Brymen, difference in reading between mA and uA scale - Page 1
Forum section: electronics / testgear

author: MarioBros69
author_rank: Contributor
timestamp: August 13, 2023, 08:15:55 am
content: Hi,These days I have used a Brymen BM869s, an impressively robust multimeterBut I have found the following, I do not know if it is my ignorance or a problem with my unitI put it on the mA scale and measure the current through a 2k resistor connected to a DC 10v source, this should give 5mA, the measurement on the mA scale is very accurate 4.999 mAThe accuracy stated in the specifications for the mA scale is50,000mA 0.15%+20d 3.3mV/mANow I turn the selector and change to the uA scale (micro amps) and the measurement gives a value of 4670.4 uAWe have a difference of 230 uA between the two scales for the same circuit. If I didn't know the current of the circuit I'm measuring, I wouldn't know which of the two is correct.The precision stated in the specifications for the uA scale is5000.0 0.1%+20d 0.15mV /uAI do a little math and the result is that for the uA scale and for this measurement the error can be plus or minus 6 uA but not 230I am also aware that the internal resistance (Shunt) according to the specifications is 3.3 ohms for the mA scale and 150 for the uA scale.My question is, should the two scales show the same measurement value or is it correct that they appear differently because the internal resistance is different on each scale?Could someone with a fluke or a Uni-t perform this simple test?

author: Performa01
author_rank: Super Contributor
timestamp: August 13, 2023, 08:44:50 am
content: Quote from: MarioBros69 on August 13, 2023, 08:15:55 amOf course the measurements will differ, thats why Daves µCurrent exists - to lower the burden voltage for current measurements.The instrument measurescurrent. You are not using a constant current source but a voltage source together with a resistor instead, both with a tolerance most probably exceeding the one of the multimeter.Ohms law says: I = U/R:10 V / (2000 + 3.3) ohms = 4.99176359 mA10 V / (2000 + 150) ohms = 4.65116279 mASeems like your measurement results are reasonably plausible.

author: Fungus
author_rank: Super Contributor
timestamp: August 13, 2023, 09:04:48 am
content: Quote from: MarioBros69 on August 13, 2023, 08:15:55 amNo.Quote from: MarioBros69 on August 13, 2023, 08:15:55 amYes.That's the entire point of having two different switch positions instead of just autoranging. Each position has an internal resistance that's optimized for the range of currents that will be read there and so have the minimum influence on the circuit being measured.

author: J-R
author_rank: Super Contributor
timestamp: August 13, 2023, 09:55:39 am
content: Side note, double-check your specific DMM yourself and don't assume the manual is perfect. Resistances on my BM869s:A - 0.0276 OhmsmA - 1.758 OhmsuA - 101.543 OhmsAnd my BM789 is 102 Ohms for uA, not 200 Ohms as stated in the manual.

author: BILLPOD
author_rank: Frequent Contributor
timestamp: August 13, 2023, 01:22:45 pm
content: I thought a yA was a Yactoampere

author: Shonky
author_rank: Frequent Contributor
timestamp: August 13, 2023, 02:04:26 pm
content: Quote from: BILLPOD on August 13, 2023, 01:22:45 pmyoctoampOP: just use a lower case u in place of mu for micro. Lower case y = 10^-24 not 10^-6

author: MarioBros69
author_rank: Contributor
timestamp: August 13, 2023, 02:17:13 pm
content: Quote from: Shonky on August 13, 2023, 02:04:26 pmI'm sorry, I have already corrected it.

author: MarioBros69
author_rank: Contributor
timestamp: August 13, 2023, 02:22:47 pm
content: Quote from: Performa01 on August 13, 2023, 08:44:50 amThe truth is that I don't understand this very well..., if the measurements differ, which of the two is correct, the one on the mA scale or the one on the uA scale?How do we know then the current that circulates through the circuit when we are not measuring?

author: sonpul
author_rank: Regular Contributor
timestamp: August 13, 2023, 02:47:07 pm
content: The current that you see on the display is real and truthful.If you set 10V 5mA with a laboratory power supply, then without a resistor you will see exactly 5mA at any position of the mA or uA switch.Carefully! if you make a mistake, you can damage an expensive fuse. Just believe.

