Answer:
[tex]V = \frac{2*V_{ref}*R_{i}}{N*\sum R}[/tex]
where:
[tex] R_{i} [/tex] is the resistor where you want to know the voltage
N : the number of resistors
[tex] \sum R [/tex] the sum of all resistors in the branch
Explanation:
As you know, Flash ADC are made with a ladder of resistors this resistors are usually the same value, that means that the voltage drop in each resistor is the same, (in real life the exact value is different due to the resistors have a tolerance and differs their value one each other), to analyze this we have.
the current that flows through is
[tex] I = \frac{V_{ref} }{ R_{1}+R_{2}+R_{3}+R_{4} }[/tex]
if the R are the same...
[tex] I = \frac{V_{ref} }{4*R} [/tex]
the voltage drop across the last resistor is... taking into account that the first resistor is the one attached to the supply voltage.
[tex] V_{4} = I * R_{4} [/tex]
[tex] V_{4} = \frac{V_{ref} }{4*R}*R_{4} [/tex]
as I said before the value of the resistors are the same so...
[tex] V_{4} = \frac{V_{ref} }{4} = 0.25*V_{ref}[/tex]
because of the voltage drop is the same in each resistor, we can say that the voltage in the resistor [tex] R_{3} [/tex] is
[tex] V_{3} = V_{3} + V_{4} = 0.25*V_{ref} + 0.25*V_{ref} [/tex]
[tex] V_{3} = 0.5 * V_{ref} [/tex]
so, if Vref = 1 V
we have the schematic shown in the second image, in the first one we can see how the input signal in orange increase and the levels of the comparators change when the input signal overcome [tex] V_{2}=0.75,V_{3}=0.5,V_{4}=0.25 [/tex]
This analysis is posible because of the resistors are the same value in the case of Flash ADC, if the resistors are different to find the voltage drop, take the current and multiply by each resistor and add the voltage under the resistor that you want to know the voltage drop, something like this...
[tex] V_{3} = V_{3} + V_{4} [/tex]
