How to Choose Resistor Values for Series Regulator with Fold-Back Current Limiting?

Working with this circuit and need to know what to use as far as resistors values. Its not for any purpose so any set of values would work lower would be easier just need to know what would work and why it would work. Thanks in advance for any help.

Interesting. I haven't seen it done this way. It looks like the current rollback level is dependent on the output voltage. The higher the output voltage, the lower the current rollback level (i.e. a smaller current will excite rollback).

Also, it appears that Q2 will have to fight with the op-amp output to achieve rollback, thus stressing both components. I suggest adding a resistor between the op-amp output and the Q1 base.

As for resistor values, that depends on what rollback currents you want and at what output voltages.

This is for a college class. I get what the circuit does just don't understand why it does it. As far as the resistors go none of it matters i just need value that would theoretically work for a made up voltage. Completely stuck on this spent all weekend trying to figure it out.

This is my assessment :

I can't make out all of the component numbers so I shall define them as fillows:

R4 is the resistor between the two Emitters
R5 is the resistor between Q1 Emitter and Q2 base
R6 is the resistor from Q2 base to ground

Notice that the series combination of R4 and R5 is across the Base-Emitter junction of Q2 -- when the voltage across this series combination reaches around .65 - .7 Q2 begins to turn on and pull the base of Q1 towards the Q1 emitter voltage -- thus it begins to turn Q1 off. As Q1 turns off it causes the current flowing through R4 to decrease and thus the voltage across R4 also decreases. This causes Q2 to stabilize (i.e. not turn off as much and eventually hold somewhere in it's active region, which causes Q1 to also stabilize -- in other words, a point of equilibration is achieved which manifests as a limiting of the current at Vout.

If the load at Vout demands more current, the voltage across R4 rises and again Q2 starts to shut Q1 down and a new equilibration point is achieved. Because of the exponential collector current response with respect to base-emitter voltage of a bipolar transistor, the voltage across R4 is going to be maintained at around .65 to .7 volts -- in other words, it's not going to matter what happens to the load on Vout, the voltage across R4 is going to remain fairly constant for loads high enough to drive the voltage across R4 to that region. Because of this, the current flowing to Vout will also be fairly constant.

Now, for the role of R5. R5 in series with R6 forms a voltage divider. Since R5 is also in series with R4 in the Q2 BE leg, the voltage across R5 plays a role in the current limiting function. The current is limited to the voltage across both R4 and R5, but, the voltage across R5 is not going to change much -- only a few hundreds of milivolt, so what is it's function? To reduce the voltage needed across R4 -- in other words, A smaller voltage across R4 will excite the current limiting function. This means that there is less power loss in R4 and more power is delivered to the load. Also, R4 doesn't need to be as high a wattage and can be less expensive and will generate less heat.

Hi Justin!
I have the exactly same problem that you had, and I just wonder if you got it to work. If so, please help.
Thanks!