Re: Current consumption in rms or average form
Hi,
Sometimes you have to calculate with RMS, sometimes you have to calculate with average.
You know P = U * I.
--> If U or I is considered to be as constant you have to use average.
--> If U and I change in the same magnitude you have to use RMS. (Twice the voltage gives twice the current)
Example:
* A 12V DC supply, a 10 Ohms resistor switched on/off with 50% duty cycle.
When the resistor is ON, then the current is 12V/10 Ohms = 1.2A.
--> Now see it from the supply side. The 12V is constant, so you need to calculate with average. The average current of 1.2A with 50% duty cycle it is 0.6A.
--> Now see it from the resistor side. Neither voltage nor current is constant. And you know the more voltage you put on a resistor the more current will flow. Therefore you need to calculate with RMS. The current when ON is 1.2A. With 50% duty cycle you get 70.7% of this current for the RMS current. This is 1.2A * 0.707 = 0.848A. But the voltage on the resistor is also 12V with 50% duty cycle. Therefore you need also calculate the RMS voltage with 12V * 0.707 = 8.484V.
Now you may be confused, because supply current is considered to be 0.6A (average) and resistor current is considered to be 0.848A (RMS). But both must be the same!
But I say both are the same. And I show this by calculating the power:
Output power = 12V * 0.6A = 7.2W.
Resistor: 8.484V * 0.848A = 7.2W. --> Both are the same. True.
Also we can check if voltage and current of the resistor follows Ohm's low. R= U / I = 8.484V / 0.848A = 10Ohms.--> True.
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You see - even in the same system you once need to calculate with RMS and the other time with average.
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Calculate the (pulsed) current of a light bulb: With a light bulb if you increase the voltage then the current is also increased. --> Therefore calculate with RMS.
*
Calculate the current (pulsed) of a LED: Since the voltage across a LED is about the same with 5mA and 20mA, therefore you have to calculate with average.
==> The result of both above is, that if you want to build a PWM regulation loop for constant brightness:
* with LEDs you have to calculate with average current
* with light bulbs you have to calculate with RMS current
I hope I could clarify a bit... With this difficult item.
Klaus
Thanks, Klaus.
Another question:
Example:
* A 12V DC supply, an IC is connected with the DC supply.
If the power supply is always constant, but the current draw from the power supply is variable, not just on or off. In this case:
--> Now see it from the supply side. The voltage is constant, the current is variable. The average and rms of voltage are both 12V. The average and rms of current are different. For example, I(average)=1A, I(rms)=0.8A.
--> Now see it from the resistor side. The voltage is constant, the current is also variable. The average and rms of voltage are both 12V. The average and rms of current are different. For example, I(average)=1A, I(rms)=0.8A.
From the supply side, the rms power=12V*1A=12W, the average power=12V*0.8A=9.6W.
From the resistor side, the rms power=12V*1A=12W, the average power=12V*0.8A=9.6W.
With one( U or I) is constant, another( I or U) is variable, in this case, both average and rms current can apply to it ?
Thanks.