kelvin type Ohm Meter

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Ashkar

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Hello peoples,

I want to ask for some help from you all.
I am building a Kelvin type ohm meter with a supply current of about 1 Amp on a 50 milli ohm resistor.It would also be used to measure resistance of about 10 K also with a selector switch.
So I want to know, how do I vary current on the device that I am testing.
 

Then I think you want two current sources for measuring mOhm range and kohm range. what range exactly you need from the two selections?
 

Yes you are right buddy.
I was thinking to use 1 Amp dc supply.But I need to know how can I produce a 1 amp current for this purpose using a Lm321 1.5 Amp regulator.
Any Idea on this.
 

But selecting the current rating is not that much easy dude. What if you connect a 1 ohm 0.5w resistor in 1A current source??

you need perfect sketches about range going to measure, power supply ability, measuring meter rating.
 

Kelvin method is 4 wire and eliminates contact resistance error with outer 2 leads supplying current and inner 2 measuring voltage.

CC method with voltage sense and power limit prevents target burnout.

I think an auto-scale approach might start with 10, 100, 1000 mA where it steps down or up according to voltage threshold of >100mV & <10mV.

perhaps you might want an option for 10A for µΩ resolution.

There are many ways to create low power CC source or sink.
 

Constant current control method is the right choice to builds
a source that would produce current in the device under test.
So how should I build a source that would produce that much current with very low voltage as I would be testing resistances as low as 50 milli ohms to 10 k ohms.
 

Enabling auto scaling current source will make the equipment more complicated. as we are measuring the output voltage in the scale If we are changing the constant current value It will not be a uniform or single scale it would make more complication for an analog equipment. or there shd be some digital mechanism to indicate the scale factor like indicating leds and so on.
 

**broken link removed**

Code: 
e = Vref / R

Ie = Ib + Ic = (Ic/B) + Ic

Ic = [B/(B+1)] *Ie

Io = Ic = [B/(B+1)] * Vref / R about Vref / R
VL = Io*RL

The resistor to be measured is connected in the collector of the transistor
By changing any ine Vref or R you can change the current value. you can put a selector switch to select which zener to in the circuit to choose the current value.

If the current values too different then you can construct two CCs and you can switch their output in load point.
 

Thanks for your work Buddy,
So your circuit provide a very low voltage of about 0.001v min?
 

I want to ask you .Is the circuit shown above is capable of producing 0.001v if I place a resistor of 0.001 Ohm over collector side of the transistor?
 

Hi,

I want to ask you .Is the circuit shown above is capable of producing 0.001v if I place a resistor of 0.001 Ohm over collector side of the transistor?
Click to expand...

Yes. It is a current source. The circuit is ok and it will work as desired.
The voltage will be according ohm's law. U = R x I.

*******
To improve the circuit you can use a pFET instead of BJT. Use one with a low threshold voltage .

Klaus
 

Thanks a lot Venkadesh.
I really appreciate your help.

- - - Updated - - -

As for the pFET what should be a suitable option for driving a 1amp load.
 

One more query I need to clarify.
How does I switch the PFET bcoz I am using a 6 v Battery supply for my purpose.
 

Hi,

How does I switch the PFET bcoz I am using a 6 v Battery supply for my purpose.
Click to expand...

Indeed it is not "switching" but it is regulating.

Your PFET parameters are:
* VDS > 6V
* IDS > 1A
* Ptot > 6W (you may need a heatsink there, so use a TO220 package)
* Vgate: smaller than VBAT_min - UShunt - VOpampLow = (estimated) 5V - 0.5V - 0.5V = 4V (@ 1A)
with a too big VG you ma not reach 1A. With a too low Vgth you may not completely switch off the FET. Depends on VOpampHigh.

Try a NDP6020P. Available at farnell.

Klaus

- - - Updated - - -

Hi,

so why do you need a pfet?
Click to expand...

according the given schematic..
the PFET has one benifit: The gate current can be considered to be 0.000A, therefore the load current is very precisely VRef/R


Klaus
 

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