# Current-limit circuit for 78XX voltage regulators

I need to find the value of Rsc and R1 in the following circuit: In this circuit, Q1 is the high-current pass transistor. 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 Rsc cutting off drive when the drop exceeds a diode drop(i.e 0.7V.)

Now I want to calculate the value of Rsc and R1.

But how can I solve this problem for a linear voltage regulator?  ## 1 Answer

R1 sets the maximum current for the 78xx regulator; once its voltage drop exceeds 0.65V, Q1 starts shunting the excess current around the regulator.

Rsc sets the maximum current through Q1. Once its voltage drop exceeds 0.65V, Q2 reduces the voltage drop across R1. This turns Q1 into a constant current source, and actually forces any additional current to flow through the regulator, via Q2.

This is not protecting the regulator at all; if anything, it's protecting Q1. The load current will only be limited once the regulator goes into thermal shutdown — but by then, it may already have been damaged. A fast load transient can easily pop a bond wire before the die heats up enough to activate the protection.

Anyway, to answer your question:

$$R_1 = \frac{0.65 V}{I_{regulator(max)}}$$

$$R_{SC} = \frac{0.65 V}{I_{Q1(max)}}$$

Suppose you want the regulator to handle 200 mA before Q1 starts to assist. R1 = 0.65 V/200 mA = 3.3 Ω.

If you want Q1 to current-limit at 5 A, then Rsc = 0.65/5 = 0.13 Ω.