### Author Topic: What does the count on a non-autoranging multimeter mean?  (Read 8229 times)

0 Members and 1 Guest are viewing this topic.

#### Zero999

• Super Contributor
• Posts: 12309
• Country:
• Hero999
##### Re: What does the count on a non-autoranging multimeter mean?
« Reply #25 on: September 04, 2016, 05:30:08 am »
So how did you get to 135V DC?

Oh and one more thing - 135V AC is dangerous. But not as dangerous as a rectified 135VDC with a capacitor.

E= 1/2 CV^2 = 0.5 * 0.01 * 135^2 = 91.1256 Joules
No, you've got that backwards.

135VAC is much more dangerous than 135VDC and any capacitor charged to 135VDC is much less dangerous than a constant DC voltage source such as a battery.

Quote
Generally speaking - 50J - is considered enough for possible ventricular fibrillation. (if all of the energy ends up in you or you end up flying to the other side of the room depends on how contact came to happen and body resistance).
The energy is only dangerous if a high enough current can flow for a long enough period to cause ventricular fibrillation, which is much more likely with mains frequency AC than DC. 135VDC will give you a bit of a shock but nowhere near as bad as the 120VAC mains.

#### Brumby

• Supporter
• Posts: 8716
• Country:
##### Re: What does the count on a non-autoranging multimeter mean?
« Reply #26 on: September 04, 2016, 12:17:13 pm »
.... I'm more interested in where the 135VDC came from.

UNLESS it is a reading error on the part of the OP - and it was 13.5VDC.  That would be about right for rectified 10VAC.

#### Assafl

• Frequent Contributor
• Posts: 577
##### Re: What does the count on a non-autoranging multimeter mean?
« Reply #27 on: September 04, 2016, 03:37:24 pm »
So how did you get to 135V DC?

Oh and one more thing - 135V AC is dangerous. But not as dangerous as a rectified 135VDC with a capacitor.

E= 1/2 CV^2 = 0.5 * 0.01 * 135^2 = 91.1256 Joules
No, you've got that backwards.

135VAC is much more dangerous than 135VDC and any capacitor charged to 135VDC is much less dangerous than a constant DC voltage source such as a battery.

Quote
Generally speaking - 50J - is considered enough for possible ventricular fibrillation. (if all of the energy ends up in you or you end up flying to the other side of the room depends on how contact came to happen and body resistance).
The energy is only dangerous if a high enough current can flow for a long enough period to cause ventricular fibrillation, which is much more likely with mains frequency AC than DC. 135VDC will give you a bit of a shock but nowhere near as bad as the 120VAC mains.
I stand corrected...

It seems that the AC/DC question is still a bit undecided (at least I couldn't find a definitive researched article). During the current wars Brown (working with Edison) wanted to prove AC was more dangerous and electrocuted various animals. AC seems to have more ways into the body (e.g. capacitive coupling). And DC vs RMS AC - the peak voltage of AC is higher. On the other hand, sources state that DC causes contractions that make it more difficult to let go. Also, DC causes hydrolysis that causes poisoning. Furthermore, it is stated that AC cause issues with fibrillation neurons, somehow upsetting them and thus preventing the heart from starting (that is why defibrillators are DC and not AC?).

Of course there are millions of anecdotes including death from a 9V battery... (I will state I do not believe death from a 9V battery is possible - unless perhaps swallowing it; not that death while playing around with a 9V battery isn't possible from many reasons - being run over by a car, falling out of a window, sticking the DMM leads into the mains...).

One would think this is a simple question...

#### Assafl

• Frequent Contributor
• Posts: 577
##### Re: What does the count on a non-autoranging multimeter mean?
« Reply #28 on: September 04, 2016, 04:09:26 pm »
It's all marketing BS as far as I'm concerned.

Technically a count is the analogue value represented by a single bit on an ADC and the number of counts is equal to 2^the number of bits. If a meter has a 16-bit ADC and an analogue range of +/-1000V, then it will have 216 = 65536 counts and each count will be just over 30.5mV, irrespective of whether or not the display has adequate digits to display values that small on the 1000V range. Internally, any digital calculations will be performed using counts, rather than Volts or Amps to avoid rounding errors, then it will be converted to the correct units before being displayed.

