Electronics/Voltage Dividers

Ideal case

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 V_in 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 V_in. Now we can calculate the voltage at V_out (still assuming that no current flows through the terminal V_out). 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 outputs

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 V_out junction to GND. The equivalent resistance of the parallel resistors then is

$R_{eq}={\frac {R_{2}R_{3}}{R_{2}+R_{3}}}$ 

and the equivalent resistance of the entire circuit is

    $R_{tot}=R_{1}+{\frac {R_{2}R_{3}}{R_{2}+R_{3}}}$.  

This yields a current of

    $I={\frac {V}{R_{tot}}}={\frac {V_{in}}{R_{1}+{\frac {R_{2}R_{3}}{R_{2}+R_{3}}}}}$.

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

    $V_{out}=IR_{eq}={\frac {R_{eq}}{R_{tot}}}V={\frac {\frac {R_{2}R_{3}}{R_{2}+R_{3}}}{R_{1}+{\frac {R_{2}R_{3}}{R_{2}+R_{3}}}}}V_{in}$

Normally R₂ will be much smaller than R₃ so R_eq 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 resistance R₃ that is 10 times as large as R₂ results in an almost 10% voltage sag.

Non-ideal case - complex impedance

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
 $V_{out}={\frac {\frac {Z_{2}Z_{3}}{Z_{2}+Z_{3}}}{Z_{1}+{\frac {Z_{2}Z_{3}}{Z_{2}+Z_{3}}}}}V_{in}$ - non-ideal case