How to carry low current accurate measurements with protection?

Other Hardware General Electronics
1

Hello Guys

I would like to employ the μCurrent gold to a system I am working on to make accurate current measurements with an external ADC and Atmega2560. However, I have a problem that the μCurrent gold doesn't have a protection against overcurrent or overvoltage.

  • I need the Ammeter to be able to measure in two ranges: ±199.9uA and ±1.999A current measurement with ±0.5% or better accuracy.
  • Preferably capable of AC and DC measurements similar to the μCurrent gold.
  • Is there a way to add a protection mechanism to the μCurrent gold?
  • Or am I better off using current sense amplifier or an instrumentation amplifier?, if so how do I protect them of several amps overcurrent or around 30V voltage?

μCurrent gold schematics
μCurrent gold
AD8418
AD8428

2

Not everyone here knows what a "μCurrent gold" actually is.

Can you supply a link to this please.

That is a big ask for a system involving Arduino's.

I would say so.

Why are you so concerned about this? Is it likely your application (currently a secret you keep) could produce this?

1 Like
3

Yes. It's called a fuse.

4

Thank you, the links for the μCurrent gold were added..

I am just making a testing diy board to carry accurate measurements for some electronic circuits, and there could be instances where I mistakenly input higher currents than the range available..

5

Think I would use an AD1115 to measure the output of the AD8418.
This is a16 bit ADC so given assuming 3 bits of noise you still have 8191 values in the range 0..5V is better than a millivolt per step.

Update: In fact it is a 15 bit ADC (thanks to wawa for pointing out)

As always the proof is building a prototype to see the actual range possible.

1 Like
6

As mentioned in the article: uCurrentArticle

Overloads Fuses have been omitted from the design to ensure as low a total burden voltage as possible. Therefore you must be careful to ensure that the input is not connected directly across a supply voltage capable of providing a current that exceeds the selected range. Failure to take care here can result in a blown shunt resistor.

Not to mention there's actually no fuse to my knowledge that is available for such low current range..

7

(thinking out loud)
Amplify the current and let that trigger a "master" fuse?

1 Like
8

You're right about that. And it's hard not to influence the measure with any protection. Does the man (Dave) himself have any advice?

1 Like
9

Add (an arduino-controlled?) relay to switch between a beefy shunt resistor and the instrument only if it measures safe.

1 Like
10

Unfortunately I didn't find anyone that attempted that..

11

Perhaps I should use two separate circuits for each of the two ranges, and switch between them. Then maybe use clamping diodes for the 200uA circuit?.

I don't think the AD8418 will be able to cover the range of 0-200uA, because with 1kOhm shunt and 1uA current the voltage will be 1mV, but with 200uA, the voltage will be 4V..

image
image712×602 8.81 KB

12

Perhaps because as the fuse material gets thinner, the inherent resistance goes up!

1 Like
13

Not with the circuit you show.

1 Like
14

image
image701×489 6.67 KB

15

If you have overvoltage or reverse voltage you will first blow-out the diodes, then the op-amp next.

1 Like
16

Do you have access to equipment to calibrate your project's measurements?

2 Likes
17

You could implement an active clamp using components like MOSFETs, SCRs, BJTs, etc. However, you must be extremely cautious about leakage current, especially with anything that has electrical contact with your leads, as it could affect the measurement.

For additional guidance, consider reviewing schematics of commercial units designed to measure your secret device.

1 Like
18

No protection needed if you use an analog Hall sensor with a solenoid coil and perhaps a soft iron core to let the current create a magnetic field.

1 Like
19

Yes I do..

20

What is to be protected with a TOTALLY ISOLATED sensor?

1 Like