Help with diode dissipation - Arduino UPS - Page 1
Forum section: electronics / projects

author: Peabody
author_rank: Super Contributor
timestamp: April 21, 2020, 11:53:58 pm
content: I'm working on adding a load sharing circuit to this 18650 charger and boost converter shield:https://www.aliexpress.com/item/32870411748.htmlThe shield consists of a lithium ion charger and both 3.3V and 5V regulators. But the USB input does not directly power the load. Instead, the load is powered by the charger and the battery, which has certain drawbacks when using it as a UPS. So I want to add a load sharing circuit that will allow the USB power source to power the load and charge the battery independently at the same time when it's plugged in, then have the circuit automatically switch to battery power when USB is not connected. The additional circuit consists of the mosfet, diode and resistor in the middle of this schematic:I think as a practical matter this device is good for about 1A. So I'm depending on the very low RDSon of the mosfet to prevent any thermal issues so long as the battery is powering the load. But when USB is plugged in, the mosfet will be off, and the entire load will be sourced through the 1N4001 diode.I rigged up a test circuit with parts at hand, and the closest I could get was about 760mA. The diode Vf at that current was up to a full volt, but it worked fine and the Vf stayed steady. But it was quite hot to the touch. The situation is made more prickly by the fact that the diode would be soldered to a tiny SOT23-to-SIP adapter board that's less than a centimeter square, so very little heat would be dissipated through that.So the question is whether I can leave the diode as is, or should I change it to a 3A diode, or even a Schottky. My understanding is that the same heat would be generated in a 3A diode, but the package and leads are considerably larger, so air cooling is more effective. A Schottky of course would have a much lower Vf, and would generate less heat, but I would prefer to avoid any leakage issues if I can, and, well, I don't have one on hand.Anyway, I just wondered if anyone here has experience running 1A through 1N400x series diodes, and if so, how did they handle that 1W.

author: David Hess
author_rank: Super Contributor
timestamp: April 22, 2020, 05:50:16 pm
content: Quote from: Peabody on April 21, 2020, 11:53:58 pmRectifier diodes are common rated by forward current and not power dissipation. For the 1N400x series, this is 1 amp average forward current at an ambient temperature of 75C so you will be fine.Still, it would not be a bad idea to use a higher current diode. While not idea, 2 diodes in parallel would also improve the situation until you get get a better part.A higher current diode will have a lower forward voltage drop at the same current which will lower the power dissipation. You might also look for diodes with higher conductivity for lower forward voltage drop.

author: Vovk_Z
author_rank: Super Contributor
timestamp: April 22, 2020, 09:48:55 pm
content: Why use usual silicon diode instead of Shottky? High leakage current is only a problem if Shottky become hot (and/or voltage is high enough). But it seems to me that is not a case.

author: Peabody
author_rank: Super Contributor
timestamp: April 23, 2020, 02:03:06 am
content: Quote from: Vovk_Z on April 22, 2020, 09:48:55 pmIn my circuit (see above), if there is too much leakage, it can raise the voltage on the mosfet gate and reduce or shut off current flowing through the mosfet. So I have to make sure that doesn't happen. I just don't have any experience with Schottky diodes in this kind of situation. I don't know how much leakage current I should plan for. If USB power has been supplying current to the load through the diode, it may be hot. Then if that is turned off, the battery may not kick in properly if there's too much leakage current. So that's why I would prefer not to deal with leakage current if it's not necessary.

author: jackthomson43
author_rank: Contributor
timestamp: April 23, 2020, 02:26:17 am
content: Instead of using 1N4007, you should trySchottky Diode. That's a better option.

author: Mr.B
author_rank: Unknown
timestamp: April 23, 2020, 02:58:01 am
content: I have built very similar circuits in the past for supplying MCUs.I used a Schottky Diode - MBR230LSFT1GI used a 10K pull down resistor on the gate to kill off the effects of any leakage.The FET I used was a SI2305DS-T1-E3. These are obsolete now, but it is not hard to find very low RDSon logic level FETs.

