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I have a sensor which can produce voltage upto 10 V but can only produce 50 nA of current. I tried to harvest energy using TI's BQ25570 evaluation board but the current is too low to kickstart the board. Also, the sensor is a high impedance source.

How can I harvest energy from this sensor?

Can I boost the current somehow? Would a Darlington pair work for this purpose?

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  • \$\begingroup\$ Do you really mean "harvest"? Maybe you want to measure it, or something? \$\endgroup\$ – Ale..chenski Mar 10 at 4:07
  • \$\begingroup\$ What is the source? You can probably get more energy more easily from ambient sources. Light, heat, broadcast or cellphone or mobile radio signals. Brownian motion is probably a better source :-). \$\endgroup\$ – Russell McMahon Mar 10 at 11:46
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There is almost no power there- 50nA at 10V is 0.5uW.

Yes, you could probably find a way to harvest that 0.5uW (for example, charge a capacitor and then kick start a converter intermittently and store it in a super cap or battery) it's hardly going to be worthwhile in most cases. By comparison a 2032 Lithium button cell has 235mAh capacity which is probably around 10 years supplying 0.5uW including self-discharge. And you're not going to get 100% efficiency harvesting that 500nW, as you've observed it's hard to get more than 0%.

A Darlington would amplify the signal but it does not create energy out of the ether- it requires another power source.

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  • \$\begingroup\$ Thank you for your answer. I agree that there is almost no power. Right now, I want to try to harvest for several hours and then run a small uW circuit for a few milliseconds. Why can't the the Darlington pair be provided the voltage from the sensor? Eg : Connect sensor + to collector and also via a potential divider to base. \$\endgroup\$ – RckR Mar 10 at 2:16
  • \$\begingroup\$ If you have another power source, the base current to the Darlington can be amplified and result in more collector current. But you would need another power source for the collector. Typically for a sensor we would probably use an op-amp to amplify a signal with good accuracy. Again, the op-amp needs one or two power supplies to work properly. \$\endgroup\$ – Spehro Pefhany Mar 10 at 3:34
  • \$\begingroup\$ Current of 50 nA at 10 V implies the internal source resistance of about 200 MOhms, likely more. Regular ceramic capacitors have leakage/impedance of about 100 MOhms per 1 uF, so a 1 uF cap will be charged to maybe 3 V. OP will need something like exotic "silicon capacitors" (insulation up to 10-50 GOhms) to harvest 50 nA. \$\endgroup\$ – Ale..chenski Mar 10 at 4:04
  • \$\begingroup\$ @Ale..chenski PP film cap should be fine. >100,000 M\$\Omega\$/uF. But leakage can come from lots of other places at that level. \$\endgroup\$ – Spehro Pefhany Mar 10 at 4:37

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