Measuring tiny unknown resistance?

Discussion in 'Electronic Basics' started by kevwalsh, Jan 28, 2005.

1. kevwalshGuest

Hi,

Anyone know a simple way I can measure a small, unknown resistance? I
am guessing it is about between 1ohm and 0.5ohms, roughly. It is part
of a car, so I can't bring it inside to test it.

I have several little multimeters, but none go down anywhere near as
low as half an ohm -- the cheap analog one goes as low as about 100ohms
at the very tinies graduation, and I don't think my digital ones go
much lower than that either.

I can also put 12v to the thing and try to measure current draw, but
obviously 10A or more is too much for just sticking my little
multimeter inline with the resistance.

I have access to plenty of simple parts, resistors and such, to build
some kind of simple measuring circuit. Any tips?

Thanks,
Kevin

BTW -- it is a rear window defroster, if you didn't guess that already.
My problem is that the new rear window draws too much current (how much
I am trying to discover) and melts things, like the on-off switch.

2. Fritz SchlunderGuest

You would have some very unusual meters if they can't measure resistances of
less than 100 ohms. As digital multimeters go they almost universally (from
the very cheapest to the most expensive) have a minimum resistance scale of
200 ohms. On the 200 ohm scale the maximum resistance that can be measured
is 199.9 ohms. The minimum resistance that can be measured is limited by
the accuracy of that scale. On the 200 ohm scale, this means the minimum
resistance that can be measured is 0.1 ohm. The last digit is not very
accurate, so it could be off by a little. Measuring a resistance between
0.5 ohm and 1 ohm will not be too especially accurate, but it will likely be
accurate to about as good as within 0.1 ohm. Sometimes the meter will not
be calibrated to zero and will thus produce a fixed offset to your
measurement. You can manually compensate for this by shorting the meter
leads together on the 200 ohm resistance scale and observing the result. It
may for instance measure 0.2 ohms, but for a hard short circuit it should be
0.0 ohms. So, take you measurement of your 0.5ohm to 1 ohm resistance and
then subtract 0.2 ohms from the displayed value.

To get better accuracy and to further extend the measurement range down to
lower values you need another technique. One way is to build a 100mA
constant current source using an LM317 linear regulator. Then apply the
100mA constant current to the unknown low value resistance and measure the
voltage across it. Ohms law predicts V=IR, so for I=100mA, the voltage that
appears will be one tenth of the resistance. IE: if the meter reads 50.0mV,
then the resistance is 0.500 ohms.

3. SpajkyGuest

try to measure current thru it & voltage across it & calculate the
resistance than ...

4. GarethGuest

If you have a low value, high power resistor you could put it in series
with your unknown resistance. Apply a voltage then measure the voltage
across the know resistance and across the unknown resistance. Since the
current through both is the same, the ratio of the two resistances will
be equal to the ratio of the two voltages. Therefore:

R1 = R2*V1/V2

Where V1 = voltage across R1, V2 = voltage across R2

To get an accurate measurement the known resistor value should be a
similar order of magnitude to the unknown resistance, then you will get
sensible voltages across both resistances.

Obviously, your known resistor must be capable of handling the current.

Gareth.

--

5. Bob MastaGuest

To add to what others have posted, note that the trick to
accurate low-resistance measurements is to measure the
voltage drop across the unknown resistance with
*different leads* than those driving the current
through it. (Unlike the the conventional ohmmeter
arrangement.) So, if you use the suggestion of
a series power resistor, you can measure the
voltage across that to determine the current,
then measure the voltage across the unknown and
compute its resistance.

Best regards,

Bob Masta

D A Q A R T A
Data AcQuisition And Real-Time Analysis
www.daqarta.com

6. john jardineGuest

Stick a 100ohm resistor in series with the defroster lead. Switch on and
measure the DC voltage across the heater. A 1ohm defroster would give 100mV.
A 0.5ohm defroster would give 50mV etc.
regards
john

7. Geir KlemetsenGuest

"john jardine" <> skrev i melding

Or, like this:

o-------| R1 |--------| R2 |-----------|!' Gnd
Ucc

Ur2 (voltage across R2) = (Ucc * R2) / (R1 + R2) // The basic formula.

