# Problem with a circuit to measure of displacement

#### Mousivandhossein

Joined Dec 8, 2020
34
I designed and built a circuit to measure displacement.
I made a coil and connected a wein oscillator to the coil, which changes the voltage at both ends of the coil by moving an iron rod inside it, then by ideal rectifier, the voltage of the coil turned to DC voltage.
unfortunately, now the output of the circuit changes about 1% when the rod not changing and the rise of temperature can change the output about 4-5% for the rod to remain fixed.
Can you please help me to improve my circuit against the variation of the output? or if you know a better circuit, please suggest it.
Thanks and Best Regards,

#### Attachments

• 83.7 KB Views: 8
Last edited:

#### AlbertHall

Joined Jun 4, 2014
12,396
The oscillator should be running at about 3.4kHz.
The iron core will increase the inductance and so increase the voltage across the inductor, but the iron core will be quite lossy at that frequency and this effect will tend to reduce the voltage.
Maybe, try a ferrite rod core for the coil?
What is the inductance of L2?

#### Mousivandhossein

Joined Dec 8, 2020
34
The oscillator should be running at about 3.4kHz.
The iron core will increase the inductance and so increase the voltage across the inductor, but the iron core will be quite lossy at that frequency and this effect will tend to reduce the voltage.
Maybe, try a ferrite rod core for the coil?
What is the inductance of L2?
Yes, the frequency of the oscillator is about 3.4 kHz.
I can not find a ferrite rod to replace with iron rod.
The losses do not matter to me, only I want to have a stable output for a fixed rod iron, the output variation is my most important issue.
L2 is the coil, The coil has 2000 turns and its length is 270 mm and its diameter 9 mm.
The distribution of the windings is uniform.

Last edited:

#### AlbertHall

Joined Jun 4, 2014
12,396
The losses do not matter to me,
The losses do matter as they will affect the voltage output of your circuit.
Do you have a brass, copper, or aluminium rod that will fit the coil?
How much displacement are you trying to measure?
How long is the coil?

#### Marley

Joined Apr 4, 2016
502
So your displacement measuring device depends on the change of inductance with the position of the ferrite rod?
A good concept. But:
Not sure why you chose the oscillator frequency to be 3.4KHz. Also why the capacitor on the oscillator output is 1uF.
The impedance of the 1uF capacitor is about 48 ohms. This is in series with the output impedance of the oscillator op-amp which will also be quite low. Possibly less than 10ohms. I don't know much about the inductor but the impedance could be quite high at 3.4kHz. What is its impedance at the limits of travel?
The "potential divider" formed by the capacitor and the op-amp output on one side and the inductor on the other is dominated by the low impedance of the capacitance verses the high impedance of the inductor.
Also, the output will be affected by any drift in the oscillator frequency.
I would change to a much higher oscillator frequency. Tens of kHz at least. Make it stable. Either a crystal oscillator or derived from a crystal with a divider. This could also have the advantage that the inductor has less turns.
Have a resistor in series with the oscillator output that has approximately the same resistance as the impedance of the inductor at half travel. Then the "potential divider" of impedances will show the maximum change with displacement.
And the rod inside the coil must be ferrite. Similar to the type in old AM transistor radios. As the frequency goes up, you need to consider the effect of stray capacitance affecting the output. Might need to screen the coil with metal - but without forming a shorted turn. Wind the coil on an aluminum tube with a slot all the way down so eddy currents cannot flow.

Last edited:

#### Mousivandhossein

Joined Dec 8, 2020
34
The losses do matter as they will affect the voltage output of your circuit.
Do you have a brass, copper, or aluminium rod that will fit the coil?
How much displacement are you trying to measure?
How long is the coil?
I did not try rods made of brass and aluminum. I think ferromagnetic materials are more effective in increasing the amount of inductance. But I tested the 316 stainless steel rod and it did not work
I think brass, copper and aluminum have the same result as SS 316.

#### Mousivandhossein

Joined Dec 8, 2020
34

The picture of the coil.

