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I have small issue with my circuit that I can't wrap my head around.

I have made a custom PCB powered both by USB and battery. I put a diode between them to block battery power from flowing into USB when both are connected.

This diode doesn't work as intended because when I plug in the battery without USB power, the USB_LED that I put with a 10k resistor in series still lights up (albeit dimmer.)

There's 1.8 V across the diode in reverse and 2.2V across the LED. (The sum equals 3.9V of my battery voltage.Can you confirm this?)

I was expecting to see 3.9V across the diode in reverse since I expected the diode to work as open circuit and certainly not for the LED to light up.

Can you tell me like I am 10 years old how this is possible, with Ohm's law and all that?

I don't get it. I specifically chose a B160 diode datasheet because it was supposed to have under 200na reverse current for my battery voltage according to the datasheet.

reverse leakage graph

I also tried a 1N5817 diode and got similar results. I noticed that when I solder the diode the LED light gets stronger, which is in line with the reverse current temperature behavior that's in the graph.

-How can I calculate the leakage current with Ohm's low with diode and LED voltage values?

  • How does the LED light up with that small of a current?
  • What diode should I use to have nano amp level reverse current ?

Here's the simplified power supply schematic with the 3.3V converter on the right:

enter image description here

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  • \$\begingroup\$ I edited my post with the schematics. The Diode is a B160 from Diodes inc bought on LCSC : lcsc.com/product-detail/… I don't think it's fake because my other IN5817 diode does the same thing. The led is a standard green led : datasheet.lcsc.com/lcsc/… \$\endgroup\$ Commented Jan 21 at 10:19
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    \$\begingroup\$ Instead of looking for esoteric diodes, hard to find, characterise, in short a life long source of trouble, you'd better design a robust circuit that does not exhibit this behaviour. A simple resistor straight on parallel with the LED is what I suggest, it won't be difficult to dimensions accommodating a maximum reasonable leakage without lighting up the LED \$\endgroup\$
    – carloc
    Commented Jan 21 at 12:11

5 Answers 5

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Schottky diodes have a lot of reverse leakage, and modern LEDs can emit visible light without much current. 1N5817 has more reverse leakage than 1N5819 (but also less forward voltage when conducting). If you replaced it with a 1N4004 it would have less leakage but higher forward drop, as per the Shockley equation.

A simulation shows typical current of 30uA+ at room temperature, which is plenty to get visible light.

The Vishay datasheet 1N5817 guarantee is only that it is less than 1mA at 25°C and 10mA at 100°C! The Diodes Inc. B120 datasheet you linked guarantees a but slightly lower 0.5mA at 25°C. So it's certainly well within spec.

In conclusion, I don't think there's anything wrong with your circuit, it's behaving as expected.

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  • \$\begingroup\$ I would like to know how do you simulate such things ? Would be really helpful to better understand things \$\endgroup\$ Commented Jan 21 at 10:35
  • \$\begingroup\$ Is it possible to qualify LEDs can emit visible light without much current? As noted in the question, when the B160 diode is operating at ambient temperature +25°C with 1.8 V reverse voltage, only expect of the order of 200 nA of reverse current. I.e. what is the nominal minimum current for a LED to visible to the human eye? \$\endgroup\$ Commented Jan 21 at 10:36
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    \$\begingroup\$ You can do it directly on this site using Circuitlab. I used LTspice, which is free. You may get different results depending on what model was used. \$\endgroup\$ Commented Jan 21 at 10:37
  • \$\begingroup\$ @ChesterGillon I would not expect 1uA or less to be visible in normal lighting. In a darkened room, a fraction of that may be visible. There's a certain amount of reflection inside the LED package anyway that will mask tiny bits of light unless it's pulsed. But the easiest way to quantify it is to deliberately pass that much current through that particular LED and observe it. Similarly the current can be measured. Of course it's also possible that OP has an issue with leakage on the PCB for some reason, but it would have to be pretty bad. \$\endgroup\$ Commented Jan 21 at 10:52
  • \$\begingroup\$ @ReickenBack I know tinkercad.com does circuit simulation too now, for free. No idea how accurate it is though. \$\endgroup\$
    – Criggie
    Commented Jan 22 at 20:09
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So after all there's nothing wrong with my circuit and the led draws very little current. Since it might be useful to other beginners, I got an idea, instead of a simulation, i measured it with my Power profiler kit and it's quite interesting. Here's the little circuit i made with the same components. First, with the diode forward biased. With the 10k resistor we get 150 ua current draw as expected

