I don't understand diode ratings

lbenson

Senior Member
and the datasheet doesn't help me.

Looking at, for instance, the 1N5817, which is rated 20V, 1A: https://www.mouser.com/datasheet/2/389/1n5817-954386.pdf

I want to be able to power a dollar store LED light, 3-AAA powered, from 5V from mains or battery sources (18650 holder module), connected through schottky diodes. Per my LCD current & voltage module, it draws 1.5A at 5V. Because the voltage is so much less than 20V, can I use a 1N5817 (or 18 or 19)? Must I use 2A diodes, like SR260?

How do you de-rate (or up-rate) a diode based on actual voltage?
 

premelec

Senior Member
The voltage is not so much relevant as the current which heats the diode junction - so use a higher current diode or in a pinch a few 1 amp units in parallel may work out... [though the current through them will not be balanced... though distributed]
 

AllyCat

Senior Member
Hi,

The reverse voltage rating is largely unimportant, but must of course be high enough for the application, and the highest rating of a particular "family" may carry a slight penalty in terms of cost or performance. Schottky diodes have a lower forward voltage drop, but personally I've been rather "disappointed" with their forward drop in practice (and they generally have a higher reverse leakage). The 1N5817 for example appears to have a not much lower forward drop than a "normal" diode at 3A (but then it is rated at 1A).

So, as indicated above, what is the maximum forward drop your application can tolerate, and will the current overheat the diode? At 1.5 A it looks as if the 1N5817 may drop around 0.45 v and dissipate 700 mW, raising the junction temperature by 70 degrees C (probably acceptable). Note that silicon diodes have a forward tempco of around -2 mV/C which will reduce the forward drop, but if you put two diodes in parallel, it means that one may take much more than its share of half of the current.

Don't forget, if rectifying a sine wave (e.g. mains/line voltage), then the diode reverse rating must be peak-peak (i.e. around 3 times the nominal rms) and the current conduction angle is quite small (just at the peak of the sine wave) so the peak current is much higher than the average. Don't be fooled by a "headline" non-repetitive current rating (which is what it says).

Finally, I believe that really low voltage drop "diode" applications (e.g. in low-voltage SMPS) now use "synchronous" rectification, i.e. a transistor or FET which is actively switched on when required. That can achieve a "forward" drop equal to the Vce or Vds saturation voltage, perhaps 100 mV or less.

Cheers, Alan.
 

premelec

Senior Member
FWIW Linear Technology [part of Analog Devices] has a line of "ideal diode" ICs that can drive an external MOSFET to give way less power losses than Schotky diodes at high currents. They are a bit expensive and more complicated than a 2 lead diode but well worth using in many circumstances - especially higher currents. .
 

lbenson

Senior Member
Thanks. All I'm wishing to do is to turn on this 24-LED module for a few minutes at a time upon the detection of motion, so there will be no rapid cycling as with PWM. The diodes are strictly to isolate the 2 5-V sources, mains power supply (5V wall plug) and 18650 boost module for battery back up.

Ambient plus 70C should typically be well less than the 150C maximum.

Intuitively (but not necessarily correctly) it seems to me that watts ought to be the limiting factor, but I don't directly see that in the datasheet.
 

AllyCat

Senior Member
Hi,
Intuitively (but not necessarily correctly) it seems to me that watts ought to be the limiting factor, but I don't directly see that in the datasheet.
The maximum junction temperature is specified as 150 degrees C and a junction rise of 100 degrees/watt. So the limit is "about" 1 watt. But the "ambient" might be >50 degrees (e.g. inside an enclosure) or conversely, the thermal coefficient might be better than 100 degrees/watt. For example, much of the heat may be conducted out through the leads, particularly if they are short and terminate on a large area of PCB copper track (acting as a heat sink).

Cheers, Alan.
 

lbenson

Senior Member
So the limit is "about" 1 watt.
Ouch. 5V @ 1.5A for my LED is 7.5 watts. Am I totally off base in looking at the 1n5817? Or should I be looking at "voltage forward" of perhaps .45V (max), and thus .45V @ 1.5A or .675 watts and all good?
 

inglewoodpete

Senior Member
Ouch. 5V @ 1.5A for my LED is 7.5 watts. Am I totally off base in looking at the 1n5817? Or should I be looking at "voltage forward" of perhaps .45V (max), and thus .45V @ 1.5A or .675 watts and all good?
While 5V @ 1.5 amps is 7.5 Watts, you have said that the forward voltage of the diode is (around) 0.5 Volts. So, 0.5V x 1.5A = 0.75 Watts

As an aside, to assist a diode to dissipate heat, keep the leads as long as possible and raise the diode body off its circuit board. If space is limited on the board, bend the diode's leads in a loop (up around and then down through the circuit board) to maximise its capacity to radiate heat through the leads.
 