#### Mousivandhossein

Joined Dec 8, 2020
34
So your displacement measuring device depends on the change of inductance with the position of the ferrite rod?
A good concept. But:
Not sure why you chose the oscillator frequency to be 3.4KHz. Also why the capacitor on the oscillator output is 1uF.
The impedance of the 1uF capacitor is about 48 ohms. This is in series with the output impedance of the oscillator op-amp which will also be quite low. Possibly less than 10ohms. I don't know much about the inductor but the impedance could be quite high at 3.4kHz. What is its impedance at the limits of travel?
The "potential divider" formed by the capacitor and the op-amp output on one side and the inductor on the other is dominated by the low impedance of the capacitance verses the high impedance of the inductor.
Also, the output will be affected by any drift in the oscillator frequency.
I would change to a much higher oscillator frequency. Tens of kHz at least. Make it stable. Either a crystal oscillator or derived from a crystal with a divider. This could also have the advantage that the inductor has less turns.
Have a resistor in series with the oscillator output that has approximately the same resistance as the impedance of the inductor at half travel. Then the "potential divider" of impedances will show the maximum change with displacement.
And the rod inside the coil must be ferrite. Similar to the type in old AM transistor radios. As the frequency goes up, you need to consider the effect of stray capacitance affecting the output. Might need to screen the coil with metal - but without forming a shorted turn. Wind the coil on an aluminum tube with a slot all the way down so eddy currents cannot flow.
Thank you very much for your good tips.
now I change the oscillator frequency to 20 kHz and I try to calculate the inductance of the inductor and then I will put a resistor series with the output of op-amp and check the output stability.

I will tell the results here.

#### Marley

Joined Apr 4, 2016
502
If you use a solid metal rod inside the coil the eddy currents induced in the metal will destroy the inductance. A ferrite rod has magnetic properties but does not conduct. You can buy ferrite rods. Or, find an old AM radio and get one from that.

#### Mousivandhossein

Joined Dec 8, 2020
34
I want to convert the ratio between voltage and rod position to a curve and find the exact position of the rod by fitting a function to the curve based on the output voltage.

To calculate I want to use a microcontroller.

#### ericgibbs

Joined Jan 29, 2010
19,118
hi,
What change in inductance to expect over the range of rod insertion.?
E

#### Mousivandhossein

Joined Dec 8, 2020
34
If you use a solid metal rod inside the coil the eddy currents induced in the metal will destroy the inductance. A ferrite rod has magnetic properties but does not conduct. You can buy ferrite rods. Or, find an old AM radio and get one from that.
In your opinion, the eddy current losses can change the output with the fixed iron rod?
When the iron rod is not in the coil, I see the output has a little variation, it is my important issue.
No changes have been made to the circuit yet.

#### Mousivandhossein

Joined Dec 8, 2020
34
hi,
What change in inductance to expect over the range of rod insertion.?
E
Hi,
I have not yet measured the range of inductance changes.

Last night I tested the circuit, the DC output changed from 1.265 V for the rod was out of the coil and 3.924V for the rod quite inside of the coil.

when the rod was out of the coil the output voltage changes from 1.259 V to 1.272 V.
when the rod was quite inside of the coil the output voltage changes from 3.915V to 3.930 V.

#### ericgibbs

Joined Jan 29, 2010
19,118
hi M.
This is a basic LTSpice simulation 5mH and 10mH
20kHz sine at 5Vpk
E

#### Attachments

• 45.4 KB Views: 4

#### Sensacell

Joined Jun 19, 2012
3,505
Better results to be had with "Linear Variable Differential Transformer" configuration - LVDT for short.

#### Mousivandhossein

Joined Dec 8, 2020
34
Better results to be had with "Linear Variable Differential Transformer" configuration - LVDT for short.

First I decided to instruct the device according to LVDT, but the range of work of it was limited and a wider range of coils was needed to get an acceptable result.
However, I have not to experience with LVDT.

#### Sensacell

Joined Jun 19, 2012
3,505
What is the performance you need?

Linear Range...
Response time...
Accuracy...
Linearity...
Physical size...
Power consumption...

#### MrChips

Joined Oct 2, 2009
31,088
That's a rather ingenious way to measure displacement.

Why not use a linear resistance coil or track or even a 10-turn pot with a rack and pinion or a flex band. A voltage divider circuit is a lot simpler, more linear, accurate and stable.

You could even use the same coil as variable linear resistor by scraping off the insulation and putting a wiper on it.

#### Mousivandhossein

Joined Dec 8, 2020
34
What is the performance you need?

Linear Range...
Response time...
Accuracy...
Linearity...
Physical size...
Power consumption...
My request is based on priority:
1- Accuracy (+-0.5 mm accuracy and stable output when the rod fixed)
2- linearity (however I can linearize the output with a microcontroller)
3- Physical size (I have about 300 mm space for measuring 250 mm)
4- power consumption (max 2 W)
5- response time (settling time about 0.25 sec is good)