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Low power mode in MSP430G2553

Haresh G
Intellectual 685 points
Other Parts Discussed in Thread: MSP430G2553, MSP430G2452, MSP-FET, ENERGYTRACE

Hi,

 I need to measure the Voltage for MSP430G2553 in LPM0 to LPM4.

  • 0
    Guru 74080 points

    I think you want measure the current when the device is in low power mode. Just hook a multimeter in series with the supply of the micro and the micro's Vcc pin. But make sure you do not measure any other currents that do not belong to the current consumption of the MSP. And the MSP does not drive anything when in low power mode - a LED that is switched on with a pin of the MSP before low power mode will remain on in it. Of course this LED-current has nothing to do with the power sonsumption of the micro, although it runs through it.

    Dennis

  • 0 in reply to Dennis Eichmann
    Intellectual 685 points
    Hi Dennis,

    I want to display the low power mode current consumption value in lcd. So how can i make it work
  • 0 in reply to Haresh G
    Guru 74080 points
    You mean the G2553 should measure it's own current consumption and then control a display itself to show it there?
  • 0 in reply to Haresh G
    Guru 12050 points

    The current just that the mcu is using or the whole board including the lcd?
    Unless you are using the mcu pins to source current to anything like a LED or MHz switching of a mosfet it will be in the uA
    If you construct your software to use lpm3 90% of the time the current will be to small to worry about

  • 0 in reply to Tony Philipsson
    Intellectual 685 points
    Hi,
    Im going to use msp430g2553 and msp430g2452 in a same board.

    I want to make the msp430g2553 in low power mode and Vcc of msp430g2553 is connected to analog i/p of msp430g2452.

    LCD is connected to msp430g2452 and i want dispay the current consumed by msp430g2553 in that lcd.
  • 0 in reply to Haresh G
    Guru 12050 points

    Measuring in the 1-4uA range requiring a 1K resistor shunt, the problem with that is if the device needs short burst of more power the 1K is causing brownout.

    Putting a 100uF cap between the shunt and mcu should work for short burst of power requirement and the reading you get from shunt will still be in long term correct.
    Just get one of the below gain amp and connect to one of G2454's analog pin.

    50x and 100x from TI
    http://www.mouser.com/Texas-Instruments/Semiconductors/Amplifier-ICs/Current-Sense-Amplifiers/_/N-6o48gZscv7?P=1yzxnagZ1z0zls6

    200x Shunt Amp from SiLabs
    http://www.mouser.com/ProductDetail/Silicon-Labs/TS1102-200EG5/?qs=sGAEpiMZZMvEn2pkGav3bLZ4YKjqCinR%252bvxYw%2f4eEQ0sB0tDWuybBw%3d%3d

  • 0 in reply to Haresh G
    Guru 74080 points

    Expected Tony's reply here, but it isn't visible...

    Look:

    In LPM4 the device draws less than 1uA (@ 2.2V supply). The ADC of the MSP430G2452 has 10bits of resolution and an internal reference voltage of 1.5V - this gives you a resolution of about 1.5mV. To get any signal from a current that is as small as 1uA you would need a shunt resistor of 1.5k - and that gives you one! digit (expecting no noise to be there and an ideal ADC, which it isn't). Now do some math and calculate which resolution you want to have - your shunt resistance will grow with that. And the bigger it gets the more voltage drop it produces when measuring current in LPM0 for example. Imagine we would set the full-scale value to 60uA, you would need a resistor of 25k, resulting in a voltage drop of 1.5V. So this is no option!

    You will definitely have to amplify your signal. But keep in mind that you only have 10 bits of resolution. With a current span from 0uA to 100uA (to have some overhead) your resolution is about 0.1uA (still thinking ideal). Sourcing an external reference into your micro doesn't make sense, too because it has to be 1.4V minimum, as well.

