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Frank Milburn's Blog Even More on Current Sources and a Kelvin (4-Wire) Milliohm Meter
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  • Author Author: fmilburn
  • Date Created: 3 Oct 2018 6:27 AM Date Created
  • Views 6748 views
  • Likes 9 likes
  • Comments 51 comments
  • current source
  • milliohm measurement
  • op amps
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Even More on Current Sources and a Kelvin (4-Wire) Milliohm Meter

fmilburn
fmilburn
3 Oct 2018

Introduction

This is the third post on the development of an inexpensive but reasonably accurate meter for measuring resistance in the milliohm range. In the first post a simple current source was described that created a 10 mA current  across a resistor that allowed the voltage drop to be measured using a multimeter and the resistance calculated.  A number of helpful suggestions were received and I ordered additional components based on that feedback.  In the second post a block diagram for the instrument was introduced and initial measurements were made with a microcontroller using the built-in ADC.  Some, but not all of the ordered parts have been received now and this post will update progress as I don't want John's popcorn to get stale.

 

A Change to the Design Objectives

I originally specified that the current to the DUT would not be greater than 10 mA.  Testing to date has indicated that meeting the desired accuracy will be difficult without amplification of the voltage difference across the DUT which adds some complexity and cost.  Accordingly, the specification is being changed to 100 mA across the DUT.

 

Component Status

First, I have to admit to making a mistake in the orders.  The MCP6N16 instrument amp comes in three versions with different minimum gain.  I wanted the version with minimum gain of 1 and ordered the version that has a minimum gain of 100.  Doh!  Always read the datasheet carefully.  For now I am substituting the MAX9619.  I also ordered a precision LDO voltage source from the TI store which has not been shipped yet.  Usually they are pretty quick. The volt meter I plan to use is still in shipment from China.

 

100 mA Current Source

This is the revised circuit, the only real changes being the addition of a MOSFET to handle the increased current and a new precision 0.1% 10 ohm resistor to set the current.  I am using an inexpensive ANENG multimeter to measure voltage but it does agree well with my bench meter.

image

And here are the results:

image

The tests are being performed the same way as previously using a coil of wire that has been center tapped.  The measured resistance of the full length of wire is 0.092 ohms as seen on top while the measured resistance of half the length is 0.046 ohms - exactly half.

 

Next Steps

The inexpensive voltmeter needs at least 4.5 V to operate so I will probably use either 4 x 1.5 V AAA batteries or USB power and a precision voltage source to set the current.  If I decide to use a microcontroller instead of a voltmeter then a 3V3 LDO will be used to power that.  The parts for Kelvin probes are on order.  Progress depends on the postal service now...

 

Past Posts on this Topic

More on Current Sources and a Kelvin (4-Wire) Milliohm Meter

Testing Current Sources for a Kelvin (4-Wire) Milliohm Meter

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Top Comments

  • fmilburn
    fmilburn over 7 years ago in reply to shabaz +5
    Hi Shabaz, An update... I read the datasheet thoroughly and set up the LM334 as a temperature compensated current source as described in the datasheet. A IN4148 was substituted for the diode they used…
  • shabaz
    shabaz over 7 years ago in reply to fmilburn +5
    Hi Frank, That's very interesting! Is it figure 15 in the datasheet? There is a modification suggested here: https://www.electronicdesign.com/analog/what-s-all-lm334-stuff-anyhow that looks like it could…
  • fmilburn
    fmilburn over 7 years ago in reply to shabaz +5
    Shabaz,, Yes, using the circuit in Figure 15 with 1% resistors and the diode noted above. I am at the limit of the resolution of my multimeter and there was some bouncing back and forth so am not sure…
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  • shabaz
    shabaz over 7 years ago

    Hi Frank!

     

    I'm wondering if it is possible to use a dual-stage design, i.e. separate the current source from the final signal output stage, i.e. use an op-amp for the voltage amplification across the load, but perhaps with a lower-current source, since 100mA might make it more difficult, because the reference resistor will get warm and maybe drift. That op-amp you mention (in a min gain of 10 version) could also be used for the second stage.