Turning the formula to get the value of R2 is quite easy:
UR2( R1 + R2 ) = Ucc * R2
( (UR2 * R1) / R2 ) + UR2 = Ucc

// This is the important stuff
So finding R2 is to calculate this formula:
R2 = (UR2 * R1) / (Ucc - UR2)

Hope this helps : )

8. Bill BowdenGuest

Anyone know a simple way I can measure a small,
You could use a 12 ohm resistor is series with the 12 volt battery
so you get about 1 amp of current. It will heat up with about 12 watts
of heat, so you need a large resistor. Or, use a tail lamp bulb
which is a about 2 amps. Then use your multimeter to measure the
exact current. Also measure the voltage across the unknown resistance
and divide this by the known current.

So, if you read 0.6 volts across the resistance, and the current
is 1 amp, the unknown resistance will be 0.6/1 = 0.6 ohms.

-Bill

9. Geir KlemetsenGuest

"Bill Bowden" <> skrev i melding

You could use a 12 ohm resistor is series with the 12 volt battery
so you get about 1 amp of current. It will heat up with about 12 watts
of heat, so you need a large resistor. Or, use a tail lamp bulb
which is a about 2 amps. Then use your multimeter to measure the
exact current. Also measure the voltage across the unknown resistance
and divide this by the known current.

So, if you read 0.6 volts across the resistance, and the current
is 1 amp, the unknown resistance will be 0.6/1 = 0.6 ohms.

Disagree. When he measures 0.6 volts across the resistance, there is no
longer 1 amp through the circuit any more because the total resistance will
be more then 12 ohms. The higher the "unknown" resistance is, the higher
error on the measurment.

If using my formula (erlier post, same thread) the resul will be:
"unknown resistance" = (0,6V * 12ohm) / (12V - 0.6V) = 0.63 ohm

10. Bill BowdenGuest

Disagree. When he measures 0.6 volts across the
I don't quite follow the logic. It doesn't really matter what
the total resistance is, or the battery voltage. We only need
to know the current through the circuit, and the voltage across the
unknown resistance. If we measure 1 amp through the circuit, and 0.6
volts across the unknown resistance, the value of the resistance will
be *exactly* 0.6 ohms, unless you are considering the error introduced
by the meter resistance, in which case you need to know the meter
resistance, which wasn't given.

-Bill

11. Geir KlemetsenGuest

That's also possible to find.

If you couple two resistors in series and measures the voltage over one of
them, say f.ex that you have R1 and R2 in series and measure the voltage
over R2. If you know the exact resistance for both R1 and R2, then the
formula for resistance in the meter is as follow:

Rmeter = (UR1 * R1 * R2) / ( Ucc*R2 - UR1*R1 - UR1*R2 )

12. Bill BowdenGuest

Rmeter = (UR1 * R1 * R2) / ( Ucc*R2 - UR1*R1 - UR1*R2 )

Yes, very good formula. But the meter resistance measuring
voltage is quite high, probably several megohms, so it
has little effect in parallel with 1 ohm or less.
Probably less than .001%

I would imagine the series meter resistance measuring current
would be of more concern, since the resistance in series
is very low and the meter resistance could be a significant
part of the total.

I think only one known resistor is needed to figure this out,
but I haven't worked it out.

-Bill

13. Don KellyGuest

Ohmmeters are notoriously lousy for measuring low resistances. While the
nominal range may allow readings to 0.1 ohms, it is likely that at this
level the error may be, in fact, much higher - esesentually making the
reading no more than a continuity test. Such things as a thin oxide coating
on the terminals . can blow the reading to hell as the ohmmeter will not
have the "oomph" to break down this layer as happens in normal use. You
need to be able to supply enough voltage (something near nominal.) to drive
a reasonable current - and measure V and I. Someone mentioned use of a known
resistance in series (check its power rating ) and measuring the voltage
drop across this (getting, by the way, the current fromV/R) and that across
the unknown. This works (and in fact is related to the way most multimeters
handle current measurements)
Use of a such a shunt (low resistance) in series with the load, and
measuring the voltage across the shunt is a lot cheaper than making a
constant current source in the 5+ A range. 100ma won't do it for proper
measurement of the resistance of a 12V device drawing 12 to 24A.