Electronic – Measuring very low current

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.

Your meter is correct. With a 3.7 V battery and a 10 ohm resistor, Ohm's law (\$V = I\cdot R\$) tells you that the current should be 0.37 A.

The 1260 mA you mentioned is probably 1260 mAh (milliamp hours). This tells you the capacity of the battery, not the current draw. For example, if you drew 100 mA from the battery continuously, it would last 1260 mAh / 100 mA = 12.6 hours.