# Measuring current drops voltage?

We have some pcb in which contains an MCU ,and some WiFi module . Usually at start, the board takes 70-100mA with 3v3 . It gets its current from the UNO board connected to a mac. It works great like that .(if you wonder how the UNO can get this current,me too)

Problem starts when we try to measure current. If we take the fluke device, and put it in serial to the voltage of the pcb to measure current, what happens is that the MCU starts to get resets, and the WiFi module voltage is drops .nothing is works than.

We have tried to change the current scale on the Amper-Meter , from 20m to 20A. Nothing helped.

*if we measure current of other stuff that consumes lest current, it DOES measure and show the right current. The problem occurs only with the pcb that contains a WiFi module that takes 70-100mA.

connections are: 3v3 from arduino - > Ampermeter plus->Ampermeter minus->board 3v3 .

• Open your manual and find the burden voltage. Then do the math. – Ignacio Vazquez-Abrams Jun 28 '15 at 9:36

This is the "burden voltage" of an ammeter - when a 'shunt resistor' is used to measure the voltage drop due to current flow, then the voltage seen by the circuit being powered is lower. Multimeters, because of their wide sensing range, tend to have relatively high shunt resistors, which causes the circuit being powered to see significantly less voltage, and so some things don't work right.

I'd recommend putting a very low value resistor in series with the 3.3V rail that's powering the wifi module (say, 100mohm), and then measure the voltage across that resistor & do the Ohm's Law calculation to work out the current.

• thanks, i could see that putting it in 20A scale, will than enable the measurement,but i wouldn't see anything than because the currents are low.. – Curnelious Jun 28 '15 at 9:38
• Either get a low-value precision resistor (to make the maths easier; you need to know the resistor value as precisely as you want your measurement to be, and say one that's 0.1049 ohms means it's harder to do in your head), or you can use something like eevblog's microcurrent which has the precision resistor and an op-amp which does the maths for you. – Pete Kirkham Jun 28 '15 at 11:14
• @Curnelious, check out Dave Jone's uCurrent; it was designed for precisely this purpose. On one side you inject the current (with very, very low shunt resistance) and on the other side you measure a proportional voltage. – sherrellbc Jun 28 '15 at 14:24