Voltage divider law

174

04:47:07

Without load
Vout = ( R2 / ( R1 + R2 )) x Vin
Iin = Vin / ( R1 + R2 ) 
With load
Vout = ( R2 / ( R1 + ( 1 / (( 1 /R2) + ( 1 / Rload)))) x Vin
Vout = ( R2 / ( R1 + (R2 || Rload) )) x Vin
Iin = Vin / ( R1 + ( 1 / (( 1 /R2) + ( 1 / Rload))))

published 9 years ago

thebugger

9 years ago

It's easier to measure the current through the 1k resistor and calculate the voltage drop via ohm's law U=I.R. So without load it draws 3.33mA then U=1000x0.0033= 3.33V then the center point will be 10-3.33≈6.67V. Then you can find the bottom resistor value again with ohm's law. R=U/I R= 6.67/0.0033≈2K. You can replicate the experiment with a load. A load will just act as a parallel resistive connection to the 2k resistor.

Robert_Kidd

9 years ago

Useful to have the equations if you are designing on paper without using a simulator or other means of measuring. I'm not sure your middle equation is quite right - at the very least there is a bracket missing since the number of opening and closing one is different. My eyes are crossed now so I'll have another look at it later lol.

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