Easy and cheap milli-Ohm-meter

[Boncuk]

Hi,

I found that circuit in Elektor.

It is quite accurate if the voltage regulator's output is exactly 6V. Use 1% or 0.1% resistors to improve accuracy.

Board size is 2.1X0.95"

Boncuk

EDIT: Faulty circuit removed - corrected circuit added

 

[Grossel]

Diode D1 will make the circuit unusable. I guess D1 should be moved to the right side of resistors R1-R4 to protect from overvoltage.

Another thing is that this is nothing else than a smple voltage divider, so you still have to calculate resistor value by hand. Voltage out won't be linear.

From my point of view, this is just crap.

 

[Boncuk]

Diode D1 will make the circuit unusable. I guess D1 should be moved to the right side of resistors R1-R4 to protect from overvoltage.

Error corrected.

Another thing is that this is nothing else than a smple voltage divider, so you still have to calculate resistor value by hand. Voltage out won't be linear.

Nope! You don't have to calculate manually. With the 120Ω resistors 1mV across RL corresponds to 10mΩ, with 1.2KΩ resistors 1mV it corresponds to 100mΩ. Ohm's law is linear!

From my point of view, this is just crap.

You can't expect a high end measuring device for a few Cents. I guess it's good enough to match emitter resistors.

If you first measure the voltage drop of the testing cable and thereafter the total voltage drop with a resistor just subtract the value of the cable from the result.

Boncuk

 

[Mikebits]

From my point of view, this is just crap.

No need to be insulting...

 

[Grossel]

No need to be insulting...

Right. Sorry for that, Boncuk. Must learn to not post when I'm in bad mood

What i SHOULD say is that a constant current source that have a diode as load would give a linear voltage out as measured resistor value changes.

It is almost true that a voltage divider is linear, but only in a very small section of possible output values it will be approximately linear. But, to get a true linear output, a true constant current source is a better choice.
A 7806 can easilly be used as current source. See page 9 **broken link removed**.

relevant search on net

 

[Boncuk]

Hi Grossel,

I didn't feel insulted at all. The ciruit is not my design (as already mentioned, Elektor - Klaus Berholdt).

As accuracy depends almost only on the precision of either voltage or current source (and of course accuracy of the resistors and last not resistance of the switch) it doesn't make a big difference in using current or voltage.

Boncuk

 

[mneary]

The topic title is clear. It is a milli-ohm meter.

It is almost true that a voltage divider is linear, but only in a very small section of possible output values it will be approximately linear.

When used to measure resistances up to 0.6 ohms, the "current source" provided by 600 ohms is still within 0.1% of linear.

When this circuit is used as intended, the diode will not be conducting.

 

[Boncuk]

When this circuit is used as intended, the diode will not be conducting.

It's only there to protect the DMM from overvoltage in case of failure.

 

[Vizier87]

what are the cases when you need to know ohms in the mili range? For now I could only think of transmission lines....

 

[Mikebits]

what are the cases when you need to know ohms in the mili range? For now I could only think of transmission lines....

Measuring current maybe?

 

[Boncuk]

another case might be when paralleling BISS (Breakthrough In Small Signal) transistors.

Check out the datasheet of the PBSS5630PA transistor (NXP).

Boncuk

 

[Vizier87]

Interesting.

Key features of low VCEsat (BISS) transistors are:
• Low collector-emitter saturation voltage and collector-emitter resistance
• Low power dissipation and consumption
• High collector current capability related to required board space
• High current gain at high collector currents

There's no wiki about it.

 

[Boncuk]

There's no wiki about it.

Study the data sheet as posted and compare.

Here it's again for your ease.

Boncuk