Where to find .5% Metal Film Resistors?

I have a 'digital panel meter' I found in my 'score' of electronics gear, and I want to incorporate it into a cana-kits variable bench power supply kit that was also in the cache of gear.

It is model PM-188BL: https://www.jameco.com/webapp/wcs/stores/servlet/ProductDisplay?langId=-1&storeId=10001&catalogId=10001%3FId%3D-1CID%3DHPCATALOG&productId=146835.

It states that in order to measure up to 200v (choice of 200mV, 20V, 200V, 500V) I would need to add a voltage divider consisting of a 10k and a 9.99M resistor. It also states that they should be 1/2W 0.5% metal film resistors.

To me, that seems awfully precise components for a display that costs about $15.

I've searched a bit, but haven't found anything promising yet :frowning:

Just put two 1% resistors in parallel for an effective tolerance of 0.5%.

OK, just kidding. You can easily find 0.5% resistors but they're most likely going to be surface-mount:

Or just go with regular 1% resistors and accept a little bit more error in the voltage of the power supply.

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That will probably work, however it's just for the display of the voltage, not the generation itself. The Cana kit is adjustable from 0-30vdc and from 2mA to 1.5A. I just want a digital readout for the output it is delivering.

Beware that many of those digital panel meters apparently can NOT measure the same supply used to power them...

Funny, that someone is talking of a 9.99 thing with .5% tolerance..

Ok, yes... Even with as small of a tolerance as .5%, 9.99 and 10 are the same, however the 'manual' specifies a 9.9M to read 20V, a 9.99M to read 200V and a 9.999M to read 500v... If it doesn't matter, why would they write it as they did (translation notwithstanding)?

Even technical writers need some comic relief from time to time....

So a 10M .5% would be fine for upwards up 500V? (only need to 30V).. Think I could scrape by with a 1%?

Assume you have a divider made from 10k+10M with +/-1% tolerance. This would mean
lowest V: 9.91k + 10.1M => 9.91/10.10999 = 0.980 mV/V
highest V: 10.1k + 9.9M => 10.1/9.9101 = 1.019 mV/V

This calculation will unnoticeably change when you use a nominal 9.99M resistor rather than a 10M

In both cases this is noticeable, as you do not want to read 1.02 V when it is 1.00 V. This is why in first class equipment 0.1% resistors are used, prize p. pc $1.

However this is the very worst case, statistically your errors might cancel:
9.91k+9.9M => 9.91/9.90991 = 1,00001 mV/V
10.1 + 10.1Mk => 10.1/10.1101 = 0,99900 mV/V

Welcome to the world of instrumentation. The old rule of thumb was your test equipment should have 10X the accuracy specifications of the device you are trying to validate, test or calibrate.

Lefty

The old rule of thumb was your test equipment should have 10X the accuracy specifications of the device you are trying to validate, test or calibrate.

How do you test/validate/calibrate the test equipment?

How do you test/validate/calibrate the test equipment?

Couple of different ways. Send it out to a calibration service that performs such services ($$$). They will check/adjust the calibration and certify the accuracy for a certain period, usually a year. They will have used traceable and certified standards to perform this service.

Keep a separate voltage or resistance 'standard' at hand that you can compare to the test instrument's reading. Send the 'standard' out for validation to a calibration service if required.

It really depends on the industry one is in and what legel obligations that cover their work or products. It can be an expensive and time consuming process and quite the pain in the ass to stay current at times. Hobbyists have no such requirment but really should check their test equipment from time to time by some means, even if just comparing two meters against each other (however which one is correct?). :wink:

Lefty

(however which one is correct?) simply buy a 3° meter, and 99% of the time it's also different. then use a mid of the three ;D

however the 'manual' specifies a 9.9M to read 20V, a 9.99M to read 200V and a 9.999M to read 500v... If it doesn't matter, why would they write it as they did

Because to write 10.000 instead of 9.999 would imply that the meter display had 5 digits where as in fact it only has 4.