author: IanB
author_rank: Super Contributor
timestamp: August 13, 2023, 03:32:41 pm
content: Quote from: Shonky on August 13, 2023, 02:04:26 pmOr even better, use the actual µ character for µA.Quote from: MarioBros69 on August 13, 2023, 02:22:47 pmBoth are correct. The truth is that inserting a current meter into a circuit will affect the current. So such measurements have to be taken with care.QuoteWith difficulty. One possibility is to find an existing resistor in the circuit and measure the voltage drop across it. So in your case, you have a 2 kΩ resistor. Measure the voltage across that and use Ohm's law to find the current. For increased accuracy, measure the resistance of the resistor out of circuit so you have a better estimate of its value.Another possibility is to use a higher current scale on the meter, such as the mA scale or even the A scale. Or use a device like Dave's µCurrent.

author: BeBuLamar
author_rank: Super Contributor
timestamp: August 13, 2023, 03:34:25 pm
content: Quote from: MarioBros69 on August 13, 2023, 02:22:47 pmBoth readings are correct. They indicate the actual current that flow thru the circuit but neither is the current flowing thru the circuit without the meter. The meter causes the current to flow slightly less than if there is no meter.

author: TimFox
author_rank: Super Contributor
timestamp: August 13, 2023, 04:32:15 pm
content: Note that the "burden" resistance, which is the actual resistance between the two ammeter terminals, is higher than the "shunt" resistance, where the voltage due to the applied current is measured.(Not necessarily a large difference, note J-R's measurements above.)In your circuit, the total resistance seen by the voltage source is the sum of your external resistor plus the burden resistance.

author: MarioBros69
author_rank: Contributor
timestamp: August 13, 2023, 05:58:10 pm
content: Quote from: IanB on August 13, 2023, 03:32:41 pmShouldn't the tester compensate the measurement based on its internal resistance to show the actual current flowing through the circuit?Something like..., if my internal resistance on the uA scale is 100 ohms I add 200 uA to give the actual current going through the circuit when it is not being measured.

author: wasedadoc
author_rank: Super Contributor
timestamp: August 13, 2023, 06:40:08 pm
content: Quote from: MarioBros69 on August 13, 2023, 05:58:10 pmIt is impossible for a meter to calculate the required correction because the size of the "correction" depends on the impedance and voltage of the source of the current. For example both a 10 Volt source with 10k resistor and a 1 Volt source with a 1k resistor would produce 1 mA current when the meter is not in circuit. Inserting a meter with 100 Ohm shunt has a different and larger effect in the 1V and 1k case. The meter cannot know which of the two cases or any of the infinite number of other possibilities (X Volts and X Ohms) to compensate for.

author: MarioBros69
author_rank: Contributor
timestamp: August 13, 2023, 07:58:55 pm
content: QuoteThanks for the explanation, now I understand.What is paradoxical is that it is the expensive meters that come with the uA scale and we find that the uA scale of a meter is generally less accurate than the mA scale since its internal resistance is usually higher.

author: TimFox
author_rank: Super Contributor
timestamp: August 13, 2023, 08:42:26 pm
content: You still don’t understand.In a battery-resistor-meter circuit, the burden resistance of the meter reduces the actual current into the meter.What you measure is what is actually flowing, not what would flow into a short circuit.

author: MarioBros69
author_rank: Contributor
timestamp: August 13, 2023, 09:57:23 pm
content: Quote from: TimFox on August 13, 2023, 08:42:26 pmYes, I think I understand it, when making the measurement I have added a resistor to the circuit (which is the internal resistance of the meter) therefore the measured current is lower than the real current

author: TimFox
author_rank: Super Contributor
timestamp: August 13, 2023, 10:18:35 pm
content: What do you mean by "real current"?Your meter is reading theactualcurrent through its terminals, which is lower than that which would flow into a short circuit, since your meter is not a short circuit.

author: siealex
author_rank: Regular Contributor
timestamp: August 13, 2023, 11:18:44 pm
content: You haven't seen a cheapo DT-830B (or another 7106-meter) with 1 kΩ input resistance on the 200 μA range.