     

    Any of the current source ideas could be used for the first stage, but I think the LM334 would be attractive (but means your first stage would not be an op-amp, so depends if this is a hard limitation or not : ) because the power in the resistor is really small (since the internal reference inside it is 68mV), i.e. if a 6.8ohm resistor is used, power dissipated in the resistor is 0.68mW.  Then for the desired 1mohm to 10 ohm range, if the meter is 0-2V (as an example) then a gain of 20 is needed (i.e. compatible with the min. gain of 10 version). Or for an op-amp version, it could be the same as the circuit you have now of course (with the reference resistor value changed). These are just some ideas, maybe unnecessary if the resistor won't drift much (or if you find in the end that less than 100mA is fine too, for the particular multimeter you use).

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  • fmilburn
    fmilburn over 7 years ago in reply to shabaz

    Hi Shabaz,

     

    An update... I read the datasheet thoroughly and set up the LM334 as a temperature compensated current source as described in the datasheet.  A IN4148 was substituted for the diode they used.  I came up with nonstandard resistor values so I put the closest values I had in.  The initial test was about 2% off the desired 10 mA current (not unexpected based on what was in the datasheet) so R1 was adjusted by adding another resistor in parallel until within 1%.  Voltage was varied over the range of interest and here are the results for the full length of test wire:

    image

    Things look good above 2.5V input (the planned instrument will have at least 3V) but notice there is a slight rise in the measured current at 5V.  This may be due to temperature from increased voltage which the datasheet warns of or just increase of temperature as it heats up in time in still air.  I left it on for an hour at 5V and from that point on it was stable although I am unable to make measurements with the resolution I would like.

     

    I added another tap in my test wire at quarter length.  The results are:

    Full Length:  0.092 Ohms

    Half Length: 0.045 Ohms

    Qtr Length:  0.022 Ohms

     

    The accuracy appears to fall off as it nears the bottom of the multimeter range but all is as expected.  I will look into adding amplification next.

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  • fmilburn
    fmilburn over 7 years ago in reply to shabaz

    That is really interesting.  If I remember from the datasheet example Vf have fair impact on the accuracy.  Again, amazing how much complexity there can be behind such a simple circuit if you dive deep enough.  I am going to spend this evening playing with the circuit a bit more.

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  • shabaz
    shabaz over 7 years ago in reply to fmilburn

    I think you're right, one of the formulas needs the Vf entered I think.. what transistor are you planning to use? I know the 2N3904 is more popular in the US, whereas in the UK the BC547 is very common. But I have 2N3904 here too, so if that's the one you plan to use, then I'll use the same one here too!

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  • fmilburn
    fmilburn over 7 years ago in reply to shabaz

    I have the 2N3904 and will get started with that.

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  • fmilburn
    fmilburn over 7 years ago in reply to shabaz

    Hi Shabaz,

     

    I set up a spreadsheet to calculate R1 and R2 and input the values you determined empirically (nice tests by the way).  Here is an example of input and output:

    image

    Note that the calculated R2 / R1 is 7.8 and not the 10 from the datasheet.  I got pretty close to these values by using two resistors in series:

    R1 = 15 + 2.2 = 17.4 measured

    R2 = 100 + 33 = 134 measured

     

    I started with the diode I already had in the circuit, IN4148.  I measured the current through it to be 4.23 mA and the voltage drop to be 687 mV which is lower than the Vf your diodes are showing.  The  current out Iset (which ideally would be 10 mA) was 9.83 mA.  I plan to redo the calculations for the actual Vf and / or change out the diode but thought I would post this in case I don't complete it tonight.

     

    Since it is sensitive to Vf and maybe other stuff, I will try a trimmer pot to see how well that works also.

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  • fmilburn
    fmilburn over 7 years ago in reply to shabaz

    Hi Shabaz,

    Just noticed something....  The tempco in your test drops as the current goes up, but in the TI chart it goes up.  Am I misunderstanding?

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  • fmilburn
    fmilburn over 7 years ago in reply to shabaz

    Hi Shabaz,

    Just noticed something....  The tempco in your test drops as the current goes up, but in the TI chart it goes up.  Am I misunderstanding?

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  • shabaz
    shabaz over 7 years ago in reply to fmilburn

    Hi! The TI chart rises as does the data in the table, I think maybe the chart I drew may be confusing things, it was just showing Vf (not tempco) just so we could see how the selected Vf would also change depending on what current setting was used.

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  • fmilburn
    fmilburn over 7 years ago in reply to shabaz

    OK thanks - it is getting late :-)

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