The circuit is powered with 3V a battery and contains a black box device who's power consumption we're measuring with a shunt. The rated required voltage for the device is 3V. Whit a 1 Ohm shunt, the voltage drop on the shunt is 50mV meaning that the most of the voltage drop is caused by the black box device. Image name: Case 1.

However, the device power consumptions needs to be measured in an environment where we can't physically place the measuring equipment so we have to use long wires that add 10 Ohms of resistance in the circuit. Image name: Case 2


simulate this circuit – Schematic created using CircuitLab

After the installation we were expecting to measure a decreased current and to probably conclude that doe to additional resistance this doesn't represent the power consumption of the device. However, we've experienced that the current in the circuit has increased almost twice. The only explanation for this is that the voltage drop has for somehow decreased.

Is it a reasonable assumption that the device has detected a voltage decrease on its ends and switched to a lesser resistance mode to be able to consume more power? The reasoning for this assumption is that given that the device is battery powered, it is prepared for a voltage drop and has a mechanism to cope with it.

  • 4
    \$\begingroup\$ Welcome to EE.SE. You need to supply a lot more information than this if you expect help. Sensor type, model and link to datasheet; a schematic - use the button on the editor toolbar. Explain what you expect and what you get. You can see the circuit. We can only imagine based on what you tell us. Put all the information in your question and not in the comments. \$\endgroup\$
    – Transistor
    Jun 8, 2016 at 13:05
  • \$\begingroup\$ What are you measuring? DC current? \$\endgroup\$
    – winny
    Jun 8, 2016 at 13:46
  • \$\begingroup\$ @winny Yes DC current \$\endgroup\$ Jun 8, 2016 at 13:53
  • 1
    \$\begingroup\$ Can you add a schematic? \$\endgroup\$
    – winny
    Jun 8, 2016 at 13:55

1 Answer 1


Your black box is a constant power load (over a limited voltage). When the loop resistance increases with 10 ohm, the voltage over the black box drops and the current increases due to the assumed DC/DC inside it increases the duty-cycle to keep the output voltage inside it constant. More current = more voltage across your 1 ohm shunt.

Does your current consumption increase by about 20% or so?

  • \$\begingroup\$ It increases by double, from 50mA to about a 100ma \$\endgroup\$ Jun 8, 2016 at 14:44
  • \$\begingroup\$ @Alan What's the 3 V battery actual voltage, and can you measure the differential voltage across your black box in both test cases? \$\endgroup\$
    – winny
    Jun 8, 2016 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.