author: J-R
author_rank: Super Contributor
timestamp: August 14, 2023, 01:00:17 am
content: To be clear, typical bench and handheld DMMs don't technically measure current directly, they measure the voltage drop across a known precise resistance (shunt) and calculate the current using Ohm's law. The reason the burden voltage is a "problem" is because these DMMs have a limit to how low of a voltage they can detect. Since V=IR, the resistance must be made larger if the current is smaller in order to maintain a voltage range that can be detected/measured. The DMM manufacturers make a design choice for each current range depending on the capability of the DMM. That is why lower-end DMMs typically have a higher burden voltage. (The 121GW uses an amplifier in some ranges similar to what the uCurrent does in order to reduce the burden voltage as much as possible.)The absolute minimum voltage the BM869s can detect is 0.001mV (500,000 count mode). The lowest current it can measure is 0.001mA and 0.01uA. Based on the shunt values I provided earlier, we can see that 0.001mA results in a value of 0.001758mV and 0.01uA results in a value of 0.00101543mV. So both of these are clearly chosen to line up with the best the BM869s can do, 0.001mV. (10A shunt calculates to 0.00276mV.)Another way to measure current is with a Hall Effect sensor found in DC clamp meters, but most are going to bottom out in the low mA area. Forget uA or nA.Speaking of mA/uA/nA, I'll maintain there is no huge reason to need the uCurrent or similar since you can simply use your own known resistor as the shunt. For example, the uCurrent uses a 10k resistor for the nA range, so assuming you need to measure 1nA, the voltage drop across your shunt would be 0.01mV which the BM869s can manage well enough. It becomes trivial at say 50nA, which is 0.5mV.

author: Monkeh
author_rank: Super Contributor
timestamp: August 14, 2023, 01:19:06 am
content: Quote from: MarioBros69 on August 13, 2023, 09:57:23 pmNo, the measured current is the real current. The current without the additional resistance in circuit is different.

author: EEVblog
author_rank: Unknown
timestamp: August 14, 2023, 05:34:54 am
content: Quote from: MarioBros69 on August 13, 2023, 02:22:47 pmBoth are correct.The act of putting your current meter in series changes (in this specific case) the currect because the meter has a shunt resistor in it.QuoteYou always have to know what you meter is doing to your circuit and whether or not it is going to change the circuit current or voltage.

author: MarioBros69
author_rank: Contributor
timestamp: August 14, 2023, 08:46:50 am
content: Quote from: TimFox on August 13, 2023, 10:18:35 pmMaybe I expressed myself wrong, by real current I meant the current that flows through the circuit when I am not measuring

author: MarioBros69
author_rank: Contributor
timestamp: August 14, 2023, 08:48:46 am
content: Quote from: J-R on August 14, 2023, 01:00:17 amIn your previous post I see that you have both the BM869s and the BM789 according to your point of view, which of the two is more accurate when measuring current?Thank you for this excellent clarification

author: Fungus
author_rank: Super Contributor
timestamp: August 14, 2023, 10:06:17 am
content: Quote from: MarioBros69 on August 14, 2023, 08:48:46 amDo you mean which has the lowest burden voltage?It's specified in the manuals/datasheets of those meters.(One of them is better on the uA range and the other is better on mA and A - you can't win!)

author: J-R
author_rank: Super Contributor
timestamp: August 14, 2023, 10:20:00 am
content: Oh boy, BM869s vs. BM789...Brymen claims better accuracy from the BM789, 0.075% + 20d from lower ranges, vs. the BM869s at 0.15% + 20d. In my testing, I found them to be virtually identical. For example, a quick test at 50mA, the BM789 was 6 counts high, while the BM869s was 4 counts low.The higher number of nominal counts with the BM789 (approx. 66,000) vs. the BM869s (approx. 53,000) seems like an interesting difference, and the calibration point for the BM789 is mid-scale (so 3V for the 6V range, for example) vs. top of the scale for the BM869s (5V for the 5V range).In my testing of voltage, I found that the BM869s had some error mid-scale, mostly with 500,000 count mode (for example, around 2.5V), and the BM789 had some error at the top end of the scale (5V). This is somewhat to be expected as each DMM has to extrapolate perhaps the proper values from a single calibration point (plus zero). So there is not a clear winner unless you pick a specific calibration point, such as 5V. In that case, you could adjust the BM869s to be spot-on at 5V, while the BM789 would have a hard time keeping up since the nearest calibration point is 3V.If you are looking for high accuracy, the solution is probably a bench DMM...

author: J-R
author_rank: Super Contributor
timestamp: August 14, 2023, 10:31:23 am
content: Shunt measurements on my BM789:A - 0.0212 OhmsmA - 2.474 OhmsuA - 102.5 OhmsAnd for easy comparisons, my BM869s:A - 0.0276 OhmsmA - 1.758 OhmsuA - 101.543 OhmsWould be nice if some other forum users could measure their BM789/BM869s and post their results. I wonder if they changed the specs or if the manual is truly just completely wrong...

author: Fungus
author_rank: Super Contributor
timestamp: August 14, 2023, 10:34:16 am
content: Quote from: J-R on August 14, 2023, 10:20:00 amThe 789 has higher burden voltage so it can give more accurate readings (there's a bigger voltage to measure).

