Very low DC current measurement
habib bouaziz-viallet
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
PeteS
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
Paul Burke
> 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 100渙 (TI specs)
>
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
habib bouaziz-viallet
Hi Pete,
Considering your idea, i drew a little schematic to clarify it.
http://cjoint.com/data/dxnAJRjWNb.htm
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
John Larkin
<ha...@mynewserverposfix.com> wrote:
>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 100渙 (TI specs)
>
>Any methods ?
>
>Many thanks, Habib
>betula.fr
Why not just use a DVM? The better handheld Flukes resolve microamps.
John
Luhan Monat
> 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
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.
--
Luhan Monat (luhanis 'at' yahoo 'dot' com)
"The future is not what it used to be..."
http://members.cox.net/berniekm
Rene Tschaggelar
> On Wed, 23 Mar 2005 11:47:18 +0100, habib bouaziz-viallet
> <ha...@mynewserverposfix.com> wrote:
>>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 ?
>
> Why not just use a DVM? The better handheld Flukes resolve microamps.
That may depend on the allowable burden voltage, meaning the
voltage loss.
Rene
--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net
p...@slingshot.co.nz
John Larkin
>Transimpedance amp?
Analog microammeter?
John
Ban
> John Larkin wrote:
>
>> On Wed, 23 Mar 2005 11:47:18 +0100, habib bouaziz-viallet
>> <ha...@mynewserverposfix.com> wrote:
>
>>> In order to estimate power consumption and Lithium battery life
>>> cycle. I need to measure supply DC current drained by an MSP430
>>>
>>> Any methods ?
>>
>> Why not just use a DVM? The better handheld Flukes resolve microamps.
>
> That may depend on the allowable burden voltage, meaning the
> voltage loss.
>
> Rene
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-----<----+-|+/ | R1 R1
I | |/ |
| ___ |
+-|___|--------+
R
(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)
--
ciao Ban
Bordighera, Italy
Rob Gaddi
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.
Ban
> Ban wrote:
>>
>> This circuit has *no* voltage loss and can be placed in either leg,
>> as long as the compliance is met.
>>
>>
>
> 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.
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.