I very much doubt if there's any calculation involved in a low cost DMM. It's a simple dual slope ADC (probably an ICL[edit:7106] or derivative). After applying the scaled input to the integrator for a fixed period, the display value is simply the number of actual hardware clock counts required to return the integration capacitor voltage to zero. The number of counts the meter has / displays is simply represents the length of the counter chain implemented in the chip..

Probably OT in terms of the OP's current questions anyway.
I am not implying that the above post denigrates the dual-slope architecture - but I believe the ICL7106 architecture deserves a bit of credit... It survived the test of time even better than the NE555 and uA741 did.

Simple or not the 7106 architecture is selected because its accuracy and precision depends solely on very few components (mainly the reference voltage generator). Variations of it are used in the cheapest Chinese \$10 DMM to HPs 3458A 8.5 digit DMM. It is actually pretty cool that to calibrate the 3458A one needs only 2 calibration references - a ne voltage and resistance references (real accurate ones indeed...).

The HP journal shows just how simple the architecture really is - and how with a bit of care and creative design one can factor out most uncertainties: http://www.hpl.hp.com/hpjournal/pdfs/IssuePDFs/1989-04.pdf ...

(The 7106 (if still available) is probably a very cheap chip now. But not long ago - 1980's and 90's it was a fixture in the electronics components lists in Elektor backpage Ads as one of the most expensive chips in the book. Along with the strange GI chips that did video games (pong)...)

#### retiredcaps

• Super Contributor
• Posts: 3257
• Country:
##### Re: What does the count on a non-autoranging multimeter mean?
« Reply #29 on: September 04, 2016, 04:21:01 pm »
(The 7106 (if still available) is probably a very cheap chip now.
The history of how the 7106 came about from someone who was there to see it.

http://www.eevblog.com/forum/testgear/old-fluke-multimeters/

The following users thanked this post: Assafl

#### Gyro

• Super Contributor
• Posts: 4149
• Country:
##### Re: What does the count on a non-autoranging multimeter mean?
« Reply #30 on: September 04, 2016, 06:48:16 pm »
I am not implying that the above post denigrates the dual-slope architecture - but I believe the ICL7106 architecture deserves a bit of credit... It survived the test of time even better than the NE555 and uA741 did.

Indeed a wondrous device (Intersil were good at those) which revolutionised the availability of affordable DMMs to the masses, as well as most discrete panel readouts. Yes it is still cheaply available, as is the LED driving 7107 which is still used in many bench PSUs. Dual slope architecture is still a (probably the) mainstay of voltage measurement, with follow-on triple, and more, slope architectures mainly being introduced to increase measurement rate (with a bit of dielectric absorption correction thrown in).
Chris

"Victor Meldrew, the Crimson Avenger!"

#### PA4TIM

• Super Contributor
• Posts: 1116
• Country:
• instruments are like rabbits, they multiply fast
##### Re: What does the count on a non-autoranging multimeter mean?
« Reply #31 on: September 04, 2016, 08:28:51 pm »
Those 7106 and 7107 are still for sale, ordered one about 5 years ago from Conrad to repair a Fluke 8020 (the lcd version) and a few weeks ago I had to replace a led version in a Megger BT51 milliohm meter. They were also used in "universal" panel displays. I used one of those in my ESR meter and an other in a capacitor reformer/leakage tester to read the voltage (for current  I used an analog null-meter)

For the contender to  the Darwin awards contest, good job, this is the way to go for the first place.  Experimenting with this kind of parts without enough knowledge.
You do not want to be near an exploding 10mF cap. But to be fair a good 65V cap probably can blow at twice the voltage but the risk is not as high as people think. Most caps are formed at twice the voltage during manufacturing. If the voltage gets to high and breaks through the oxide layer it will warm up but most times voltage also drops. If this condition stays to long, the temp can go up and then it could pop but most times it ends up as a dead short. It also can fry the electrodes that connect to the pins and in that case become an open. But there is the possibility it gets so hot, so fast that it can blow up. Do not experiment like you did with this kind of parts.

Feeding it with a reversed voltage is very very very dangerous, above the "zener" level of the oxide layer and with acces to enough current it will explodes within seconds.
www.pa4tim.nl my collection measurement gear and experiments Also lots of info about network analyse
www.schneiderelectronicsrepair.nl  repair of test and calibration equipment