hello guys....this is my first post at the forum...so hopefully ill get to learn a lot from you guys..my question consists of two parts...ill just focus on the first part as of now..
i have built a barebone arduino...consisting of LM7805 regulators,resistors, capacitors,crystal clock, etc...i have also interfaced RF module 433Mhz..so wireless transmittion for the barebone arduino...first part of my question is i need to measure how much current the barebone arduino is consuming..i know measuring current with multimetre can be a bit tricky as compared to measuring voltage...bascially i added one point of shunt resistor to GND of LM7805..in series with multimetre...to pin of 5V of LM7805..would that method be ok to know the overall current consuming in chip...oh and soldering of all smd components..on pcb..are already done...
if pics would help...i can send pic tommrow..thank you...
To measure current, you need to put the meter in mA mode and put it in series with the power source to the 7805 input.
Not sure what your shunt will do, the description doesn't sound correct.
Place the multimeter (current mode) across a switch in the supply lead. Once all is working
open the switch and the meter measures the current. If you need to switch
current range, close the switch, change the range, open the switch - that way the
chip won't reset or power-cycle due to range-changes.
This matters when measuring small currents as the low-current range on the meter
uses a large shunt resistance which is unable to carry enough current to reboot
the processor, but which is needed to measure tiny sleep-mode currents.
The shunt resistor on ground of 7805 makes no sense to me - the ground pin
must be at ground potential or the output voltage could be too high.
A 7805 will totally dominate the power consumption of the board in sleep mode,
there are special low-quiescent-current regulator chips available - look for the
magic word "micropower" in datasheets.
Microcontrollers in sleep mode are typically 5 to 0.1 uA, 7805 quiescent
current is 3mA, about 1000 times bigger.
wassey:
bascially i added one point of shunt resistor to GND of LM7805..in series with multimetre...to pin of 5V of LM7805..would that method be ok to know the overall current consuming in chip...o
That makes no sense at all. You're measure how much current goes through a resistor to ground.
Good for confirming Ohm's law. Useless for finding out how much current your circuit uses.
wassey:
bascially i added one point of shunt resistor to GND of LM7805..in series with multimetre...to pin of 5V of LM7805..would that method be ok to know the overall current consuming in chip...o
That makes no sense at all. You're measure how much current goes through a resistor to ground.
Good for confirming Ohm's law. Useless for finding out how much current your circuit uses.
this is the logic i assumed...the chip has resistance connected in combined series and parallel..let me call that combiined resistance R(t)...that R(t) would be in SERIES..with the shunt resistor which i added..lets call it R(s),so the current in series remains same...so if i work out the current across shunt resistor..since that is in series with the overall resistance of barebone uno..so the current would be the same...wouldnt that be right....
CrossRoads:
To measure current, you need to put the meter in mA mode and put it in series with the power source to the 7805 input.
Not sure what your shunt will do, the description doesn't sound correct.
the reason y i used i a shunt resistor was because as we all know..we cannot directly measure current through multimetre..it has to connected in series..so what i thought was if i can measure current across LM7805...which is stepping down voltage from 9V to 5V..in my case..so yeah if i measure the current coming out of LM7805..then i would know how much current is going in the the chip...
so since i couldnt measure current directly with multimetre..i added a shunt resistor..connected in series with multimetre..with one probe of multimetre..connected with resistor..and another probe of resistor connected to GND..of LM7805..and the one point of resistor..connected to 5V pin of LM7805 voltage regulator..so in that case i would know how much current in the chip..wouldnt that be right??
cant upload the pic..because size is too high...here is the link of the components i have used..havnet used FTDI connector...havent used any leds..didnt use any switch..no pin headers..no 220ohm resistor..and thats pretty much it ...
and below is the schematic diagram...i have also added interfaced RF module..not mentioned in diagram..
CrossRoads:
To measure current, you need to put the meter in mA mode and put it in series with the power source to the 7805 input.
Not sure what your shunt will do, the description doesn't sound correct.
yep my bad guys...now i have figured out that i was doing it wrong....adding a shunt resistor with multimetere across 7805...would just measure the current across the resistor...anyways i have measured current through the power source...with multimetre.across the input of LM7805.and iam getting 8.2mA...that i think seems about right..considering on average...barebones take up around 6mA..iam guessing my LM7805...is consuming a lot of power....
Grumpy_Mike:
That measurement sounds too low.
That schematic has no decoupling capacitors on the arduino chip!
That schematic has no decoupling capacitors on the arduino chip!
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it has decoupling capacitors across LM 7805...and it has also decoupling capacitors on the the chip as well...22pF..sorry couldnt understand your point??
