So just received my uCurrent, I was very excited to measure how much my battery powered device really consumes in sleep mode. But before that I decided to test it on resistors. Here is my testing setup:
1AA battery half full, happened to be about 1.064V great, makes calculations easier and set of 100k, 1M, 10M resistors.
I created the test circuit and measured the current with my multimeter on uA range. I got expected results.
Then I measured it using the uCurrent on uA range, got about the same results, so far so good.
Then I switched to the nA range and when I measured current through 1M resistor I got about 770 nA, but it should be around 1000nA, when I measured the current through 10M resistor I got about 33nA, should be 100nA which is waaaaaaaay outside the specs.
Is this expected behavior? Did I get a faulty unit? Is my methodology faulty? There was some oily liquid coming from the switches on uCurrent, I'm not talking about flux on the other side of the board, I'm talking about the tiny gaps in the switches, some liquid is oozing from it. Could this interfere with my measurements? Can you advice me how to trouble shoot this?
What about your cables and connectors? are they adequate and clean enough?
it is very easy to loose some nanoamps along the way
You should check with the designer of the uCurrent device
You should check with the designer of the uCurrent device
hmmm... i wonder if he is a member of this forum..?
Apparently the OP's unit is not hunky-dory
You could try switching your DMM to read AC volts to see if something is oscillating or if you're picking up a lot of noise because of the high impedance of your test circuit. It should be close to zero if everything is fine.
What?
Battery is FLAT, ANY current draw will cause the battery volage to collapse.
Try again with a decent battery.
What?
Battery is FLAT, ANY current draw will cause the battery volage to collapse.
Obviously not with 1 Megohm load.
What?
Battery is FLAT, ANY current draw will cause the battery volage to collapse.
Try again with a decent battery.
I agree it's probably sensible to try with a better voltage source, but if that's the cause wouldn't you expect it to be better at 10M, not worse?
maybe you are running into the issue of dissimilar metals in your wiring creating a voltage and upsetting the current?
Did you measure the Voltage directly across the input of the µCurrent?
The internal shunt is 10k and should result a voltage drop of 1.0mV, if the real current is 100nA; otherwise 0.330mA if the current is in fact 33nA.
EDIT: I'm talking rubbish: what I said here does'n make sence
2nd try: Did you measure the nA shunt resistor? It should be 10k.
Did you add any extra multimeter to the µcurrent device to check it? (and forget its influence?)
If the battery is a simple dry type (not alkaline), than the battery can well be the problem - it's really dead then. With an alkaline battery down at 1 V you may still draw 100 mA, so not a problem.
People think measuring nA is trivial, it is not, many things can influence the measurement.
Have you measured the 10k sense resistor on the nA range (with switch on another range), I'd be incredibly surprised if it's anything other than exactly 10k.
A battery is a poor voltage source, particually afer being loaded, as the battery voltage will rise back up slowly after being loaded.
The description of the measurement circuitry is mediocre!
Maybe the TO may offer a schematic or better: pictures for the different setups?
Anyhow, if I assume right, and if the TO first has put the 1V in series with 100k, that would produce 10µA.
If the DMM then read about 10µA, and so did the µCurrent adapter, the battery would not be the root cause:
If 1M and 10M were put in series, then the 1µA and 100nA currents could not flatten the battery more, than a 10µ had done!
Therefore, something else went wrong, maybe the TO added something in the current loop, which has additional resistance on the order of 300k for the 770nA, and 20M for the 33nA
Frank
Obviously not with 1 Megohm load.
You would be surprised.
That voltage would drop by 70% to cause what OP sees?
I think the OP is introducing a lot of noise into the uCurrent and probably hitting one or both rails on the opamp. It's clipping the noise.
That's why I wanted to see an AC voltage reading.
That voltage would drop by 70% to cause what OP sees?
While battery cells may be thought of as "voltage sources," they do in reality have finite source resistances. One particular way to model the issues involved, is to consider the fact that
1. current is the flow of charge per unit time (dq/dt).
2. with a perfect zero-flow, a static quantity of charge can produce a particular EMF
3. If you measure static EMF and then work to flow an available charge, some amount of work is dissipated as heat.
So suppose the battery is practically discharged, yet still reads 1.000V some how. If there is only a means for about 500nC/s of charge to flow, the presence of 1Mohm will get you a change in voltage reading down to 0.5V. Incidentally, this infers that the source impedance is on the order of 1Mohm anyway. Then consider the fact that the means of transmission is through dissimilar metals in the drycell, but as the thing ages, the electrodes get consumed. What might have been 1m^2 of contact surface may now be only 1mm^2, this too would be a form of source-resistance increase. There are many mechanisms which make a used battery a piss poor choice of standard.
Another way to think about this, is someone buying a 9V battery and sticking it on a $4k Keithley 7.5 digit DMM and bitching that the Keithley is a POS because it measures the battery at 9.2105V and not 9.0000000V.
