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i am looking for sensor supply circuit for a sensor. The sensor requires bipolar +-6V to -15V DC and 10 mA power consumption at 6V DC (6V*10mA=0.06mW). Also power supply of this sensor requires less than 100Vrms/root(Hz) noise level. So i can use a low noise adjustable regulator for it (for example lt1763 adj) but i want to know what is the best way for it? Is there any better or sensible idea? Also i want to add a current limiting curcuit not to exceed to 10 mA. I thought a 10 mA polyfuse can be useful for protection. But you can advice better way to me. Finally How a sensor supply circuit should be? Can anyone tell me the method of this design? King regards, Cem

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  • \$\begingroup\$ Please post the sensor's data sheet or a link to it \$\endgroup\$ – EM Fields Dec 30 '15 at 11:53
  • \$\begingroup\$ Hi sir i am adding its link and datasheet right below. Thank you for quick reply. I am wating your response. \$\endgroup\$ – Cem Dec 30 '15 at 12:49
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The sensor requires a power supply with an output voltage anywhere between \$\pm\$ 6 volts and \$\pm\$ 15 volts, and the sensor will draw 10 milliamperes from the supply if it's a \$\pm\$ 6 volt supply. The data sheet doesn't specify the current needed by the sensor at other input voltage levels.

The type of power supply will probably depend on what the supply input needs to be, (AC mains, DC supply, battery, etc.) and if you can define that (edit your question instead of leaving a comment) we'll be able to get you closer to your goal.

You shouldn't have to current limit the supply outputs, and if you need to do something on the supply input that'll depend on what kind of supply it is.

Here's a link to a nice little dual supply, and here's the first page:

enter image description here

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  • \$\begingroup\$ The sensor datasheet says on pg3 that the noise spec is a PSU requirement, not an output spec. I doubt that this DC-DC switcher will meet it without filtering. \$\endgroup\$ – brhans Jan 2 '16 at 12:48
  • \$\begingroup\$ @brhans: Right you are. Good catch and answer edited. Thanks. :) \$\endgroup\$ – EM Fields Jan 5 '16 at 18:21
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Datasheet or Part number of the sensor would be helpful to give a better answer. But broadly speaking:

Usually sensors also have a quiescent current rating, stated in their datasheet. This is the current that is consumed for most of the time by the sensor, i.e "ready to sense, but nothing to sense state". (10ma is definitely not your quiescent current rating). This is a major design consideration for sensors. Your design requirements will also be affected by standby time and active time. i.e for how long does it stay in the active "sensing" state.

For your rated voltage needs, LDO's are usually a better option to power your sensors. And for rated current, there are also current limiting diodes you can try. read the below article from TI, It answers to all your design requirements. Also It would be a better solution for your future supply design needs.

https://e2e.ti.com/blogs_/b/powerhouse/archive/2015/02/21/how-to-get-more-out-of-your-sensor-with-the-right-ldo

Edit: Just saw the link. You need to design the power supply considering this as an op amp IC. In your question you state "6v * 10ma" but the fact is you require a dual supply voltage, i.e at minimum +6 at +vss and -6 at -vss. And the current consumption they refer to in the datasheet is not the max current, its just the current your IC would consume if it were given a rail of +6 0 -6.

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  • \$\begingroup\$ Thank you for your quickly reply. After you said i recognized 10mA was Quiescent current of this sensor :) I am adding link datasheet of sensor link \$\endgroup\$ – Cem Dec 30 '15 at 12:48

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