author: EEVblog
author_rank: Unknown
timestamp: August 14, 2023, 10:55:04 am
content: Quote from: MarioBros69 on August 14, 2023, 08:46:50 amIf it's a large current mA to A then you can use a clamp meter to measure current without disturbing your circuit. But they aren't very accurate like what you can get with a proper current meter.As I said, adding a resistor (shunt resistor it's called in a current meter) in series with the circuit you are measuring may change the current in that circuit. It's up to the user to not only be aware of that, but also know if it will impact your circuit or not and by how much.My Silicon Chip article:https://alternatezone.com/electronics/ucurrent/uCurrentArticle.pdf

author: J-R
author_rank: Super Contributor
timestamp: August 14, 2023, 07:14:06 pm
content: I don't see enough of a difference between the shunts used in the BM869s and the BM789 to outright state the BM789 is more accurate, and as I mentioned in my testing I did not see a visible indication that the BM789 was any better than the BM869s. If the uA shunt was truly 200 Ohms like the manual states, then that would be a bigger difference compared to 100 Ohms.

author: Kim Christensen
author_rank: Super Contributor
timestamp: August 14, 2023, 07:28:45 pm
content: Many people have already explained it, but this effect is true for measuring anything in the real world. It's calledthe observer effect.

author: EEVblog
author_rank: Unknown
timestamp: August 14, 2023, 11:35:45 pm
content: Quote from: J-R on August 14, 2023, 07:14:06 pmMine are both identical:

author: Fungus
author_rank: Super Contributor
timestamp: August 15, 2023, 05:55:25 am
content: I just realized my Brymen BM857s has better current measurement specs than the BM869s.(it's probably no better in practice, but...)

author: LazyJack
author_rank: Frequent Contributor
timestamp: August 16, 2023, 08:58:05 pm
content: Quote from: MarioBros69 on August 14, 2023, 08:46:50 amYou still don't understand. The meter is part of the circuit. If you are not measuring and substitute the meter with a short then it is a different circuit with different current. Please draw the circuit for yourself and using Ohm's law calculate the current for the different scenarios.

author: sonpul
author_rank: Regular Contributor
timestamp: August 17, 2023, 05:25:16 am
content: I think he understands this. He doesn't understand why you can't make relative measurements like the REL button. Subtracting the resistance of probes, input capacitance, etc. It looks like he would like to automatically offset the display using Ohm's law.

author: wasedadoc
author_rank: Super Contributor
timestamp: August 17, 2023, 02:49:34 pm
content: The effect of the shunt is the current equivalent of the Ohms/Volt loading effect that one needed to be mindful of when using an analogue multimeter to measure voltage. Depending on the resistance of the voltage source the current through the meter (needed to move the needle) causes a voltage drop. Now that DMMs routinely have 10 MegOhm input resitance the efect is less of an issue but it is still something to be aware of.

author: BeBuLamar
author_rank: Super Contributor
timestamp: August 17, 2023, 03:19:57 pm
content: Quote from: wasedadoc on August 17, 2023, 02:49:34 pmThe 10 Megaohm input resistance is irrelevance in the case of measuring current. It's the shunt resistor that matters.

author: wasedadoc
author_rank: Super Contributor
timestamp: August 17, 2023, 04:37:56 pm
content: Quote from: BeBuLamar on August 17, 2023, 03:19:57 pm@BeBuLamar. Try reading what I wrote more carefully.

author: MarioBros69
author_rank: Contributor
timestamp: August 17, 2023, 05:57:47 pm
content: Quote from: wasedadoc on August 17, 2023, 04:37:56 pmIn the video that appears a few comments above, what does the calibration device do to compensate for the internal resistance of the Brymen?When the Brymen is connected to the device, its internal resistance is also being put in series with the calibration device.

author: IanB
author_rank: Super Contributor
timestamp: August 17, 2023, 06:01:43 pm
content: Quote from: MarioBros69 on August 17, 2023, 05:57:47 pmThe calibration device produces a known fixed current using a regulator.

author: TimFox
author_rank: Super Contributor
timestamp: August 17, 2023, 06:15:40 pm
content: If you don't have a calibrated constant-current source, you can always connect a calibrated ammeter in series with the ammeter being tested and supply DC through a resistor to that series connection.

author: EEVblog
author_rank: Unknown
timestamp: August 17, 2023, 11:53:05 pm
content: Quote from: MarioBros69 on August 17, 2023, 05:57:47 pmIt's called a Contant Current Source. It doesn't matter what the load resisatnce is, it will push a constant current into that load, even a short circuit. The only limit is the maximum compliance voltage.