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Very low DC current measurement

There are a number of ways.

One of the ways I have done this (over 6 decades of current, incidentally) is to put the load current path as the programming side of a current mirror.

The other side of the mirror will have the same current (within the tolerance of the device - see the ADI MAT-0x series for excellent matched mirrors, or perhaps choose a Wilson mirror - you should really only need this if you have large variations of programming current or supply voltage- a Wilson mirror is designed to negate Early effect, a variation of bipolar transistor gain proportional to Vce. Use one if the effective Vce of both sides of the mirror will differ significantly).

This output current may then be amplified, converted etc., to your heart's content for the measurement system.

Cheers PeteS

Reply to
PeteS
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Why not just use a DVM? The better handheld Flukes resolve microamps.

John

Reply to
John Larkin

A 1k resistor in parallel with a shotkey diode makes a useful bench tool. Put in series with the power feed for the device you are testing, it allows for both 'full power' operation and for measuring microamps for low power measurements.

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Luhan Monat (luhanis 'at' yahoo 'dot' com)
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Reply to
Luhan Monat

Just stick a 1k resistor in series with the battery and measure volts across it. The 0.1V or so dropped shouldn't make much difference to the consumption of the MSP430.

If it has brief phases of high power consumption, measure those separately with a different resistor.

Paul Burke

Reply to
Paul Burke

Hi All,

In order to estimate power consumption and Lithium battery life cycle. I need to measure supply DC current drained by an MSP430 micro which is as low as 100µA (TI specs)

Any methods ?

Many thanks, Habib betula.fr

Reply to
habib bouaziz-viallet

Le Wed, 23 Mar 2005 03:31:19 -0800, PeteS a écrit :

Hi Pete,

Considering your idea, i drew a little schematic to clarify it.

formatting link
Is that conform about what you think ?

In that schematic the two Vce's differ, what is Wilson I-mirror ? so far, i have never hear about.

Many thanks, Habib betula.fr

Reply to
habib bouaziz-viallet

Analog microammeter?

John

Reply to
John Larkin

That may depend on the allowable burden voltage, meaning the voltage loss.

Rene

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Reply to
Rene Tschaggelar

This circuit has *no* voltage loss and can be placed in either leg, as long as the compliance is met. ___ +-|___|------+ | R | I | |\ | o----->----+-|+\ | ___ ___ | >--+----)-+-|___|-+-|___|-+ | +-|-/ | | | R1 | R1 | | | |/ .-. | | | === | | | | | | | GND | | | |R | | | | | '-' | | | |\ Ue=0 | | | | +-|+\ Uout | +-------+ | | | >--+----o | | | | | +-|-/ | | | .-. | | | |/ | Uout=2*R*I | | | | | | | | | | | |R | | | | | | |\ '-' | | | | | +-|-\ | | | ___ | ___ | V | >--+----+-)-|___|-+-|___|-+ o-----

Reply to
Ban

But, unless there's something I'm not seeing, does require a seperate, higher voltage source than what is being monitored. As a bare minimum, if you're measuring the current from Vcc, the power to the opamps will have to be Vcc + Vopamp_clearance + Imax * R.

Reply to
Rob Gaddi

snip circuit

That is exactly what *compliance* means. If you put the circuit into the ground leg, there is almost no voltage across it and you could also power the circuit from the same supply as the DUT. Nice is the bipolar capability for small ac currents.

--
ciao Ban
Bordighera, Italy
Reply to
Ban

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