Hemi345

Senior Member
As an aside, to assist a diode to dissipate heat, keep the leads as long as possible and raise the diode body off its circuit board. If space is limited on the board, bend the diode's leads in a loop (up around and then down through the circuit board) to maximise its capacity to radiate heat through the leads.
(y) I like to stand them up on end and "u" one of the leads over and down along side of it. Saves PCB space and adds a little heat dissipation.
 

AllyCat

Senior Member
Hi,

An advantage of through-hole components! : If the diode is mounted horizontally, touching on the PCB then the heat dissipation will be compromised (I don't think the data sheet specifies if this is the test condition for 100 C/W, or not). Extending the leads and raising the body a little will help the heat convection and radiation, but I like the vertical mounting idea because the whole circumference will be "exposed" to the air/surroundings.

From the data sheet, the (black) surface area of the body is about 2.4 * PI * 4.6 = 35mm2
and a 10mm length of a (silver) lead is about 0.8 * PI * 10 = 25mm2.

The body material is chosen to be thermally conductive (and black to enhance radiation), but won't be as (thermally) conductive as the copper lead(s) ! Some data sheets actually include heatsinking data for PCB pads; A 10mm square "flood fill" pad will give 100mm2 of radiation/convection area, and perhaps on the "cool" side of the PCB. ;)

Cheers, Alan.

PS: Well, that covers the three primary methods of heat transfer, I'll leave you to research the fourth and others. :)
 

hippy

Technical Support
Staff member
Seems to me a diode-mixing solution for power supplies -
Code:
5V >--------|>|---.-----.
                  |     |
        .---|>|---'    _|_   ~
      __|__           _\ /_  ~
       -.-              |    ~
        |               |
0V >----^---------------'
 

lbenson

Senior Member
Lance, what is the real purpose for using the diode in the first place?
Seems to me a diode-mixing solution for power supplies
Correct, hippy. Ok, Tex, you asked for it.

It may be too xmas-tree-like, belts and suspenders plus an additional belt.

I want to have a motion sensor plus DS18B20 outside the door to my porch, and inside another DS18B20 and 60-LED dollar store light, monitored by a picaxe 14M2 with connection to my network using an ESP8266 D1 mini module (sending UDP messages).

The problem is, what happens if mains goes out (I've looked a many aspects of this problem over the last dozen or more years). The D1 mini draws a fair amount of current for a few seconds as it is turned on with a low-side switch when the picaxe has something to report. There's a nice Wemos 18650 battery shield ( https://www.ebay.com/itm/Micro-USB-Wemos-ESP32-18650-Battery-Shield-V3-ESP-32-LED-for-Arduino-Raspberry/183458100976 ) (dual battery module for me), but as Andreas Spiess says here (
), and as my testing has shown, if the battery is run down, it won't recharge if the draw is too great upon the return of mains power.

I propose 3 power sources--5V from a 2A mains power supply, 5V from the 18650, and 4V5 from 3-AAs. These are diode-mixed and run through a 5V-3V3 DC-DC converter. I monitor each of the 5V sources and variously throttle back on reporting if mains or 18650 is absent. If only 4V5 is present, maybe only transmit daily "I'm alive" messages and motion alarms and no LED.

If mains returns after the 18650 is drawn down low enough to switch off, there should be no draw on the 18650 battery, so the low-current charge should work until the battery develops enough voltage for the high-current charge to start up.

Here's the circuit (for some reason, all my icons at the top of this window are greyed out, so I can't insert links or images as I have in the past--only attachments) (ignore the 2-pin 3V3 battery connector on the left--it's gone from my plan, but not yet from the PCB, so I can't remove it without there being some traces removed which I want to keep):
14Annex_sch2.jpg
And the dollar store light, which draws about 1.5A @ 5V.
dollar store LED.jpg
It's a nice, bright light. A different type burned out for me after a few minutes at 5V. A different one of this type has been always on for me for two months, so it seems fine at 5V.

Suggestions for improvements will be much appreciated.
 
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Hemi345

Senior Member
Do you operate the 5V light while on 18650 power?

