I have a circuit designed on PCB, and components (resistors, capacitors, diodes, Microcontrollers etc.) are all surface mount. I want to find the total current consumed by this circuit. I have traced the schematic of the circuit completely, and the circuit and MCU operate on 3.3 Vdc. The supply 3.3Vdc is generated through AC 240rms, then half wave rectifier, then filter capacitor and finally 3.3V regulators.

There is also a port left open for the battery but no battery is attached. I want to make circuit portable so that it can work without direct supply that is why I need current consumption of circuit to estimate the battery specifications.

I can't break any connections to measure the current using DMM, because its layered PCB and I am afraid I might damage it.

Is there any other way to find the total load that circuit has or current it consumes??

Note: MCU has the normal operating current of 2.4-5.4 mA.


closed as too broad by Chris Stratton, Elliot Alderson, Finbarr, Dave Tweed Dec 7 '18 at 14:25

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    \$\begingroup\$ If the circuit is line powered, the exact current really doesn't matter that much. You can get a rough idea with an AC mains meter you plug it into. If it's too small to measure with that, then it's also small enough that you really don't need to know exactly what it is, and if you still want to badly enough, then you can afford to buy a cheap hot air tool and sacrifice one copy of the design to measurement or learn to use it well enough that you'd don't even damage that copy. Milliwatts matter when you run on batteries, not on the mains. \$\endgroup\$ – Chris Stratton Dec 5 '18 at 4:46
  • \$\begingroup\$ When people speak of idle mains-powered appliances unjustifiably wasting power, they're talking about fractions of a watt large enough to measure with a plug in meter. \$\endgroup\$ – Chris Stratton Dec 5 '18 at 4:50
  • \$\begingroup\$ If I can measure (which I can) the current from AC mains, then it would mean that this will be the current consumed by a circuit (DC circuit working at 3.3 Vdc)? \$\endgroup\$ – BetaEngineer Dec 5 '18 at 4:51
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    \$\begingroup\$ No, the power drawn at the mains input will be a little more than that consumed by the DC side. But you've failed to state a justifiable reason why you need to know, so there's no argument as to which current matters until you can make a sound argument why either current matters. \$\endgroup\$ – Chris Stratton Dec 5 '18 at 4:52
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    \$\begingroup\$ Pretty obviously you can't install a battery until you find a way to power the circuit with DC. Once you've done that, measure the current consumed from the DC supply. Feeding the existing mains supply from an inverter powered from batteries is probably a grossly inefficient idea, if the overall power consumption is as low as you are implying it is or low enough for the plug in meter not to give a fair representation. \$\endgroup\$ – Chris Stratton Dec 5 '18 at 4:53

Standard way to measure currents on PCB where all components are already integrated (AC, transformer, secondary coil, diode bridge, regulator) is to make a narrow trace cut between the regulator and the rest of the circuit, and measure the current either by a fixed-range DMM, or placing a current-sense resistor (say, 0.1 Ohms SMT 0805) across the cut, and measure small voltage across it. Then you either leave the sensing resistor to continue to operate your device, or restore the trace continuity with a solder bridge.

(Suggestion: you can find various small-ohm chip resistors on all broken hard drive electronics or old dead CD-ROMs.)

  • \$\begingroup\$ That’s also one of the reasons why 0ohm resistors were invented. Many of my designs have a few of those at that precise location. \$\endgroup\$ – Edgar Brown Dec 5 '18 at 5:07
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    \$\begingroup\$ @BetaEngineer, if you want to use the battery input, it is safe to assume your regulator as linear (LDO), and use the battery voltage about 2V above your 3.3 V power rail. 4 AA battery should do the job. In this way you don't need to cut anything, just measure the current from 6-V battery. \$\endgroup\$ – Ale..chenski Dec 5 '18 at 5:09
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    \$\begingroup\$ @EdgarBrown, yes, zero-ohm are good things, in industrial world it is called DFT, "design for test". Whoever fails to do this upfront on first board iteration go out of business quickly. \$\endgroup\$ – Ale..chenski Dec 5 '18 at 5:11

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