# Measuring mA (debate solution needed)

Discussion in 'Electronic Design' started by James Lerch, Aug 4, 2007.

1. ### James LerchGuest

So,

A friend and I are having a polite debate.

He's calculating the current in a circuit my measuring the voltage
drop across a 20ohm resistor. (Current is ~30mA range for refernce)

I think best approach is to put a mA meter in series with circuit and
read the meter. He disagrees and said his method of measuring the
voltage drop across the resistor is more accurate.

His claim is the mA meter adds resistance to the circuit and under
reports the results. I agree in theory with his statement, but
disagree on the magnitude of the error.

I claim the amount of error in both the ability to measure 0.6vdc and
the error in the certaitnty of his "known" resistance is of a larger
concern.

For the record, were just simple tinkerers, using simple (but not
junk) tools, working remotetly from each other. No one's life or
livelyhood depend on the results.

If a little background will help, here's the circuit.

5vdc -> led -> resistor -> led -> pin on shift register -> gnd

Leds drop ~2v each (rated at max of 30mA current)
Resistor is 1/4 watt carbon film 20ohm +/- 5% variety
Shift register on resistance is rated at 6.5ohm at 50ma at temp of 25c

I agree with him that he should use a 33ohm resistor for safety sake,
but would like debate resolution over using a mA meter

So, what's the collective opinion on this?
--
Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction,testing, and coating site)
http://lerch.no-ip.com/ChangFa_Gen (My 15KW generator project)

"Anything that can happen, will happen" -Stephen Pollock from:
"Particle Physics for Non-Physicists: A Tour of the Microcosmos"

" Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent. "

2. ### James LerchGuest

The 20ohm resistor is In circuit,

5vdc -> led -> resistor -> led -> pin on shift register -> gnd

Leds drop ~2v each, spec sheet below, these are LY T776 S2 leds
http://catalog.osram-os.com/catalogue/catalogue.do?act=showBookmark&favOid=000000000002c54b0003003a

Resistor is 20ohm

Shift register drain on-resistance is rated at 6.5 ohms @ 50ma and
25c.

Asides from suggesting he measures with DVM on mA scale, also suggest
he goes to 33ohm resistor as he's right on the edge of max current for
the Osram Leds. The is for a christmas decoration display panel with
approx 4k leds on it Would suck for him if he ran it one night and
the next night half the leds were dead or substantially dimmer than
before..

--
Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction,testing, and coating site)
http://lerch.no-ip.com/ChangFa_Gen (My 15KW generator project)

"Anything that can happen, will happen" -Stephen Pollock from:
"Particle Physics for Non-Physicists: A Tour of the Microcosmos"

" Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent. "

3. ### RST Engineering \(jw\)Guest

Check the specs on the multimeter. SOme of the cheapies from China drop
half a volt right off the crack of the bat, no matter the current scale.
Some drop voltage as a function of current.

Jim

4. ### Guest

The 20 ohm resistor is MUCH more than a current meter will put in the
circuit. Is the 20- ohm already there, or is he inserting it?

5. ### Spehro PefhanyGuest

Measuring the voltage across an existing resistor of known value is
generally more accurate, provided your voltmeter does not shunt the
resistor too much (figure 10M for a modern cheap multimeter on a 2V
range and virtually infinity on a 200mV range, neither of which will
affect a 20 ohm resistor much. You can also measure the resistor and
get closer if the value is not known. Chances are the mA meter will
insert a resistance of 1 ohm on the 200mA scale, which could affect
the current a bit. With 2 LEDs in series, depending on type, you
probably don't have enough voltage to get a well controlled current
from 5V.

Best regards,
Spehro Pefhany

6. ### GarethGuest

Understanding measurement errors can get quite complicated, and the fact
that you understand that there are errors rather than just writing down
the number on the meter is a good start.

If it is a digital meter it will probably measure current by measuring
the voltage across a small, known resistance, so the accuracy of the
meter when measuring a small voltage is probably not an issue. On the
other hand, if it is a simple analog moving coil meter the meter
actually responds to current and the voltage has to be converted to
current internally (e.g. by a series resistor).

As you rightly say, both methods will have errors, and if you really
want to know which is best you will probably have to do some math.

For the voltage across the resistor you have:

1) Error in the resistor value - probably quite significant if you just
go be the nominal value. Lower if you measure it, but there will be
some error in this measurement too.

2) Current flowing through the meter - when you put the meter across the
resistor you will have the resistance of the meter in parallel with the
resistor which will reduce the total resistance. In this case I would
expect this to be negligible unless you have a very poor meter.

3) Voltage measurement accuracy of the meter.

For the direct current measurement you have:

1) Change in resistance due to series resistance of the meter

2) Current measurement accuracy of the meter.

If you have the specifications for the meter you should be able to find
its accuracy and series resistance and work out which sources of error
are the most significant.

Personally, I would probably measure the current across the resistor
because it is easier than breaking the circuit and inserting the current
meter.

--

7. ### ChuckGuest

voltage, of course intended

across the resistor
With one DMM doing a series current
measurement, you can use a second DMM to
measure the voltage dropped across the
first one and calculate the actual
series resistance the first one
introduces. Might be interesting.