A voltage supervisor might be a good solution to control the 3.3V regulator. That will keep the 3.3V load turned off during the initial charge of the 18650 battery until the voltage reaches a high enough level that the supervisor will turn the 3.3V back on. That should eliminate the need for the AA batteries.

TPS3840 Voltage Supervisor - page 29 has all the threshold voltages you can choose.
 
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lbenson

Senior Member
Do you operate the 5V light while on 18650 power?
Yes, but this is what requires the dual 18650 module, since otherwise the battery protection kicks in after a few minutes and you don't get 5V again until you have removed and re-inserted the batteries.

A voltage supervisor might be a good solution to control the 3.3V regulator. That will keep the 3.3V load turned off during the initial charge of the 18650 battery until the voltage reaches a high enough level that the supervisor will turn the 3.3V back on. That should eliminate the need for the AA batteries.
The AA batteries with low-current procedures should allow the picaxe to run for a fair while if the 18650 shuts down. I'll run tests to see how long. The AAs aren't there to enable to the 18650 to recharge once mains comes back on--diode-mixing the mains 5V with the 18650-module 5V should prevent any drain on the 18650 upon reappearance of mains 5V, so it should charge--another thing to test.

You could also diode mix the enable pin on the 3.3V regulator so the PICAXE keeps it on even if the supervisor wants to shut it down (to squeeze out every last bit of till the battery protection kicks in).
Maybe. I'm still trying to avoid fiddling with surface-mount parts. Mulling things over after a nap gave me other things to try.
 

Hemi345

Senior Member
The AA batteries with low-current procedures should allow the picaxe to run for a fair while if the 18650 shuts down. I'll run tests to see how long. The AAs aren't there to enable to the 18650 to recharge once mains comes back on--diode-mixing the mains 5V with the 18650-module 5V should prevent any drain on the 18650 upon reappearance of mains 5V, so it should charge--another thing to test.
Maybe. I'm still trying to avoid fiddling with surface-mount parts. Mulling things over after a nap gave me other things to try.
Ahh, I see. You just want to continue getting updates even after the 18650 power is depleted. I'm using four AA batteries to power a few projects that open and close my window blinds (using servos) and check in twice a day via an ESP8266. The batteries last right at a year, I think you'll be fine during power outages unless the regulator you're using isn't very efficient.

SOT23-3 isn't really hard to solder by hand. A little dab of flux and a small-ish tipped iron makes quick work of it. I'll send you my Eagle library with a bunch of footprints I've made if that's what's holding you back. :)
 

lbenson

Senior Member
I'll send you my Eagle library
Thanks--I have and use your picaxe library. But that's not what holds me back from SMD. I have plenty of parts libraries, and have soldered up a number of boards which have SMD parts, but if I'm designing it, I may as well treat myself and use just through-hole as long as I can find appropriate parts.
 

Hemi345

Senior Member
I may as well treat myself and use just through-hole as long as I can find appropriate parts.
I hear that. Luckily, it seems like pretty much everything that you could want or don't enjoy soldering can be found in a breakout board on Ebay.
 

techElder

Well-known member
5V from the 18650
Lance, how are you getting the 5V from these? Seems like if you are already monitoring their voltage you could be using a better switch to have a sequential 18650 solution. Perhaps even eliminating the 3.3 regulation stage?
 

Hemi345

Senior Member
He'll need the 3V3 reg for when the device is running on 3 AA after the 18650s are depleted. Or use only 2 AA for backup and then he could use the 3V provided by the 18650 module during normal mains/18650 power?
 

lbenson

Senior Member
how are you getting the 5V from these [18650s]?
The Wemos 18650 modules shown in the video have 5V out, through USB or headers (if you solder them in), and 3V3 through headers. The circuit on these modules will drop the voltages upon depletion of the battery or over-current. For the circuitry on the modules to restart the 5V supply after battery depletion upon restoration of mains, the current draw on the 5V must be minimal until the battery is sufficiently recharged. (That's what the video says--I'll be testing that. Every UPS I've tried has had the same fault (from my perspective)--needing manual intervention to restart after battery depletion.)

(It looks like the problem with this 18650 module could be fixed by diode-mixing the 5V output with the 5V charging supply and the 5V boost output onboard. This would eliminate the recharge problem (if I understand it correctly). Perhaps an updated version will fix this.)

My current plan is to run everything off of the 5V (down to 4V5 if mains and the 18650 give out), feeding that to the ESP8266 through its 5V pin and only turning on the ESP8266 with a low-side switch on 0V when needed for transmission.
 
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