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Resistor test question from a newbie

J

Jo

Jan 1, 1970
0
If I wanted to test a 1M ohm resistor to make sure it's working, would
I set my digital multimeter to 20M? The next lowest is 200k. Here's a
stupid question - will setting it to 20M cause damage to the resistor?
 
M

Michael Black

Jan 1, 1970
0
If I wanted to test a 1M ohm resistor to make sure it's working, would
I set my digital multimeter to 20M? The next lowest is 200k. Here's a
stupid question - will setting it to 20M cause damage to the resistor?
You won't be able to measure the resistor unless you put the meter
on a higher scale.

If you set the meter to the 200K range, then all you'll get is a full
scale reading, since a 1M resistor is 5 times as large as 200K.

Given that, explain how a higher setting could possibly do damage
to the resistor?

Michael
 
B

Bob Engelhardt

Jan 1, 1970
0
Jo said:
... digital multimeter ... cause damage to the resistor?

There is no setting on a DMM, or VOM, that will damage any resistor.

Bob
 
A

amdx

Jan 1, 1970
0
Jo said:
If I wanted to test a 1M ohm resistor to make sure it's working, would
I set my digital multimeter to 20M? The next lowest is 200k. Here's a
stupid question - will setting it to 20M cause damage to the resistor?

Study this page about ohm meters. It will give you an understanding about
what is happening
when you measure resistance. It does use an analog meter but you can just
assume your digital
meter is measuring the voltage across the 500 ohm resistor referenced on the
page. If you know the
voltage across the resistor you can calculate the current. If you know the
current and the total
voltage you can calculate the total resistance and then your unknown
resistance.
http://www.allaboutcircuits.com/vol_1/chpt_8/6.html

Mike
 
T

Tom Biasi

Jan 1, 1970
0
Jo said:
If I wanted to test a 1M ohm resistor to make sure it's working, would
I set my digital multimeter to 20M? The next lowest is 200k. Here's a
stupid question - will setting it to 20M cause damage to the resistor?+
The meter setting is the highest value the meter can measure.
If you think this can damage the resistor then you need more instruction on
the use of millimeters.
No problem giving you the info here but please look into a multimeter more.

Tom
 
D

David L. Jones

Jan 1, 1970
0
Jo said:
If I wanted to test a 1M ohm resistor to make sure it's working, would
I set my digital multimeter to 20M? The next lowest is 200k.

Yes, you set it to the next highest range.
Or simply use an auto-ranging digital multimeter and it does it all for you.
Here's a stupid question - will setting it to 20M cause damage to the
resistor?

No.
The current the meter outputs is so low it won't damage any resistor on any
range.

Dave.
 
D

David L. Jones

Jan 1, 1970
0
John Larkin said:
What's that? A very small meter?

Don't be silly, it's a meter that measures milli's!
Only problem with the damn things is that you have to multiply the result by
1000 to get the real value :->

Dave.
 
T

Tom Biasi

Jan 1, 1970
0
John Larkin said:
What's that? A very small meter?

John
OK, I missed that. My fingers didn't type what my brain wanted. ;-)
 
N

Nobody

Jan 1, 1970
0
Others have answered your question... use the 20 M
range. However, suppose your meter didn't have
any range bigger than 200k? You can still measure
your 1M resistor. First get a resistor in the
100k-200k range and measure it, then connect the
1M resistor to be measured in parallel. (Note, as
another post mentioned, not to touch the leads
during either reading.)

Now use the "resistors in parallel" equation in
reverse: Normally, Rpar = (R0 * Rx) / (R0 + Rx).
Rearranging, Rx = (R0 * Rpar) / (R0 - Rpar)

So if R0 = 100k and the parallel combo Rpar is
90k, the unknown is (100k * 90k) / (100k - 90k)
= 900k.

The main problem with this approach is that any uncertainty (i.e.
tolerance) in the known resistance creates a much greater degree of
uncertainty in the result.

E.g. if R0 = 100k +/- 5% (i.e. 95k-105k) and the parallel combo is 90k,
the unknown is between:

(95k * 90k) / (95k - 90k) = 1710k
(105k * 90k) / (105k - 90k) = 630k

This is so inaccurate as to be practically useless.

[With a 10% resistor, i.e. 90k-110k, it would be *completely* useless; the
90k lower bound means that the upper bound on the unknown would be infinity.]

With a 2% resistor, the figures are more reasonable, but still quite
inaccurate:

(98k * 90k) / (98k - 90k) = 1102k
(102k * 90k) / (102k - 90k) = 765k

=> 933k +/- 18%.

With a 1% resistor, the inaccuracy is tolerable for some purposes:

(99k * 90k) / (99k - 90k) = 990k
(101k * 90k) / (101k - 90k) = 826k

=> 908k +/- 9%.

Using a larger resistor (e.g. 180k, or 220k if that produces an in-range
result) would reduce the inaccuracy further.
 
B

Bob Engelhardt

Jan 1, 1970
0
Nobody said:
The main problem with this approach is that any uncertainty (i.e.
tolerance) in the known resistance ...

I think that you missed this part: "... get a resistor in the 100k-200k
range and _measure it_ ..." (emphasis mine). Since the OP is using a
DMM, his uncertainty will be small. Even cheap ones are +-1%.

Bob
 
D

David L. Jones

Jan 1, 1970
0
Nobody said:
The main problem with this approach is that any uncertainty (i.e.
tolerance) in the known resistance creates a much greater degree of
uncertainty in the result.

E.g. if R0 = 100k +/- 5% (i.e. 95k-105k) and the parallel combo is 90k,
the unknown is between:

(95k * 90k) / (95k - 90k) = 1710k
(105k * 90k) / (105k - 90k) = 630k

This is so inaccurate as to be practically useless.

That's why you MEASURE it first, as Bob said.
I think you missed that point entirely.
Any cheap-o DMM will be good to around .5% or so.

Dave.
 
N

Nobody

Jan 1, 1970
0
I think that you missed this part: "... get a resistor in the 100k-200k
range and _measure it_ ..." (emphasis mine).

Er, yeah, I did miss it.

The error "magnification" will still occur, but that's not going to be a
problem with DMM levels of accuracy.
 
F

Fred Abse

Jan 1, 1970
0
The older Simpson-type VOMs would source over 100 mA on their lowest
ohms range, and could fry germanium transistors and diodes. They
applied 1.5 volts behind 10 ohms or so. That would dump around 100
milliwatts into an external 10 ohm resistor,.

There were more tunnel diodes fried that way than ever actually failed in
service.
 
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