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I have to perfomr battery characterization(power and current consumption measurement) of an embedded system module. I have embedded PCD and an Battery pack PCB.Therefore I am planning to include the following PCB in the attached figure for current measurment. The battery pack PCB has 12 batteries divided into 3 sections. If you see the molex connector,it has the following in configurations. Pin a - 3 V Pin b -7.5V pin c - 7.5V pin d - ground

I am planning to perform current measurement by getting the embedded PCB to work in different modes according to functionalites.When the embedded PCB is in sleep mode, the current consumption is supposed to be as low as 150Ua and when not in sleep mode, it consumes upto 30mA.

My questions are. 1. Does the following circuit diagram help me in finding and performing current measurement? 2. Do I require any additional components in the circuit. 3.For measurement, I have a Tek MS0 3014 oscilloscope and a Keithley DMM(6 1/2). How can I design a PCB to easily probe in and find adjacent current and voltages.

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  • \$\begingroup\$ possible duplicate of Electronic design with current sense resistor. The circuit diagram looks alright, though. \$\endgroup\$ – Nick Alexeev Oct 28 '12 at 0:26
  • \$\begingroup\$ Thanks. What about the values of Rsens and Rout used.Is it appropriate to measure current varying from 150uA to 30mA or more. I also need to know how to include pins or something on the pCB so that I can easily probe in with oscilloscope of DMM for current measurement. \$\endgroup\$ – john Oct 28 '12 at 0:59
  • \$\begingroup\$ What is the maximum voltage drop you can have? \$\endgroup\$ – abdullah kahraman Oct 28 '12 at 9:03
  • \$\begingroup\$ The idea is to measure current varying from 200uA to 30mA.Thanks. \$\endgroup\$ – john Oct 28 '12 at 12:41
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    \$\begingroup\$ Geez, didn't realize this was the Photography Stack Exchange \$\endgroup\$ – Chris Laplante Oct 29 '12 at 14:19
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You are trying to make high side current measurements. In order to achieve this, you have selected ZXCT1009 high-side current monitor, which is available in a very small footprint; a 3-pin SOT23.

A high-side current monitor consists a sense resistor and a differential amplifier. These two and some other helping circuity creates a current sense amplifier.

A sense resistor is used to create a voltage drop between its terminals, then, a differential amplifier will amplify the voltage that is dropped on the sense resistor and create an output that is proportional to the current passing through the sense resistor.

For example, if you have a sense resistor of 0.1 \$\Omega\$ and you have a current of 1A passing through, then you have a voltage drop on the sense resistor of:

\$V_{drop}= R_{sense}*I_{sense}=0.1*1=0.1 V\$

So, there is a 0.1V voltage drop that is going to be fed into the differential amplifier. Say that this differential amplifier is configure with a gain of 10. If we input 0.1V, it will give us 1V as output voltage. If you take a look at the whole thing now, we have 1V output for a 1A of current that is passing through our sense resistor.

Coming back to you requirements, you need to sense the current from 150uA to 30mA. These are small currents! Let's use our 0.1 \$\Omega\$ sense resistor and calculate how much current will drop on it for 30mA:

\$V_{drop}= R_{sense}*I_{sense}=0.1*0.03=0.003 V=3 mV\$

With the differential amplifier that is configured as x10, the output is 30mV. This is a very low output voltage, and it is going to be lower when you lower the current.

Let's look at the output characteristics of the IC you have selected:

From page 3 of the datasheet:

Vsense vs. Iout

As you can see, for a 10mV of voltage drop on sense resistor, output current is about 100uA. If you connect a 100 \$\Omega\$ resistor on the output, it will give you;

\$V_{out}=I_{out}*R_{out}=100*10^{-6}*100=10mV\$

So, for a 10mV of voltage drop on the sense resistor, we get a 10mV of output voltage with these values. What should our sense resistor be, to have a voltage drop of 10mV with 150uA:

\$V_{sense}=I_{sense}*R_{sense}=10*10^{-3}=150*10^{-6}*R_{sense}\$

\$R_{sense}=66 \Omega\$

But why do we care so much about the voltage drop? Because the voltage drop will affect the voltage that embedded PCB sees on the power rail. The voltage drop with the 66 \$\Omega\$ sense resistor and 30mA current will be about 2V. That means your 3V rail will see only 1V as supply voltage.

A solution would be to have different ranges where you switch sense resistors in order to switch ranges. You can use simple mechanical switches or you can use MOSFETs if you use a microcontroller, like so, component selections may vary:

Range selection with MOSFETs

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  • \$\begingroup\$ Thanks for the detailed explanation.What would be the best way to measure current as low as 165uA. I need to put the embedded system in sleep mode to perform this measurement.Any way without causing voltage drop?? \$\endgroup\$ – john Oct 29 '12 at 13:51
  • \$\begingroup\$ @john A 100 \$\Omega\$ resistor will drop 16.5 mV at 165uA, which is OK and is well within +- 5% of 3V. \$\endgroup\$ – abdullah kahraman Oct 29 '12 at 17:04

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