8
\$\begingroup\$

The Keysight 34465A Digital Multimeter is specified as a 6½ digit multimeter.

However, the meter is actually a 1,200,000 counts meter, since the meter wraps at 1.2, so if measuring a 1.19 kOhm resister, the reading is 1.190,000 kOhm, and if measuring a 1.21 kOhm resistor, the reading will be 01.210,00 kOhm.

Usually the ½ in specification of a multimeter means that it wraps as 2.0 (see EEVblog #26), for example a 3½ digit multimeter has 2000 counts, and can do reading up to 1999, so if measuring a 1.9 kOhm resister, the reading is 1.900 kOhm, and if measuring a 1.2 kOhm resistor, the reading will be 02.10 kOhm.

Based on this, I would think the Keysight 34465A Digital Multimeter should be specified as less than 6½ digits, for example log10(1200000) = 6.08 digit multimeter.

So, is a multimeter with 1,200,000 counts actually a 6½ digit multimeter?

\$\endgroup\$

3 Answers 3

Reset to default
14
\$\begingroup\$

The 1.999999 is already a lie. A 6.5 digit meter should have \$10^{6.5}\$ = 3162377 "counts".

So, extending the lie, meters that have 3999 counts are described as 3-3/4 digit, when in fact they're barely 3.5 digit.

Lies, damn lies, and marketing. Maybe a class action suit is in order if you can claim you were hoodwinked by their fake "marketing digits".

To answer your title question, the meter in question is a 6.08 digit multimeter if it has 1,200,000 counts.

\$\endgroup\$
7
  • 1
    \$\begingroup\$ Lies, damn lies and "alternate reality" are syn's. \$\endgroup\$
    – D.A.S.
    Commented Dec 23, 2018 at 19:25
  • 1
    \$\begingroup\$ Or "half digit" isn't a precisely defined concept. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Dec 24, 2018 at 3:12
  • 3
    \$\begingroup\$ A half digit is well-defined in information theory. It is especially useful in arithmetic coding. \$\endgroup\$
    – Nayuki
    Commented Dec 24, 2018 at 7:12
  • \$\begingroup\$ Expressed in another way, a multimeter with 1.2 million counts has 20 bits of precision (similar to a 20-bit ADC on a sound card), or 121.6 dB of dynamic range. \$\endgroup\$
    – Nayuki
    Commented Dec 24, 2018 at 7:13
  • \$\begingroup\$ Heck, even saying it should measure to 5, i.e. 3.5 count as meaning to 5000 (0000-4999) as in a linear interpolation, would make more sense than this scheme. \$\endgroup\$
    – The_Sympathizer
    Commented Dec 24, 2018 at 14:47
5
\$\begingroup\$

I think designing a more than 6 digit meter is a pretty impressive feat.

If you get off on precise numbers, then you're right, sure, go rain on their (publicity department's) parade.

It's interesting that a device intended for extreme precision should fall down in the precision of how it's described.

\$\endgroup\$
2
  • \$\begingroup\$ Do you have a reference for definition of digit in this context, is there a requirement for the span to be specific range (i.e. can the half digit be in the decimal with lower top range)? To me it seems probable that someone took the effective bit resolution of the ADC and publish the effective decimal digits. It is disconcerting that Keysight of all folks does not have their definition in the datasheet. \$\endgroup\$
    – crasic
    Commented Dec 23, 2018 at 18:33
  • 2
    \$\begingroup\$ The only fair definition of digits is \$log_{10}(count)\$ but everyone lies, 1999 count is closer to 3.3 than 3.5 \$\endgroup\$
    – Jasen Слава Україні
    Commented Dec 23, 2018 at 18:56
2
\$\begingroup\$

The additional overrange digit referred to as a 1/2 digit.

The overflow resulting resolution ratio with range and its name is insignificant anyways.

What really counts is accuracy.

That 1/2 digit typically shows only the values 0 or 1.

There may be some exceptions that are designed for +/-2.99 ... such as resistance overflow.

However anyone can take credit for it's definition, that's also insignificant. It is just common wisdom.

I once had a 10 1/2 digit digital counter too but resolution is insignificant usually compared to accuracy.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.