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### Topic: Arduino Ammeter (Read 3071 times)previous topic - next topic

#### aman92

##### Jan 11, 2013, 05:59 pm
I have been working on a project which measures inductance. It basically uses a moving coil ammeter to measure the average current and correlates that with inductance. Now I want to replicate the function of the moving coil ammeter on my Arduino Uno i.e. measure the average current but can't seem to find a suitable way of doing this. Any suggestions?

#### jackrae

#1
##### Jan 11, 2013, 06:13 pm
Use a hall sensor, sample current at intervals over a fixed period of time, say a couple of seconds, and add up the respective values. Then divide the totalised current by the number of samples take.  The resultant figure is the "average" current.

#### aman92

#2
##### Jan 11, 2013, 06:57 pm
Yes but does it give the same average current as the ammeter? This current is the current through an inductor and the  average is of the exponentially charge and discharge curve.

#### aman92

#3
##### Jan 11, 2013, 06:57 pm
Here is the circuit which I am trying to implement

#### aman92

#4
##### Jan 12, 2013, 06:56 pm
I really need to complete this project within a deadline, so any kind of suggestions would be greatly appreciated

#### dhenry

#5
##### Jan 12, 2013, 07:18 pm
A few ways but the simplest (but not so accurate) is to use an inductor (the smaller ones that look like resisors), 1mh or be more than sufficient, in place of the coil meter.

Put your arduino's adc input to the transistor's collector. The voltage across the inductor is proportional to the current, and is an average if the inductor is of sufficient size.

BTW, the circuit can be greatly simplified by your arduino.

#### dhenry

#6
##### Jan 12, 2013, 07:18 pm
BTW, the only purpose the inductor performs is to "choke" the current, thus providing an average.

#### aman92

#7
##### Jan 12, 2013, 07:33 pm
Won't measuring the voltage across the inductor give it's max current instead of average?

#### dhenry

#8
##### Jan 12, 2013, 09:02 pm
The voltage across an inductor is proportional to di/dt.

The basic principle in that circuit is to measure the change in current across Lx.

#### aman92

#9
##### Jan 13, 2013, 08:01 am
So this average current is the same when calculated as e^-R*t/Ldt ?

#### PeterH

#10
##### Jan 13, 2013, 05:02 pm

The voltage across an inductor is proportional to di/dt.

For an ideal inductor, yes. A real physical inductor is likely to have resistance and capacitance too. (How significant these are would depend on the inductor in question.)
I only provide help via the forum - please do not contact me for private consultancy.

#### dhenry

#11
##### Jan 13, 2013, 07:14 pm
Quote
So this average current is the same when calculated as e^-R*t/Ldt ?

The basic operation of the circuit, with the coil meter, is this:

1) when Q1 is turned into a diode (S2/S3 closed and S1/S4 open), the inductor is charged up. During this time, the coil meter is by-passed by S3.
2) when Q1 is turned "off" (chocked off maybe a better word, when S2/S3 open and S1/S4 closed), the inductor continues to flow current, but at an exponentially lower levels (the inductor being discharged). The current will flow through the coil meter. Due to its inertial, you are measuring the average of that current, which is a function of the inductance -> why it can be used as an inductance meter.

Using an inductor in place of the coil meter and measure its current does the same, as the voltage across the inductor follows the same di/dt that's determined by the DUT.

Alternatively, you can use a resistor in place of the coil meter. The voltage across that resistor is proportional to the current, and you can average that, with a diode + r/c network: put a 1k resistor in place of the coil meter (can be lower). From the top of that resistor, run a r//c (10k // 4.7u, or there abouts. The resistor can be even higher but not smaller), from the bottom of the r//c pair, run a diode (1n4148 or the likes) to the bottom of the 1k resistor. So this diode + r//c is paralleling the 1k resistor.

Over a small period of time, the bottom of the r//c pair will accumulate a voltage (differential to the power rail) that is proportional to the inductance of DUT.

#### aman92

#12
##### Jan 14, 2013, 05:44 pm
Didn't understand the working of the second circuit but what value of inductor should I use in the first method? And at what frequency should I sample the voltage levels?

#### aman92

#13
##### Feb 06, 2013, 12:18 pm
Can you explain to me the use of the r//c network?

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