Help understanding oscilloscope shunt resistor readings - Page 1
Forum section: electronics / beginners

author: jacau
author_rank: Contributor
timestamp: March 26, 2022, 07:01:34 am
content: I would grateful for some advice on my situation.I have a 3v lithium battery micro-controller project that wakes for 300 microseconds (accelerometer threshold reading) and then sleeps for 1 second and loops.Not including the MCU filter caps, there is a total of 2000uf of capacitors on VCC (which I think might be high but I have a 4G SIM module).My multi-meter reads 4mA when the MCU is awake and around 80 uA when sleeping.My min/max/avg on the multi-meter never showed any spikes like I was expecting so I thought that maybe the sample rate was too slow to detect these.I borrowed an oscilloscope that can do 10MS/s and used a 1760ohm shunt resistor. Battery was 3.02 volts. Voltage drop over 1 hour was a very consistent 109-113 mV (min max). Using an online voltage divider calc I worked out the sleep resistance was around 48Kohm (inc shunt resistor) - or about 60 uA while in the sleep/wake loop.I was expecting to see regular spikes after the wake period from the increased current consumption but the oscilloscope was also consistent at that very high sample rate.I'm not sure if I've used this capacitor discharge calculator properly but it indicates I would have very little voltage drop based on a 4mA draw for 300uS which would explain the the oscilloscope readings.I figure the capacitors are providing the initial current but why do they seem to draw such a consistent current from the battery and not show periodic spikes? Have I done something wrong in my test setup?Thanks

author: Arseny
author_rank: Newbie
timestamp: March 29, 2022, 02:48:11 pm
content: QuoteAll pulsed current is drawn from your large capacitor, because impedance of this capacitor for pulsed current is much less than impedance of measurement shunt. So you wont get pulsed current reading on oscilloscope in such setup.Trace of voltage drop over shunt and traces of current flowing through capacitor and shunt are shown on attached image. Steps on voltage trace are only 600 uV and it will be hard to see them on an oscilloscope.You could try to decrease capacitance value for this measurement or insert shunt after bulk capacitor.

author: jacau
author_rank: Contributor
timestamp: March 31, 2022, 09:58:20 am
content: Thanks very much Arseny for your fantastic reply; that really helps me understand what I was seeing!So would this graph from all about circuits explain why a smaller capacitor would show the pulse? The further from Cmax the greater the charge current?I think my capacitance at 80uA would give 30+ seconds of power without a battery so I take it that 300us 4mA pulse means C is barely changed from Cmax and Imax is therefore very low?Thanks again!

author: Arseny
author_rank: Newbie
timestamp: April 01, 2022, 05:38:27 pm
content: This picture is not exactly relevant for this problem in my opinion because you are not fully discharging capacitor during your current pulse.But yes, your statements re true.Lets calculate some example:Vin = 3V, R1 = 1 kOhm, C = 2 mF, i0 = 80 uA, i1 = 4 mA, dt = 300usInitial condition :i0 = 80 uA,All this current flows through R1 because ideal capacitors are open circuit at DCVc = Vin - i0*R1 = 2.92VWhen current pulse starts, voltage on capacitor starts to changeCapacitor voltage drop will be dV = I * dt / C = (4mA - 80uA) * 300 us / 2mF = 588 uVWe consider that almost all pulsed current is flowing through capacitor because it is large and voltage over C is almost constant, so voltage over R is also almost constant and hence current is almost constant.See the first attachment.If capacitance is small, calculation is more complicated, so I just use simulation. See the second attachment.

author: jacau
author_rank: Contributor
timestamp: April 02, 2022, 03:02:53 am
content: Those two simulations really helped - as the small capacitor is drained, the current through the resistor proportionally increases to supply the 4mA current to the MCU.So the impedance (ESR?) of the small cap must be much larger than the 750ohm load which is why it is only re-charged after the MCU is back asleep and the green voltage line on the first simulation stays low because Imax of the cap is so low that it requires much more than a second to get back to Cmax (if ever).I really appreciate your time and thorough explanation