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I am trying to track down a ~10uA current leak in my circuit.

My PCB has this 10uF ceramic decoupling capacitor between 1.8V and GND.

I cannot see it's ESR value in the datasheet so I do not know what leakage current I should expect. I also cannot see a minimum voltage listed to check if it is suitable for 1.8V.

I am using a CurrentRanger, a multimeter and an oscilloscope to measure current. When I physically removed the capacitor, the current dropped 10uA.

The datasheet for the LIS3DH sensor that it is attached to, says it should be an aluminium capacitor.

What leakage current should I expect from this capacitor?

Any help is greatly appreciated.

enter image description here

SamGibson
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JasonEdinburgh
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  • If you used water soluble flux and didn't clean it thoroughly that can lead to leakage, so make sure you check this too. – bobflux Oct 23 '20 at 11:00
  • There will be no minimum voltage for an MLCC. || Al cap leakage will be higher to much higher than a ceramic. – Russell McMahon Oct 23 '20 at 11:40
  • Consider trying using a doppler RADAR module for hearbeat sensing. Current drain is liable to be too high for your application. Prices can be stupidly low. – Russell McMahon Oct 23 '20 at 11:42
  • You can't normally calculate leakage current from ESR. –  Oct 23 '20 at 11:48

1 Answers1

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I misread the question. I had in mind that you wanted to replace an aluminum capacitor with an MLCC.

What you are considering is replacing an MLCC with an aluminum capacitor.

Don't.

The leakage of the aluminum capacitor is almost guaranteed to be higher.

ESR isn't what defines the leakage current.

ESR (equivalent series resistance) is like putting a resistor in series with an ideal capacitor, while leakage is like putting a resistor in parallel with an ideal capacitor.

The datasheet you linked gives an insulation resistance of 50 ohms F. Take 50, divide by 10 microfarads (as farads) and that's the resistance of the insulation. That works out to 500000 ohms for your 10uF part.

At 1.8V, that's about 3.5 microamperes.

You need to find an MLCC with a higher insulation resistance.

The one I picked before for comparison says 100000 Mohm or 1000 ohms F, which ever is larger." For a 10uF part, that'll be 100000 Mohm because the ohms F is smaller. The leakage current is then far below 1uA.

You can reduce the leakage current by using a better MLCC. Aluminum capacitors don't get much better.

Incorrect answer. Leaving it here for the comparison of MLCC to aluminum capacitors.

Picking the datasheet for an aluminum capacitor at random, I find a maximum of 3 microamperes. That's in the same ballpark as your 10 microamperes so it looks you have a reasonable measured value.

Picking a datasheet for a 10 microfarad MLCC capacitor at random, I don't find leakage directly given. It does give a value for "insulating resistance," which I take to be related to leakage current. At 100000 megaohm, the leakage will be lower than the aluminum part at any rated voltage.

The MLCC won't necessarily give you the same capacitance, though. You may need to use a 20 uF or more rated part to guarantee the same capacitance at your operating voltage. Ok, you are working at 1.8V, so the loss may be lower than usual. You should still consider it, if only to look at the datasheet voltage derating curve and think "nah, doesn't matter."

JRE
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  • Thank you this is superb! I have several (different) capacitors on the board so I guess this might add up to the leakage current I am seeing. I allowed PCBWay to choose standard caps and I didn't inspect the BOM enough to realise there would be these problems. (My first board) – JasonEdinburgh Oct 23 '20 at 10:08
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    If 10 microamperes of leakage is a problem, then you've picked yourself an interestingly difficult task for your first PCB. Good luck. – JRE Oct 23 '20 at 10:12
  • Yes I'm really excellent at vastly underestimating how difficult things will be. – JasonEdinburgh Oct 23 '20 at 10:18
  • Paying me to check your stuff is a bad plan. I'm a hobbyist. I do this stuff for fun in my spare time. Many of the questions I answer here teach me something. I didn't know about ohms F until I ran into it in the datsheets while answering your question. – JRE Oct 23 '20 at 10:23
  • The heartbeat detector is pretty much an abandoned thing. It reaches many of the goals I set for the project, but failed the major one - simple enough for anyone to strap on and use. It would work OK as a heart rate monitor for a person lying in bed, but it won't reliably work for a jogger (which was the original use case.) – JRE Oct 23 '20 at 10:26
  • Strangely the use-case would be someone lying in bed! but the sensor might not be close enough to their heart to get a good signal to noise ratio. Very interesting though. I will do a test with a stethoscope. And thanks for taking the time to help me. I almost got the answer from the Murata FAQs but I didn't quite get the final figure. I will spend considerably more time checking the datasheets of the caps etc before I order the next pcb. – JasonEdinburgh Oct 23 '20 at 10:35
  • The heartbeat counter requires a microphone in physical contact with the chest. It won't work through the air. – JRE Oct 23 '20 at 10:39
  • Also not a problem, the device will be held against the skin. I was thinkin of using a tiny MEMS microphone. – JasonEdinburgh Oct 23 '20 at 11:33
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/115430/discussion-between-jasonedinburgh-and-jre). – JasonEdinburgh Oct 23 '20 at 11:46
  • I think there was a mistake in the calculation. I think the Insulation resistance was 10 times higher. 50 ΩF/10 μF = 5 MΩ. See the second method here https://www.murata.com/en-global/support/faqs/products/capacitor/mlcc/char/0039 – JasonEdinburgh Nov 01 '20 at 13:02