# Electronics/Voltage Dividers

### Ideal caseEdit

Consider the illustration below. Assume initially that no current is flowing in or out of the terminal marked Vout. In this case, the only path for current is from Vin through R1 and R2 to GND. The equivalent resistance of this configuration is R1+R2 since these are resistors in series. From Ohm's law the current flowing through both resistors is

    $I = \frac{V_{in}}{R_1+R_2}$.


Also from Ohm's law, we know that the voltage drops across the resistors are I*R1 and I*R2 respectively. A quick check shows that the sum of the voltage drops across the resistors adds up to Vin. Now we can calculate the voltage at Vout (still assuming that no current flows through the terminal Vout). In this case the voltage is just

    $0V + IR_2$


where the 0V is the voltage at GND. If we substitute what we calculated for I, we obtain

    $V_{out} = \frac{R_2}{R_1+R_2}V_{in}$.


This is the voltage divider equation for the ideal case where no current flows through the output. Another way of saying that no current flows through the output is that the output has infinite resistance. A quick mental check using $I=V/R$ shows that the voltage calculated divided by infinity equals zero

### Non-ideal case - finite resistance outputsEdit

In the non-ideal case, we need to consider the resistance of the output. If we assume that the resistance of the output is R3 (and it is connected only to GND), we need to modify our analysis as follows. Now we have two resistors in parallel from the Vout junction to GND. The equivalent resistance of the parallel reistors then is

$R_{eq}=\frac{R_2R_3}{R_2+R_3}$


and the equivalent reistance of the entire circuit is

    $R_{tot}=R_1+\frac{R_2R_3}{R_2+R_3}$.


This yields a current of

    $I=\frac{V}{R_{tot}}=\frac{V_{in}}{R_1+\frac{R_2R_3}{R_2+R_3}}$.


Now we multiply this by $R_{eq}$ calculated above to obtain the output voltage:

    $V_{out} = IR_{eq} = \frac{R_{eq}}{R_{tot}}V=\frac{\frac{R_2R_3}{R_2+R_3}}{R_1+\frac{R_2R_3}{R_2+R_3}}V_{in}$


Normally R2 will be much smaller than R3 so Req will be approximately equal to R2. Keep in mind that a resistance R3 that is 100 times as large as R2 results in a voltage sag of about 1% and that a reistance R3 that is 10 times as large as R2 results in an almost 10% voltage sag.

### Non-ideal case - complex impedanceEdit

Both voltage divider equations hold for complex impedances. Just substitute Z's for R's and do the complex arithmetic. The resulting equations are just the following:

 $V_{out}=\frac{Z_2}{Z_1+Z_2}V_{in}$ - ideal case
$V_{out}=\frac{\frac{Z_2Z_3}{Z_2+Z_3}}{Z_1+\frac{Z_2Z_3}{Z_2+Z_3}}V_{in}$ - non-ideal case