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I have a circuit with a very-low-power Jennic JN5148 module with microcontroller and 2.4 GHz radio, and some low-power sensors.

I have to measure the supply current of all these components, in an interval of around a second and with a resolution of about 100 uA. These currents may have a maximum value of about 30 mA for the Jennic module, and slightly under 1 mA for the other components.

I should measure these currents simultaneously and at a frequency of about 10 ksample/s, and i need at least 4 channels.

The other requirements are to use as more as possible instruments over building amplifiers and so, and to perturb at least as possible the supply of components. Actually, the requirement is NO COMPONENTS and ONLY INSTRUMENTS.

Does anybody has an idea about the best fitting solution? (i think that i've explained all but tell me if it lacks something)

EDIT: Found this that could be a solution, but can you help me understanding what is the perturbation that it adds to the circuit?

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  • \$\begingroup\$ Even if I don't have a solution I missed the specification your desired accuracy (which is in a lot of cases different from the resolution, think for example about noise). \$\endgroup\$ – 0x6d64 Jan 4 '12 at 15:49
  • \$\begingroup\$ Well i think that in this case the answer is the same because i don't worry so much about noise, and I just need to detect if the current absorbtion (i hope it's written right) is in the normal range or not; so i need only a rough (with respect with the small value) measure of it. \$\endgroup\$ – clabacchio Jan 4 '12 at 15:54
  • \$\begingroup\$ Similar question electronics.stackexchange.com/questions/14879/… \$\endgroup\$ – Toby Jaffey Feb 22 '12 at 13:20
  • \$\begingroup\$ But it's not a new question, I've just updated it and my requirements were different as I needed to measure the power during the operation, not just the total energy. \$\endgroup\$ – clabacchio Feb 22 '12 at 13:27
  • \$\begingroup\$ @clabacchio Indeed, I wasn't suggesting this is a duplicate, just that you might find some useful info in the other question's answers \$\endgroup\$ – Toby Jaffey Feb 22 '12 at 13:48
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So you need to measure supply current at 10 ksamp/s from 100µA to 30mA, which is 300:1 range.

That by itself sounds doable enough. Even a 10 bit A/D built into a microcontroller is enough resolution if the signal is amplified properly. 10 kHz sample rate is also quite doable. In fact, I'd want to sample faster than that and do a little low pass filtering and decimation in the micro. 100 kHz sample rate isn't even pushing it for something like a PIC 24H. At 40 MIPS that would leave 400 instructions/sample. That is much more than needed for a little low pass filtering and background bookeeping, so that checks out fine too.

The real question is what does the power feed look like and to what extent can you break into it. Are the units under test powered with LDOs? That would be useful, since a small current sense resitor before the LDO wouldn't effect the unit under test power voltage at all. You'd have to subtract off the LDO current, but that is doable. By putting the current sense before the LDO, you can afford to have it drop a little more voltage since the LDO will make sure the UUT still sees the same supply voltage. This of course assumes there is enough input voltage headroom to play with.

If you have to put the current sense directly in line with the UUT, then you have to carefully consider voltage drop versus sensitivity and therefore ultimately signal to noise ratio. Maybe 1Ω is reasonable. That would only drop 30mV max, which wouldn't effect most devices much at all. You'd need a differential amplifier and a overall gain of 100 so that 0-30mA results in 0-3.0V, which is just about the right target for a processor running at 3.3V. Various folks make such diff amps or specifically high side current sense amps. If this is a one off, I'd start with Analog Devices. A 10x diff amp with 1 MHz gain-bandwidth shouldn't be hard to find. That would need to be follwed by a ordinary 10x amp before the micro, again with 1 MHz gain-bandwidth being adequate. You could try doing the whole thing with a single 100x diff amp, but the gain-bandwidth product should be at least 10 MHz so the choices will be more limited.

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  • \$\begingroup\$ Very complete and precise answer, thank you, but by now i'm required to use a solution that involves as least as possible circuitry design and tuning. Therefore the best solution by now seems to be the high-side current amp, but i have to look if there is an external-powered one, because i want to add possibly no load to the UUT. \$\endgroup\$ – clabacchio Jan 4 '12 at 16:49
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    \$\begingroup\$ @clabacchio, re-read the third paragraph of the answer. If you can put the sense resistor between the output and feedback nodes of an LDO, there's no impact on voltage supplied to the UUT (although the effective dropout voltage of the LDO is increased slightly). \$\endgroup\$ – The Photon Jan 4 '12 at 17:19
  • \$\begingroup\$ If @clabacchio is interested in measuring microamps, he probably doesn't want the LDO ground current added into the measurement. \$\endgroup\$ – markrages Jan 4 '12 at 22:43
  • \$\begingroup\$ @markrages: Perhaps, but he's only interested in 100uA and the no load state can be measured and digitally subtracted later. Still, you're right, it's a issue to consider. \$\endgroup\$ – Olin Lathrop Jan 4 '12 at 23:34
  • \$\begingroup\$ LDO ground pin current can vary based on load and voltage drop. But it shouldn't be too hard to source an LDO with under 100uA ground pin current. \$\endgroup\$ – markrages Jan 4 '12 at 23:41
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enter image description here

Here's a circuit that I used in a test fixture to measure current.