That voltage would drop by 70% to cause what OP sees?
While battery cells may be thought of as "voltage sources," they do in reality have finite source resistances. One particular way to model the issues involved, is to consider the fact that
1. current is the flow of charge per unit time (dq/dt).
2. with a perfect zero-flow, a static quantity of charge can produce a particular EMF
3. If you measure static EMF and then work to flow an available charge, some amount of work is dissipated as heat.
So suppose the battery is practically discharged, yet still reads 1.000V some how. If there is only a means for about 500nC/s of charge to flow, the presence of 1Mohm will get you a change in voltage reading down to 0.5V. Incidentally, this infers that the source impedance is on the order of 1Mohm anyway. Then consider the fact that the means of transmission is through dissimilar metals in the drycell, but as the thing ages, the electrodes get consumed. What might have been 1m^2 of contact surface may now be only 1mm^2, this too would be a form of source-resistance increase. There are many mechanisms which make a used battery a piss poor choice of standard.
Another way to think about this, is someone buying a 9V battery and sticking it on a $4k Keithley 7.5 digit DMM and bitching that the Keithley is a POS because it measures the battery at 9.2105V and not 9.0000000V.
A lot of "smart" words. How this is relevant to this particular case? Please show me a battery which increases it's internal resistance by an order of magnitude with increasing the load resistance. But anyway please show me a real battery which shows 1V on multimeter (with its internal resistance) and then voltage sags by 30% under 1M load and by 70% under 10M load. I've seen batteries with very high Internal resistance BTW.
So please read again and explain this behavior with internal resistance of the battery:
1AA battery half full, happened to be about 1.064V great, makes calculations easier and set of 100k, 1M, 10M resistors.
I created the test circuit and measured the current with my multimeter on uA range. I got expected results.
Then I measured it using the uCurrent on uA range, got about the same results, so far so good.
Then I switched to the nA range and when I measured current through 1M resistor I got about 770 nA, but it should be around 1000nA, when I measured the current through 10M resistor I got about 33nA
I haven't seen anyone mention it but.... why not wire the DMM or whichever ammeter you were using in series with the uCurrent?
That way, they both measure the current at the same time. Ruling out differences between burden or other residual voltages that can affect the overall reading.
I haven't seen anyone mention it but.... why not wire the DMM or whichever ammeter you were using in series with the uCurrent?
Most DMMs are not accurate to <.1 µA
A lot of "smart" words. How this is relevant to this particular case? Please show me a battery which increases it's internal resistance by an order of magnitude with increasing the load resistance. But anyway please show me a real battery which shows 1V on multimeter (with its internal resistance) and then voltage sags by 30% under 1M load and by 70% under 10M load. I've seen batteries with very high Internal resistance BTW.
So please read again and explain this behavior with internal resistance of the battery:
1AA battery half full, happened to be about 1.064V great, makes calculations easier and set of 100k, 1M, 10M resistors.
I created the test circuit and measured the current with my multimeter on uA range. I got expected results.
Then I measured it using the uCurrent on uA range, got about the same results, so far so good.
Then I switched to the nA range and when I measured current through 1M resistor I got about 770 nA, but it should be around 1000nA, when I measured the current through 10M resistor I got about 33nA
If "smart words" are a bit much for you, then how about something remedial:
The amount of charge left on the battery at his first test is unknown. Apparently you can tell us. Please do.
The amount of charge left after his second test is now different and still unknown, again, you can tell us.
Once we're on to the third test, what's left on a dead battery is still unknown and is different still. If the source impedance is high, after a
dead battery has given all it can, then it is perfectly reasonable that even 1Mohm will show up a 30% sag. So for all of your "please show me," tell you what, send me the battery and I'll do the measurements. You were the first to make an assertion, the onus is on you to back it. With your response, maybe you can show us all a perfectly linear battery which happens to never change it's source impedance as it dies. If not, sod off.
If "smart words" are a bit much for you, then how about something remedial:
The amount of charge left on the battery at his first test is unknown. Apparently you can tell us. Please do.
The amount of charge left after his second test is now different and still unknown, again, you can tell us.
Once we're on to the third test, what's left on a dead battery is still unknown and is different still. If the source impedance is high, after a dead battery has given all it can, then it is perfectly reasonable that even 1Mohm will show up a 30% sag. So for all of your "please show me," tell you what, send me the battery and I'll do the measurements. You were the first to make an assertion, the onus is on you to back it. With your response, maybe you can show us all a perfectly linear battery which happens to never change it's source impedance as it dies. If not, sod off.
Yeah, now take a battery with high internal resistance. Then you will notice than you won't be able to discharge it fast because of high internal resistance and very soon voltage will return to the previous level. Moreover something like 1M internal resistance won't happen with AA batteries. Though it can be present on very old lithium batteries like CR2032, they can hold most of the capacity at the same time regardless of this.