that goes through a NCP1117-5.0 regulator. When the external power is plugged in, a selector switch IC chooses the external power in preference to the DC-DC. The part of the circuit dealing with this is below:There are problems with the regulation of 5V. The external power NCP1117-5.0 regulator gets hot (shuts down in fact) when supplied with 12V, and falls to below 4V when 230mA is drawn. This was from Digikey and it's made by OnSemi - I'm trying to avoid low-cost parts for these.https://www.onsemi.com/pdf/datasheet/ncp1117lp-d.pdfAgain, reading the datasheet, am I mistaken to expect it to give 5V at 1A with 18V supplied? Does it need much better heatsinking on the PCB?Many thanks for any help. I'll add any thing for clarification.

author: golden_labels
author_rank: Super Contributor
timestamp: August 16, 2023, 04:09:22 am
content: Note: this response was written to the original version of the post, which included a question about 0505S.Based on the datasheet, 0505S has a typical efficiency of 76% (page 1). With 230 mA supplied it must consume at least 303 mA. Without negotiation USB is guaranteed to delivery only 100 mA (or 150 mA for USB 3). It may deliver more and many hosts will not limit power. But if it’s the case your host can’t deliver more, 0505S is simply not getting enough juice.Powered from 12 V, NCP1117-5.0 has to drop 7 V. At 230 mA this is over 1.6W released at the junction. The junction-to-ambient thermal resistance is given as 108 °C/W. This gives us 173 °C above ambient. Since the maximum junction temperature is 125 °C and you are probably not running it in -50 °C environment, it’s too hot.Note 1 (page 2) gives an equation, which tells what is the maximum power for the given ambient temperature. For 25 °C it is 0.925 W. With PCB being the only means of heat dissipation, getting continuous 1 A will be impossible from any source voltage.Thermal issues aside: in both cases you reported that the current is limited to the same 230 mA and voltage drops in a similar way. I would double-check, if there is no additional trouble in the circuit. Because whatever limitations each part faces, it sounds unexpected to see both of them exhibit the same thing.

author: UHF
author_rank: Contributor
timestamp: August 16, 2023, 04:16:57 am
content: Ok, thanks very much. I think a redesign of the copper under the NCP1117 is needed and perhaps lower the maximum external input voltage. By the way, the first problem in my post, I deleted - it was the USB cable dropping the voltage.

author: srb1954
author_rank: Super Contributor
timestamp: August 16, 2023, 05:42:19 am
content: Quote from: UHF on August 16, 2023, 04:16:57 amYou are going to need quite a bit of copper (>500mm2 of 1oz Cu) to dissipate that level of power in that chip.It might be better to consider a TO-220 packaged through-hole part in preference to the SMT part. The THT package can easily dissipate that amount of power with fairly minimal heat-sinking.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: August 16, 2023, 08:40:07 am
content: Also, it will actually help you if you just use a regular diode (e.g. 1N4007) for your polarity protection rather than the MOSFET arrangement, because the voltage drop (~0.7-1V) of the diode helps reduce the input voltage to the linear regulator. That is, you spread the power dissipation between the diode and regulator.BTW, what's with the 5.6V gate zener? If you're inputting 7-12V, that'll be conducting all the time. Surely something like (MOSFET max. Vgs permitting) a 12V zener would be more suitable?Edit:I see the DMG2305UX has only ±8V Vgs. Perhaps something like a DMG2301LK with ±12V Vgs could be used instead.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: August 16, 2023, 09:13:25 am
content: Quote from: UHF on August 16, 2023, 03:22:07 amIf the current is 1A and the voltage regulator has to regulate from 18V to 5V, then that is: P = U*I = (18-5) * 1 = 13Watt to dissipate. And you need a quite sizable chunk of aluminimum to get rid of that much heat. It's much more common and efficient to use a small SMPS module if voltage drops get that big. (If you can tolerate the noise).

author: UHF
author_rank: Contributor
timestamp: August 16, 2023, 11:07:01 am
content: Quote from: HwAoRrDk on August 16, 2023, 08:40:07 amYes that's why, to protect the MOSFET.The external voltage doesn't need to be more than 7V at the most, but I want to protect from over voltage. I can't find anything simple just for that. Any ideas? Also I came up with this copper island connected with vias to the other side as an attempt at heatsinking - yes the contact with the device will be poor. KiCAD says it's a total of 628mm2.The device itself seems like the wrong type of package with no metal underneath just the legs and tab, for anything drawing more than very modest current. I'm coming around to using another regulator, but really I just need 5V up to 1A from an external PSU.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: August 16, 2023, 12:25:38 pm
content: Quote from: UHF on August 16, 2023, 11:07:01 amI wasn't questioning the existence of the zener diode, just it's voltage rating.I think if you're going to stick with MOSFET-based polarity protection, then you should change to a MOSFET with a higher Vgs and use a higher-value zener diode. You want a zener rated such that it protects the gate of the MOSFET adequately, but does not conduct in reverse bias at normal operating input voltages.Quote from: UHF on August 16, 2023, 11:07:01 amI think you should probably look at using a buck regulator instead for 1A. You can get synchronous regulators that don't need an external diode and switch at high 100's kHz or MHz that only need a small inductor, so you can probably fit it all in the same space on your PCB. Something like anAP62300.Even with that copper area, and an input voltage of only 7V, you'd still be massively overloading the NCP1117 with 1A. There is a chart, figure 32, in the NCP1117 datasheet of RθJAfor a given copper heatsinking area, and extrapolating out 600mm2still only gives about 80 °C/W. If the regulator is dissipating 2W (7V input, 1A output), that's still a 160 °C temperature rise!