Resistor test question from a newbie

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

Correct.

Nope.

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

Bob

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.

Mike

Yes, that's correct. You are selecting the range of the meter. You may ask why you can't simply use the 20M range for much smaller resistors, such as 100 Ohm. The explanation is that the wider range is achieved at the cost of a coarser resolution. If you measure a 100 Ohm resistor with the DMM set to the 20M range, the meter should have displayed

0.0001, but since it doesn't have that many digits, it would simply show zero, giving you no useable result.

No, it will not. The DMM uses a *very* small current to measure the resistance. This current is so small that no resistor will take damage from it, no matter what range your DMM is at.

One thing to remember when measuring very large resistances: Keep your hand off the measurement points. You body has a resistance in the megaohm range, and may introduce errors in your measurement. You can see this by simply touching both test tips and see how the meter indicates a resistance.

--
RoRo

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,.

John

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

What's that? A very small meter?

John

Yes, and they're easily damaged.

Yes, you set it to the next highest range. Or simply use an auto-ranging digital multimeter and it does it all for you.

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

Dave.

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.

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

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.

Best regards,

Bob Masta DAQARTA v4.51 Data AcQuisition And Real-Time Analysis

Scope, Spectrum, Spectrogram, Sound Level Meter FREE Signal Generator Science with your sound card!

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.

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

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.

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.

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

--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
                                             (Stephen Leacock)