    Maybe you think measuring the current is quite simple, but with those small values it isn't. Of course it is possible, but not by only hooking two MSPs together. Sourcing the G2553 from the G2452 doesn't mean the G2452 can measure it's current through an output pin. Furthermore there is a voltage drop in the output pin that has a dependency on the current it has to source. This will lead to different supply voltages for the G2553 when in different power modes. OK, you're in the uA-range so the drop will be minimal, but keep that in mind anyway and remember that you want to measure sub uAs.

    I don't want to allure you from your idea, but think about it again - you have to invest a little bit more effort to it. Search in the web for how to measure those small currents.

    Dennis

  • 0 in reply to Tony Philipsson
    Guru 74080 points
    There's Tony's reply again :-) Why wasn't it visible? I got the eMail-notification for that, but it didn't appear here.
  • 0 in reply to Haresh G
    Guru 227245 points
    While the CPU is in LPM, it cannot do anything. Not measure a current, nor changing an LCD display content.
    It's like asking someone 'What are you dreaming right now?'
    You can let the ADC take some voltage conversions during LPM, and if you manage to turn the supply current into a voltage (which isn't simple because any current-sensing shunt would be above VCC while the voltage you need to measure must be below VREF and well below VCC). You'll need an OpAmp for this, and most OpAmps would draw much more current than the MSP takes. Even the better (and more expensive) ones will draw currents in the MSP's range.

    A possible solution is the one TI uses in its MSP-FET. Basically, it counts the pulses of the switching regulator that powers the MSP. The regulator keeps the output voltage constant (so it is known) and the amount of energy pushed on each pulse is also constant. Only the frequency changes. So you can calculate the average current in an interval by the number of pulses in this interval. And counting pulses is something the MSP can easily do while in LPM. This method is ~10% precise and works across several decades. However, LPM4 would be difficult, as you need to have a clock running to measure the interval time and wake you up to do the math and show the result.
  • 0 in reply to Jens-Michael Gross
    Guru 46710 points

    Jens-Michael Gross said:
    While the CPU is in LPM, it cannot do anything. Not measure a current, nor changing an LCD display content.

    You don't need to measure, do ADC conversion and display at the same time. While sleeping - you just measure and store measurement in some storage (capacitor). When wake-up - do ADC conversion & display. Solution (including circuit) of getting answer to question "what you are dreaming" is in 32L152CDISCOVERY user's manual.

    Jens-Michael Gross said:
    most OpAmps would draw much more current than the MSP takes.

    What's the problem to power OpAmp directly from power supply such a way that it does not affect current on the shunt? :)

  • 0 in reply to Haresh G
    Guru 58965 points

    Haresh G said:
    ... I want to make the msp430g2553 in low power mode and Vcc of msp430g2553 is connected to analog i/p of msp430g2452....

    Measuring Vcc of the target alone is useless. You need a shunt resistor and measure the voltage drop across the shunt. If the Vss of the target is not connected to anything else, use a "low side" shunt (as oppose to a "high side" shunt) is easier to handle by the ADC. 

  • 0 in reply to old_cow_yellow
    Guru 227245 points
    Well, it is possible:
    The suggested 1k resistor between VCC and 2553, followed by a rather large cap for current peaks.
    If you measure the voltage on the 2553 VCC pin and know the voltage before the resistor, you can calculate the current by the voltage drop.
    However, resolution won't be great, and the resistor will prevent the 2553 from drawing higher currents for a longer time (you're losing 1V of VCC per mA supply current).
    A low-side shunt is not a good idea for several reasons. First, it raises the GND level of the MSP, so interaction with external circuitry will become complicated. Also, you're measuring not only the current consumption of the MSP itself but also any current it sinks from external circuitry.

    The MSP-FET uses a different method for its EnergyTrace function: here a switching regulator is used to supply the target. The energy moved by every switch pulse is known, and so is the output voltage. The switching frequency therefore gives the power rate consumed by the target and (divided by the voltage) the current. Precision isn't superior, but it works over several decades (from nA to mA) without any influence on the target supply voltage.

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