Current-limit circuits for 78XX voltage regulators

[khatus]

Hello guys i Want to find the value of R_SC and R₁ in the following circuit


In this circuits, Q₁ is the high-current pass transistor, and its emitter-to-base resistor has been chosen to turn it on at a given load current. In the circuit, Q₂ senses the load current via the drop across the resistor R_SC cutting off drive when the drop exceeds a diode drop(i.e 0.7V). Now i want to calculate the value of R_SC and R₁ But How can i solve this problem???

For linear voltage regulator

[RES]

check this application note:
http://www.ti.com/product/LM340/technicaldocuments
(direct link) http://www.ti.com/lit/gpn/lm340
on page 16 you see your example.(figure 25, top right) two formulas:
RSC = 0.8 / ISC
R1 = βVBE(Q1) / IREG Max (β +1) – IO Max

[iMo]

It will not work as a current limit.
When the Rsc creates 0.7V the Q2 starts to conduct and the excessive current will flow through 7805.
You have to put a resistor R2 into collector of the Q2.
Set R1 such you get enough current for 7805 (10mA).
For example 50ohm.
The R2 for example 50-100ohm.
Edit: I would put R3=100ohm into the Q2 base.
With Rsc 1ohm it will limit to 1A, with Rsc=4ohm to 250mA, and so on..