l200 current regulator
Take a look at this circuit. To adjust the output from zero volts you need a -1.25V (or more). referred to TP4. The external transistor you added is not necessary if the current will always be 1A max.
To limit the current you use the opamp, with a 0~0.250V reference created by R6-R7 from the +5V power. The +5V source is also referenced to TP4. The current sense resistor is R3. You can simply connect four 1Ω resistors in parallel. So you should now see how the current limit works: you set R6 and the opamp now adjusts the output to mainain the voltage across R3 equal to the voltage across R6. Since the voltage across R3 is proportional to the output current, the latter is now kept constant. R6 is used to adjust the current limit from 0-1A.
Note that if you do not have the -1.25V the power supply will only work from 1.25V to 30V and the current limit will not work properly under short circuit conditions.
As for the 240Ω resistor, the LM338 develops a 1.25V reference between the OUT and the ADJ pins. Since the current that flows into the ADJ pin causes an additional voltage drop across R1, to minimize it you select a fairly high current (5mA) through the output divider R1-R2.
So, the output voltage is in good approximation: Vout=(R1+R2)/R1*1.25-1.25, that is from 0V (R2=0) to 31.25V (R2=5K).
Good luck!