Chuck

8. ### Tom BruhnsGuest

Given that the resistor is already in the circuit, so long as you know
its resistance accurately, that will be the better way. I'm assuming
a volt meter that has something like a 10 megohm input resistance,
common in DVMs. If you insert a meter, you'll disturb things by
whatever the meter resistance is. Say it's 0.1 ohms. That's 0.5% of
the total of 20 ohms plus the meter. On the other hand, 10 megohms in
parallel with 20 ohms disturbs it by about .0002%.

For current in an LED, neither matters a whit. But measuring the
voltage across a resistor is almost always trivial compared with
inserting a meter in series with an existing circuit.

How accurately you know the value of the 20 ohm resistor would be an
issue, along with how stable it is with self-heating and stuff like
that, but again, measuring current in an LED really doesn't require
extreme accuracy. The max current rating of the LED is for rated
life; it's not going to die in an hour if you go 1% over. You can
measure the value of the resistor with the power turned off in that
particular circuit; if the LED is reverse biased by less than a volt
by the meter, the reading should be quite accurate.

I've never met a DC current meter that drops half a volt at
(practically) no current, like Jim mentioned, and hope I never do.
Even my cheapies don't do that. On AC, yeah, quite possibly, but on
DC???

Cheers,
Tom

9. ### Phil AllisonGuest

"James Lerch"

** Time you measured the internal resistance of a DMM when set to the 200
/ 300 mA range.

A typical 3.75 digit meter I have here exhibits a whooping 5.5 ohms !!!

The voltage drop at full reading = 1.8 volts !!

Most of it is due to the 500mA glass fuse employed on that range.

Your friend is right & you are 100% wrong.

........ Phil

10. ### EeyoreGuest

You mean a moving coil meter ? They are inherently inaccurate (subject to modest
calibration, linearity and measurement errors due to parallax).

Graham

11. ### BenjGuest

Phil is 100% RIGHT ON!

Since you trying to measure LED currents in a given circuit, you
really DON'T want to have the measurement changing the circuit values.
(Especially in this case where LED values are probably not well known
and perhaps flaky from unit to unit as well.)

The big mistake is assuming that meters don't change the circuit. As
you see above that isn't true. Of course they ALSO change a voltage
impedance. But here 10 meg vs 20 Ohms won't be seen.
Plus, the fact that the 20 Ohms is already in the circuit makes it an
easy way to go. In this case it'd be wise to measure a whole bunch of
these dual LEDs in the "on" state and then set the value of the 20
Ohms based on the maximum current you observed. You want to keep the
max current somewhere below 30ma. This is why you don't want to insert
extra resistance from the meter that will change the observed current.
Even more than you might think given that LEDs are non-linear
devices! My feeling would be that resistor accuracy could be an issue
(you could temporarily put 1% Rs in the circuit) but it probably isn't
important in that the LEDs will be rather variable and the idea is to
just keep the current below max values. If you set R to give the
highest current 5% lower than 30 ma, then that also takes care of the
resistor accuracy problem. Hell, make it 10% lower just to be on the
safe side. As someone already noted the current regulation in this
circuit is less than optimum already. An open collector from a higher
voltage would be better.

Next time listen to your friend!

12. ### neon

1,325
0
Oct 21, 2006
Both of you are wrong. there is not a specified +-% on the resistor i it is a 1% tollerance then the error is +- !%. most meters [analog] are 50 microamps movement and here is a shunt to read anything above that. there is meter % error full deflection. I do't see how putting an additional shunt cross the meter and reading the votage drop is any difference from reading the meter shunt diectly.

13. ### James LerchGuest

I hate being wrong, but I'd rather be wrong and learn something that
wrong and think I'm right!

Thanks Gents!

(off to develop a circuit to see just how much my mA meter changes
things)

--
Take Care,
James Lerch
http://lerch.no-ip.com/atm (My telescope construction,testing, and coating site)
http://lerch.no-ip.com/ChangFa_Gen (My 15KW generator project)

"Anything that can happen, will happen" -Stephen Pollock from:
"Particle Physics for Non-Physicists: A Tour of the Microcosmos"

" Press on: nothing in the world can take the place of perseverance.
Talent will not; nothing is more common than unsuccessful men with talent.
Genius will not; unrewarded genius is almost a proverb.
Education will not; the world is full of educated derelicts.
Persistence and determination alone are omnipotent. "

14. ### Rich GriseGuest

.
You could definitively solve this once and for all by using two meters -
interpose the ammeter, then use a voltmeter to measure both the drop
across the meter and the drop across the load.

Good Luck!
Rich

15. ### JosephKKGuest

Phil Allison posted to
sci.electronics.design:
So what make and model of piece of shit is that so i can avoid

16. ### Phil AllisonGuest

** Fluke 73 or Fluke 77 and many, many others.

Burden voltage is speced as " 6 mV per mA " on the 32mA and 320 mA
ranges.

Equates to a 6 ohm resistance.

Fuckwit.

......... Phil

17. ### JosephKKGuest

Phil Allison posted to
sci.electronics.design:
I threw away an old fluke 77 years ago. It did ok on voltage on
voltage and soso on resistance. Another overpriced brand name
mediocre meter.