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I'm wondering whether galvanometer can be used or not ?

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closed as unclear what you're asking by Nick Alexeev Mar 10 at 18:08

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    \$\begingroup\$ What do you mean by "low"? Is this dc, and if not, at what frequency? \$\endgroup\$ – Elliot Alderson Mar 9 at 16:31
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    \$\begingroup\$ One man's low is another man's high. For pA, you'd need a preamplifier before any mechanical instrument. For uA, you'd be OK with a galvo. For MA, a piece of iron hanging from a string in the general vicinity of the conductor would show a response. \$\endgroup\$ – Neil_UK Mar 9 at 16:38
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    \$\begingroup\$ @Neil_UK Under no circumstances short of the literally astronomical would anyone call MA-level currents "low", though. \$\endgroup\$ – Hearth Mar 9 at 17:35
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    \$\begingroup\$ @Hearth True, but it makes for an interesting image, doesn't it? Maybe it will help people pay attention to the difference between MA and mA. \$\endgroup\$ – Elliot Alderson Mar 9 at 18:21
  • \$\begingroup\$ Like everyone else is saying, what exactly is meant by "very low?" \$\endgroup\$ – jonk Mar 9 at 20:03
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For low current, amplifying the voltage drop across a very precise and very small sense resistor is the usual approach. A commercially available and open source implementation of that technique is the uCurrent produced by EEVBlog. It has a range switch for nA, uA, and mA.

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  • \$\begingroup\$ I've been involved in measuring very low currents and, in the process, moved away from using trans-impedance (which uses a very high impedance feedback, not a very precise and very small sense resistor) to integration means, with capacitors. Best of all has been the use of ICs designed for this purpose (Burr Brown ACF2101, DDC112, etc.) In the case of small photocurrents, I was able to get down to noise levels where I could actually see boson flocking effects showing up just slightly above the Johnson and shot noise floors (low hundreds of aA -- below fA if aA is unfamiliar.) \$\endgroup\$ – jonk Mar 9 at 20:03
  • \$\begingroup\$ @jonk you are right, transimpedance is based on a virtual ground usually, not a sense resistance amplifier. \$\endgroup\$ – vicatcu Mar 10 at 16:47

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