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I am working on extending output voltage of a fixed linear voltage regulator.Yes, I accept I could go for the variable voltage regulator.But I am in need to use a fixed voltage regulator. My query is regarding the derivation of Output Voltage (V0) given in Datasheet Can you help me to figure out how this equation rolls out?
The output voltage \$V_O\$ will be whatever the regulator voltage is plus whatever the voltage is at the top of \$R2\$.
\$V_O = V_{XX} + (I_{R2}) * R2\$
So, to determine the voltage at the top of \$R2\$ you need to know the current going through it, \$I_{R2}\$.
That will be \$I_Q\$, the quiescent current of the device plus whatever current goes through \$R1\$.
\$V_O = V_{XX} + (I_Q + I_{R1}) * R2\$
Since \$R1\$ has the regulator voltage across it, we know:
\$I_{R1} = V_{XX}/R1\$
So, substituting for \$I_{R1}\$, we get:
=> \$V_O = V_{XX} + (I_Q + V_{XX}/R1) * R2\$
=> \$V_O = V_{XX} + V_{XX}.R2/R1 + I_Q.R2\$
=> \$V_O= V_{XX}.(1 + R2/R1) + I_Q.R2\$
Note: This circuit should never be used if the input voltage exceeds the maximum input voltage, \$V_I\$, of the regulator chosen. I have seen folks try to use this method in an attempt to "shoe-horn" the device to work at higher voltages, and, although once in a steady state it will work ok, sooner or later the initial start-up, or shut-down voltages will kill the device if you do not surround it with protection devices.
Iq is a sort-of constant current, typically in the 5mA range, that comes out of the regulator.
You can derive the equation for the output voltage by superposition, for example.
