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'[OT] measuring value of .001 ohm resistors'
1999\04\14@224124 by Jim Paul

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Seems to me that unless you have some very stout currents flowing through a
resistance of this value, it wouldn't drop enough voltage across it to be
significant.   And if the current through it were small, it wouldn't matter.
Of course I don't know the whole story, but the easiest way to measure a
resistance of this magnitude would bve measure the voltage across it, the
current through it, and calculate the resistance value from those
parameters.    Or you could use op amps to amplify and scale the voltage
across it, and and use an ADC to measure the voltage
and calaulate the resistance value.   I don't know if this would suit your
purpose or not, but it might.



{Original Message removed}

1999\04\15@120126 by John Payson

|Any clever ideas about measuring value of .001 resistors?

The normal technique would be:

[1] Connect a 1 or 10 amp power supply to the resistor using aligator

[2] Using a SECOND SET of aligator clips (which mustn't touch the first)
   connect a sensitive volt meter to the resistor.

If you put one amp through the resistor, your voltage reading will be
1 volt per ohm (1mv/0.001ohm); if you put 10 amps through, the reading
will be scaled tenfold (10mv/0.001ohm).

Even at 10 amps, the power in a 0.001ohm resistor will only be 0.1watt,
which shouldn't be too much.  Note, though, that if the resistor is much
larger (e.g. 0.01ohm) the power would be 1 watt, which might be a problem.
Also be aware that if there is any significant contact resistance, the
contacts may get hot.

1999\04\15@190032 by johnb

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Scott Newell wrote:
> >Any clever ideas about measuring value of .001 resistors?

> Four-wire (Kelvin) connections should be used.
True: The Kelvin bridge (or half-bridge) is the only way that works.

The Alligator Clips (mentioned elsewhere) are available in a special
form known as Kelvin Clips. Each half of the clip is insulated from the
other half; when it is clipped onto the resistor being tested, the test
current flows in through one half of the clip, and the voltage is picked
off by the other half. It isn't essential to do it this way, but it
means that milliohm measurements are as simple to do as normal

> I think the HP micro-ohmeter uses an AC excitation and sync rectification,
> but a modern chopper op-amp might be suitable.  You'll need to flip
> polarity at the very least--the thermal EMF will be a very significant
> error source.

AC excitation is really very important, and allows the use of quite low
currents in the resistor. Suppose a square-wave (of about 1 kHz) is
applied that sends a peak current of 10 milliamps through a 1 milliohm
resistor. The voltage appearing across the resistor is 10mA x 1
milli-ohm = 10 micro-volts. This might seem too small to measure, but it
is easily amplified by a factor of 10,000 which will give a 100
millivolt square-wave output. There is no thermal drift problem since it
is an AC-coupled amplifier. Thermoelectric voltages cancel out.
The 100 millivolt output signal now has to be rectified. This is NOT
done with a diode, but with a synchronous (or phase-sensitive)
rectifier. In its simplest form, this is a transistor switch (or FET
switch) which is switched on and off by the 1 kHz source that drives the

The synchronous rectifier has very remarkable features (if properly
designed). It only rectifies signals of the same frequency that drive
the bridge. Interfering signals that are not harmonically related (such
as mains hum) are completely ignored. It therefore functions as a very
narrow-band tuned amplifier. It is capable of measuring signals buried
deep in noise, because the signal is rectified but the noise averages to

In general, if the rectifier is followed by a low-pass filter of time
constant CR, the bandwidth will be 1/CR. For example, if a 2-second time
constant is used (eg, 1 microfarad and 2 megohms), the bandwidth is 0.5

Precautions: the AC amplifier must have a very wide dynamic range if
large interfering signals are not to overload it. The excitation
frequency must be very low if the resistors being measuring are wire
wound and have significent inductance. Ideally, sine-wave drive should
be used, and the AC amplifier should function as a band-pass filter.
This is because synchronous detectors also rectify odd harmonics of the

There are a number of chips around that do this function, but there are
mainly of great antiquity. Another name for "synchronous detector" is
"double-balanced modulator". The two things do different jobs but they
contain the same circuitry. An interesting chip for modem builders is
the EXAR XR2211, which is a phase-locked FSK (frequency shift keying)
demodulator. The tenacity with which this chip locks on to modem tones
is very remarkable. But only at 300 baud, unfortunately. The EXAR Web
site has a PDF file for this chip and others.

John Blackburn,
London UK.

1999\04\15@204209 by Mark Moss

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Use a very large constant current.

BTW, one way to calibrate a sense resistor, even as high as the .1 Ohm range
is to adjust the lead length and/or the amount of solder.

Mark Moss, KC8DEI

-----Original Message-----

>Gus Calabrese wrote:
>> PicGeniuses
>> Any clever ideas about measuring value of .001 resistors?
>> Gus

1999\04\15@205008 by Sean Breheny

face picon face
This thread makes my mind beg a question:

What does a .001 ohm resistor look like?! Must have pretty big connection
terminals on it to give that low a resistance.


At 12:17 AM 4/15/99 +0100, you wrote:
>Use a very large constant current.
>BTW, one way to calibrate a sense resistor, even as high as the .1 Ohm range
>is to adjust the lead length and/or the amount of solder.
>Mark Moss, KC8DEI

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
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