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raphaelriv

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here i was reading about current comsuption of electrical devices and i came across this article by Michael E. Manna

the section i have trouble understanding is the "Current Requirements" section. My question is about the set up for measuring current, the need of a low resistance shunt (0.1W or 0.01W) to evaluate transient pulses correctly is not clear to me? a shunt as i understood can be consider as a device with very low resistance? how can this help?

below is a image of the set up...
 

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A resistor in series with the load will create a voltage drop. Then the load gets a reduced voltage and the resistor heats up.
When the value of the resistor is very low then its voltage drop is low and its heating is also low.

Resistance is measured in OHMS. Power is measured in WATTS. You said "0.1W or 0.01W" but I think you meant 0.1 ohms or 0.01 ohms.

A multimeter has a low resistance shunt inside that is used for measuring current.
 
oh i think i understand because this article refers to measuring current using an oscilloscope instead of a meter, the oscilloscope would need a shunt to be able to measure current?
 
To help your understanding, I will explain further.

To measure current in a wire, you break the wire and connect the ammeter in series with the wire.
Ammeters usually measure small currents, say 1 milli-amp.
If there is 10 amps flowing in the wire, the meter will be destroyed.
So, we use a low value resistor to "shunt" most of the current past the meter, and just let a small curent flow in the meter.

When a current is flowing through the shunt, (or any other resistor), there will be a voltage developed across the shunt.
In the example you have, that voltage is being measured by the oscilloscope.
So, if you know the resistance of the shunt, and the voltage across it, you can use Ohms Law to calculate the current.

JimB
 
re:

got it, and lets say i do this set up,

Vin = 7volts
Rshunt= 75ohms
RLoad= unknown ohms (some device)

I measure the voltage drop accross the Rshunt to be 0.052109V, thus my current is 6.947866Amps, how do I know the current comsuption of this device? if I know that the current should be more without the Rshunt.
 
If the 75 ohms shunt has 0.052V across it then the current is only 0.69milliamps, not 6.9A! 10,000 times lower.
That is the current consumption.
The Vin doesn't matter unless we know the load resistance to calculate its current.
 
raphaelriv said:
got it, and lets say i do this set up,

Vin = 7volts
Rshunt= 75ohms
RLoad= unknown ohms (some device)

I measure the voltage drop accross the Rshunt to be 0.052109V, thus my current is 6.947866Amps, how do I know the current comsuption of this device? if I know that the current should be more without the Rshunt.
You slipped your decimal point badly. 0.052109/75=0.69479mA.
To find the unknown resistance,

I=Vin/(Rshunt+Rload)

Since this soumds like a homework problem, I think you should do the math yourself.
 
re:

ohhhh gotcha, this wasn't homework honest. i just picked out some values and started doing a simulation. I just needed I=Vin/(Rshunt+Rload)
knowing this i calculated Rload and then with that value I go back and remove Rshunt and just calculate 'I' with Rload and Vin to know the current consumption of this device! great! thanks :D

what about the evaluation of transient pulses that article is talking about? is it just the magnitude of their spikes that they would display on the oscilloscope? but the oscilloscope would be measuring the voltage drop across the shunt, not current. This part i am still confused about
 
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A transient has a very short duration that cannot be seen on a meter. An oscilloscope shows transients perfectly.
If the value of the shunt resistance is selected as 0.01 or 0.1 ohms then the voltage seen on an the oscilloscope will also be the amount of current.
 
raphaelriv said:
is this because the value of Rshunt is so low as to say compared to Rload.
Yes, they form a voltage divider. The voltage across the shunt resistor is subtracted from the load resistor, creating an error. You want the error to be small.
 
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