In that schematic, Q2 will deliver current not just to the feedback network, but also to the load, meaning that it will have to conduct a lot of current (most of which goes to the load, which you don't want) to start limiting current. Try the circuit below:
This way the current limit transistor diverts current away from the load and directly towards the feedback voltage. The value of R5 can be adjusted to change the "sensitivity" of the current limit.
That's actually what you should expect. The limit transistor only needs to be slightly on to work, so it probably only needs Vbe=0.6. So your current limit will be 0.6/0.275=2.18A, and your Vout should be 2.18*3.33=7.27V. So the value of the current limit has decreased, but it should be much more stable and independent of the load. To adjust the current limit, just change the sense resistor.I tried your configuration with a 100Ω but with 3.3Ω my voltage came down to 7.2V (worse). Then I tried with 820Ω and I couldn't see a lot of difference. :s Do you have any idea what i might be doing wrong?
The value of R5 really just changes the "sharpness" of the transition from constant voltage to constant current. It's not really critical, and just needs to be lower than the impedance of the feedback divider.Can you suggest some resistor values to start with? when you suggested to move the current limit to ground what do you mean exactly please? I do not need variable current but if I can set it to a particular value forever thats fine for me.
Dear Friend
Again Hi
Use below circuit to obtain good current limiting and variable current limit ( as i know , the tl072 has 2 op amp , one of them can be used for current limit ):
View attachment 67918
R6 is load resistor . and by changing the v3 ( it is equal to the vR7) you can control your current simply . ( to select your desired current) .
Best Wishes
Goldsmith
Dear Brandon
Hi
The purpose of c1 : increasing the ripple rejection ratio ( at dc it considered open circuit , and at ripply , it will be short circuit (approx) ) thus the feed back network can sense the ripple faster than constant DC voltage.
And C2 : the current that will flow from R2 , is not stable and it will have large value of ripple . with that capacitor , i prevented from interference between the ripple and constant DC and thus the voltage of base will be very constant .
If you use darlington pair , the out put impedance of your power supply will decrease and it's out put current will increase ( it's ability to give current) BTW you can use 2 transistor ( 2n3055 in parallel with together to obtain lower dissipation).
( and if you want use parallel transistors , you should use a 0.1 or 0.27 ohms resistor in series with their emitters ( each transistor , one resistor ) To provide , the balanced pair.
Good luck
Goldsmith
Dear FvM
Hi
As i tested that circuit that i have designed it , it's procedure was , terrific . and without any problem.
Best Regards
Goldsmith
Good point, I didn't notice the problem yet. In fact the original circuit doesn't work without a dual OP supply, or current measurement shunts with at least 3 V voltage drop.I changes the op amp to an LM358N to be safe.
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