Using an analogue multimeter to measure average current - Page 1
Forum section: electronics / testgear

author: trevwhite
author_rank: Frequent Contributor
timestamp: February 26, 2013, 10:19:18 am
content: Hi all.I am working on a project that draws maybe 50mA for 10ms every 1 second. It is a processor based project where the device goes to sleep, wakes up, does some stuff and then goes back to sleep. I have done my calculations to work out the average current draw so I can determine battery life but to actually demonstrate it I was wondering what to do. I had heard someone used an analogue meter like the one shown in the link below. Apparantly the needle coil averages out the currents. I was wondering if this was true? Anyone used this technique?http://www.ebay.co.uk/itm/ROLSON-27249-Analogue-Electrical-Multimeter-Meter-Tester-Test-Leads-850636-/121046979754?pt=UK_BOI_Electrical_Test_Measurement_Equipment_ET&hash=item1c2ef654aaIf this is a good technique then can anyone recommend a decent analogue meter?ThanksTrev

author: KJDS
author_rank: Super Contributor
timestamp: February 26, 2013, 11:57:12 am
content: If the current was being drawn for 1mS every 10mS you might be ok. Analog meters are inherently slow, but I doubt they;d be slow enough to stay still for the 990mS you have whilst your device is asleep. A peak hold meter to measure the current peaks, and a scope to check the times would be easiest.

author: trevwhite
author_rank: Frequent Contributor
timestamp: February 26, 2013, 12:31:54 pm
content: Well I was wondering if the coil would average the current pulses out but as you say 1 second is quite a long time to average a 10ms pulse over. Or at least it feels like it. Maybe I just need to buy one and find out. Prefer not to buy one if it doesnt work as I am not sure I would use one for anything else. Seems a waste.

author: deephaven
author_rank: Frequent Contributor
timestamp: February 27, 2013, 12:26:56 pm
content: Add a small series resistor in the supply (ideally in the 0V lead if your system will allow it) and probe across it with a scope. You can then see exactly what current is being taken during the pulse periods and also the quiescent periods.

author: akis
author_rank: Frequent Contributor
timestamp: February 27, 2013, 07:55:43 pm
content: How about this: use an RC circuit to average the current and thus replace the mechanical parts of an analogue multimeter. I presume with careful adjustment of the RC network, the correct value to discharge the capacitor slowly, but not too slowly.I attach a simple example of such a circuit I just came up with. We use a resistor Rsense to sense the current going into the load. Ideally we would want the Rsense not to impede too much on the circuit downstream, so for example Rsense should be 1/20 or less of the lowest expected Rload.The very small, floating voltage developed across Rsense is referenced to ground as well as amplified (10 times in this example) by a simple rail-to-rail op-amp. This in turn feeds the RC circuit and in the attached example the voltage over the capacitor in mV is the translated average current in mA (so you could attach a simple voltmeter).

author: trevwhite
author_rank: Frequent Contributor
timestamp: February 28, 2013, 03:11:08 pm
content: Interesting circuit. Be interesting to get a walk through the calculations for it?

author: akis
author_rank: Frequent Contributor
timestamp: February 28, 2013, 06:41:08 pm
content: It is an idea just. Using an analogue ammeter, you rely on mechanical parts with mass and inertia to provide the "averaging" that you need, but those mechanical properties of whichever meter you use are not related to the 10ms pulse of 50mA, other than plain luck.Using an RC circuit you may be able to tweak it (the R and the C) so that the charge/discharge time of the capacitor will provide the averaging that you need. In the circuit presented above the 5*RC is 1 second (5 times RC is 99.36% charge). I was thinking that this can be adjusted so that the arithmetic average which is 0.5mA can be read as a voltage using a voltmeter straight off the capacitor.