enter image description hereenter image description here

Now if we reverse the diode, the led still emits light (although barely visible, dimmer than on my pcb) The current is around half a microamp which is very low, but it's still enough to light it up. When i put my finger on the diode, it gets warmer and the reverse leakage increases up to 2ua which makes it clearly visible. Same thing happens with the more common IN5817 Shottky Diode with approx the same values.

enter image description here enter image description here

Thank you guys for your input, I hope that my little experiment will be useful for other beginners. Now i just have to decide if i'm ok with a few microamps current and a led that shouldn't be on. I think i'll try a regular diode instead of a shottky, should be better but i don't have any at the moment. I might update my post in a few days with measurements with a new diode.

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    \$\begingroup\$ If you want to prevent the LED from producing light when there's a tiny current through it, you could try putting large resistor in parallel with it. If you're getting, say, 5 μA of current, then even a 200 kΩ resistor would be sufficiently conductive to limit the voltage across the LED to 1 V. \$\endgroup\$ Commented Jan 22 at 8:36
  • \$\begingroup\$ What software/hardware are those plots from? \$\endgroup\$ Commented Jan 22 at 21:16
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    \$\begingroup\$ Power profiler kit II from nordic semi. The software is nRF Connect \$\endgroup\$ Commented Jan 23 at 0:12
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How does the LED light up with that small of a current?

There are two things I'd like to point out. First this table in the datasheet: -

enter image description here

So, for the B160 diode, the worst case leakage current at 25°C with a reverse voltage of 60 volts is 0.5 mA. Compare that with the graph ( typical values) with my adds in green and pink: -

enter image description here

The pink line is my guesstimate of what the worst case reverse leakage might be at 25°C. If you followed it down to near zero reverse volts, you might expect to see a worst case leakage of around 50 μA and this would easily illuminate a modern LED. Compare this with the typical reverse leakage of maybe 200 nA and you can see that this device has a massive spread.

What diode should I use to have no level reverse current ?

There is no such diode; you have to compromise. I suggest you find a better device with a better spread of likely reverse leakage currents.

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  • \$\begingroup\$ Thanks now I have a better grasp of typical vs maximum values. I did not pay enough attention to the "typical" word meaning. \$\endgroup\$ Commented Jan 21 at 12:09
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Use parallel resistor to fix the problem

As other answers have detailed, the small leakage current is enough to light up the LED dimly.

Fortunately it is easy to fix by adding a resistor in parallel with the LED:

schematic

simulate this circuit – Schematic created using CircuitLab

Maximum resistance for R2 can be calculated by VF / IR where VF is the LED forward voltage and IR is the leakage current. The 100 kohm resistor shown is enough to sink up to 20 µA leakage current before the voltage will exceed the LED forward voltage.

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    \$\begingroup\$ As it's a low power board, every microamp counts. Since I don't really care about the forward drop of the diode when connected to 5v USB i will be using a BAS716 diode which guarantees a maximum of 5na reverse current. \$\endgroup\$ Commented Jan 22 at 12:17
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What diode should I use to have no level reverse current

All diodes have reverse current, you just want it low enough that the LED won't be seen as lit :)

I'd use a high-conductance silicon diode, e.g. MMBD4448. A SiC (silicon carbide) diode, like GB01SLT06, will also work - those have very low leakage at low reverse voltages (rating is 600V!), but cost about 20x as much as a high-conductance silicon diode.

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    \$\begingroup\$ There was a typo in my post, i meant na level, meaning nano amp. I will use BAS716 which is rated to 5 na max reverse current \$\endgroup\$ Commented Jan 22 at 12:14

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