The current comes in at V3 and leaves for the target at VTG.

The sense resistor is one ohm, which doesn't drop much voltage, but the transresistance of the circuit is 100 ohms. (R1*R24/R23)

It is important to use a good autozeroing op-amp for U4, because any offset voltage will cause huge errors in the output. With a good op-amp, errors are mostly resistor matching and Q1's alpha. I used an OPA2333.

R23, Q1, and U4a could probably be replaced with a ZXCT1009.

The circuit has two outputs: VCR is the unfiltered current signal. Because most low-power systems achieve their low currents by duty-cycling higher-current outputs, monitoring VCR on an oscilloscope will give you a good snapshot of the health and operation of the system. And it's easy to visually integrate under the current peaks to get an quick idea of the power budget. Here's an example of a system with a 2.4GHz radio module (I've notated the different power-using parts of the system):

enter image description here

Because of the duty-cycling, it is hard to get a good average current reading with a DMM that samples several times/second. The VC output provides a low-pass filtered view of the current signal. (The capacitor shows a + symbol, but don't use an electrolytic with its high leakage current.)

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    \$\begingroup\$ Replacing Q1 with a FET would give better accuracy. Right now the base current flows thru R24 but not R23. That's sortof like the error from LDO quiescient current you were pointing out earlier. With a FET the drain and source currents would be equal. \$\endgroup\$ – Olin Lathrop Jan 4 '12 at 23:45
  • \$\begingroup\$ I used bipolar for the voltage headroom. A fet would be better. Since BC817 alpha is guaranteed to be >0.99 it's less than 1% error. Close enough for my purposes. \$\endgroup\$ – markrages Jan 4 '12 at 23:58
  • \$\begingroup\$ Yes but why you didn't use the ZXCT1009 or even an instrumentation amplifier? The latter would give you a fully differential solution with almost no offset with the possibility to zero it. And i agree that could be a good solution, but for the specs i have i'd prefer to build just an interface, as minimal as possible, for the use of an external instrument, like a data acquisition card. (I know, it's a weird case) \$\endgroup\$ – clabacchio Jan 5 '12 at 8:48
  • \$\begingroup\$ Of course the low pass RC filter (1 Mega Ohm, 1 micro Farad) before the final OpAmp has to be changed for the requirements of clabacchio (10kS/s). \$\endgroup\$ – Curd Jan 5 '12 at 9:09
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This is not a low-cost solution, but it might actually do what you're asking for.

Consider a high-end benchtop multimeter like Agilent's 34410A or 34411A. Keithley and other vendors are likely to have comparable models available. The Agilent meter can measure 10k samples per second (50k/s for 34411A) at 5-1/2 digit resolution, and has an external trigger which would enable you to syncronize measurements between 4 meters. Readings can be logged to internal memory or streamed out via USB, GPIB, or LAN to your PC.

The downside is a list price of $1300 per channel.

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  • \$\begingroup\$ I think that this is a choice that my boss would like :) Do you know if there is an instrument like this but in a rack structure? Something PXI or so... Anyway thank you! \$\endgroup\$ – clabacchio Jan 6 '12 at 19:23
  • \$\begingroup\$ The bench meters have an available mounting kit that would let you mount them in a 19" rack. Two units lock together to fill the 19" width. \$\endgroup\$ – The Photon Jan 6 '12 at 20:02
  • \$\begingroup\$ For a modular instrument, like PXI, I'm not sure. I'd check National Instruments in addition to the vendors I already named. "Scanning multimeters" are widely available in modular form-factors; they use relays to switch multiple input channels into a single DMM-like circuit, but I don't think you'll get 10kSa/s with a scanning multimeter. \$\endgroup\$ – The Photon Jan 6 '12 at 20:04
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Just for information, we found this DC power analyzer with this Supply Measurement Unit (SMU) to be the best solution for our purpose. It's a veeeery nice toy, even if we have only two of these modules. It has a huge amount of features, like an auto-ranging functions that allows to measure up to 3A and down to about 10 nA automatically. It can log at up to 5 us/sample and for up to 999 hours. Ah, it has also USB port with dedicated PC interface.

Definitely, more than we need, except for the number of channels, but we made it to be enough :). The price is not the lowest, but we (in another lab, of course) already had one, so...If you can afford it it's definitely worth!

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