1.6 KiB
1.6 KiB
Voltage Divider
Simple Voltage Divider
This is sort of unrealistic because there is no current flowing out of the voltage divider on the right side:
Equation
\begin{flalign}
Vout & =\text{Vin }x*(\frac{R2}{R1+R2})&\\
\end{flalign}
Voltage Divider with Load
When the output of the voltage divider is connected to something the current drops on the output, as that something uses some of it.
The load is connected in parallel to R2, so we can calculate it as a parallel resistor.
The new Equation:
\begin{flalign}
Vout & =\text{Vin }x*(\frac{R2 || RL}{R1+R2 || RL})&\\
\end{flalign}
Example
Lets calculate the current in this circuit:
- We calculate the Resistance in the subcircuit (R2 and RL) as they are connected in parallel which means
\begin{align}
\frac{1}{Re}&=\frac{1}{R2}+\frac{1}{RL} &\textit{Replace Variables}\\
\frac{1}{Re}&=\frac{1}{100}+\frac{1}{150} &\textit{Add Fractions}\\
\frac{1}{Re}&=\frac{1}{60} &\textit{* Re}\\
1&=\frac{Re}{60} &\textit{* 60}\\
60&=Re \\
\end{align}
The simplified circuit now looks like this;
Now we can easily calculate the Resistance in the circuit
With the resistance we can now calculate the current inside the load circuit by using the simple voltage divider equation:
\begin{flalign}
Vout &=\text{Vin }x*(\frac{Re}{R1+Re})&\\
Vout &=5*(\frac{60}{100+60})&\\
Vout &=5*(\frac{60}{100+60})&\\
Vout &=1.875v
\end{flalign}