Circuit Diagram for Extending Output Volatage

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?

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  • 1
    \$\begingroup\$ What don't you understand in that equation? \$\endgroup\$ Commented Feb 26, 2018 at 15:04
  • \$\begingroup\$ @Leon Heller I don't know the derivation of V0. \$\endgroup\$
    – VKJ
    Commented Feb 26, 2018 at 15:05
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    \$\begingroup\$ Just use the equation. You don't have to know how it is derived. \$\endgroup\$ Commented Feb 26, 2018 at 15:07
  • \$\begingroup\$ @ Leon Heller It would be helpful if u explain as u know how it is derived. \$\endgroup\$
    – VKJ
    Commented Feb 26, 2018 at 15:08
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    \$\begingroup\$ Do not use that circuit if the input voltage exceeds the delta V limit of the device.... \$\endgroup\$
    – Trevor_G
    Commented Feb 26, 2018 at 16:14

2 Answers 2


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.


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