# What values of components to use in this 30V-2V, 3A(max) current limiting circuit?

I'm looking for a current limiting circuit for my lab bench PSU, and found this one.

What are the values for the components? Is R_sense the pot?

• There isn't nearly enough information here for us to help. What current range do you want? What voltage? – Hearth May 27 '18 at 19:12
• Sorry, I've posted them in the title, it's 30V-2V DC and 3A. – thegamebusterpl May 27 '18 at 19:23
• @thegamebusterpl Are you expecting the output voltage to be regulated by your power supply, while at the same time you have some kind of current limit knob that lets you set a current limit that will "kick in" at appropriate times? Or are you looking ONLY for a way to turn your lab bench supply into a current source at times? Different things. – jonk May 27 '18 at 20:35
• Jonk - I'm looking to have a way to limit my current in a range of 0.something to 2A, after regulating and stepping down voltage with a knob. This will function at all times. – thegamebusterpl May 27 '18 at 20:47

How it works:

• R1 biases on Q1 and the collector current starts to increase.
• When the voltage across Rsense rises to about 0.6 V Q2 begins to turn on and steal the bias from Q1.
• The output stabilises with 0.6 V across Rsense.

For 3 A max current limiting $R_{sense} = \frac {V}{I} = \frac {0.6}{3} = 0.2 \ \Omega$.

Is R sense the pot?

No. This is a fixed-current constant current source. Pots don't come in low values like that. 100 Ω would be a low value in most potentiometer series.

• You could do something with a fixed resistor in parallel with a potentiometer to trim the current, if you wanted to. – Hearth May 27 '18 at 19:30
• Give an example of the fixed R and variable R and let's do some current calculations. – Transistor May 27 '18 at 19:32
• so 1k for pot and 10ohm fixed? – thegamebusterpl May 27 '18 at 19:40
• So what happens when you want 3 A? You have to wind the pot down to 0.2 Ω (that's 0.02% of the track resistance) and you've got 3 A running through a 1/4 W pot with a max track current of $I = \sqrt {\frac {P}{R}} = \sqrt {\frac {0.25}{1000} } = 16 \ \mathrm {mA}$. As well as being possible to adjust the pot won't last a second. – Transistor May 27 '18 at 19:54
• Oh,sorry by 3A i meant the max current possible, normal operation would probobly only 1A. – thegamebusterpl May 27 '18 at 20:00