author: Peabody
author_rank: Super Contributor
timestamp: April 23, 2020, 02:46:11 pm
content: Quote from: Mr.B on April 23, 2020, 02:58:01 amNot being formally trained in this stuff, I don't know how to "do the math" on the leakage current.Your 10K pulldown resistor will sink 420µA at the maximum 4.2V battery voltage. If there is leakage, how do you calculate the resulting gate voltage. Is the diode an equivalent resistor, and do you just calculate it as a divider? Let's say the leakage current is also 420µA. Then what's the resulting gate voltage? Is it 2.1V? It seems to me that the gate voltage needs to remain very close to ground to keep the mosfet fully on when battery voltage drops to, say, 3.5V. So how do I know the pulldown shouldn't be 1K, or even 100R?Then on top of that, from the datasheet the leakage current increases exponentially (literally) as the diode heats up. But I don't know how to figure out how hot it will get at 1A or 2A.How did you figure all this out for your circuits? Or was it just doing a test circuit and seeing what happens?The mosfet I've picked is the IRLML6401. Since I'm modifying an existing board, I'll have to mount that on a SIP adapter board, and the diode will probably have to be a through-hole package. I assume I can find one of those with numbers very close to yours, but I just need to be sure that using a Schottky isn't going to get me in trouble.

author: David Hess
author_rank: Super Contributor
timestamp: April 23, 2020, 06:35:47 pm
content: Quote from: Peabody on April 23, 2020, 02:46:11 pmThe leakage current creates a voltage across the resistor as per ohms law.QuoteThe datasheet for the diode will give maximum leakage specifications at both room temperature and high temperature and you can work backwards from that for lower temperature operation. Your worry is well founded since at high temperature, it is 50 microamps for the 1N400x and 25 milliamps for the mentioned schottky diode.

author: Mr.B
author_rank: Unknown
timestamp: April 23, 2020, 08:42:25 pm
content: Quote from: Peabody on April 23, 2020, 02:46:11 pmThe resistor will only sink about 500µA when the USB supply is plugged in.The resistor will sink the combination of the diode leakage current and the gate leakage current when running on battery.

author: Peabody
author_rank: Super Contributor
timestamp: April 23, 2020, 09:38:56 pm
content: It's all still a mystery to me. If the leakage current is 420µA when on battery, then it seems to me the gate voltage would be 2.1V (ignoring for the moment any gate leakage current). And I'm not sure that turns on the mosfet enough. But then I would have to know how hot the diode will get before I know what the leakage current will be.Anyway, it looks like I was wrong about finding an equivalent DO-41 axial diode. It seems all the good stuff is SMD these days.I did find the PMEG10020AELP, which has extremely low leakage values. But the tradeoff is that Vf is much higher than the MBR230LSFT1G, almost as high as regular diode. So I guess there are no free lunches in Schottky diodes either.

author: Mr.B
author_rank: Unknown
timestamp: April 23, 2020, 11:14:15 pm
content: Quote from: Peabody on April 23, 2020, 09:38:56 pmHow can it be that high?Current flows through the 10K resistor only when the USB 5v is available.When the 5v is removed the gate of the FET is pulled low.At that point, the only current flowing through the diode is leakage current, therefore no heating effect on the diode.

author: Mr.B
author_rank: Unknown
timestamp: April 23, 2020, 11:25:53 pm
content: Here is the exact circuit I have used countless times:

author: Peabody
author_rank: Super Contributor
timestamp: April 24, 2020, 01:52:14 am
content: Quote from: Mr.B on April 23, 2020, 11:14:15 pmBut the instant before USB was disconnected, all the load current was flowing through the diode, so the diode may be very hot. The mosfet has to turn on so the battery can power the load. But if the diode is still hot enough that leakage current is high, gate voltage may stay too high.

author: Peabody
author_rank: Super Contributor
timestamp: April 24, 2020, 01:54:10 am
content: Quote from: Mr.B on April 23, 2020, 11:25:53 pmWhat would be the maximum current you would put through that circuit?

author: Mr.B
author_rank: Unknown
timestamp: April 24, 2020, 02:31:43 am
content: In this particular application, only 300mA. Generally powering ESP8266 that needs almost 300mA when transmitting.The MBR230LSFT1G diode is rated 2A continuous.The SI2305DS-T1-E3 FET is rated 3.5A continuous.The MCP1802T-3302I/OT regulator is rated 300mA continuous.However, it is your choice of components that dictates the maximum current.In your OP you mention 1A in the last line of the post.Selecting a 3.3v LDO regulator that can do 1A is trivial, but maybe not in a SOT-23-5 package as in my circuit.

author: Peabody
author_rank: Super Contributor
timestamp: April 25, 2020, 11:13:33 pm
content: I tested the IRLML6401 mosfet I had chosen, and at 1A I'm getting a 0.1V drop across the mosfet. That indicates Rds of about 100mΩ. That's double what the datasheet says it should be, but I was using a breadboard, so that may have been the source of the additional resistance. Anyway, if my math is right, that's 100mW dissipated in the SOT-23 mosfet, which I think it should handle ok.