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Basic impedance matching for a resistor divider network.

Giving the following circuit:

Vin > ---o
         |
        .-.
        | | Rx
        | |
        '-'
         |
         o-- > Vout
         |
        .-.
        | | Ry
        | |
        '-'
         |
    0v --0-- 0v

Taking into consideration that the impedance of source Vin is substantially low, following relationships can be established:

$$\frac {1}{x} + \frac {1}{y} = \frac {1}{a}$$

and

$$\frac {y}{y + x}=\frac{1}{b}$$

Where

$a$ = output impedance

$\frac {1}{b}$ = ratio Vout relative to Vin

(For example, with a given Vin of 3v and output voltage Vout of 1v, $\frac {1}{b}$ would equal to $\frac {1}{3}$)

Expressing x and y in $a$ and $b$ gives:

$x = a \cdot b$

$y = \frac {a \cdot b}{b-1}$

For example: $a=75$ Ohm and $b=3.3{\text v}$ gives:

$x = 247.5$

$y = \frac {247.5}{2.3}$

Please note that under load, when the impedance is connected to Vout, the voltage measured at the output will reduce to exactly half the voltage in comparison to when it is open.