Hero999
Banned
0.22Frosty_47 said:Thank you for explaining the circuit to me. At least now I have some understanding of the circuit. I figured that the best way for me to achieve 3A of current is to use 0.22 Ohm Rsc resistor because its available to me.
That would be fine but finding those resistor values could be difficult, although I suppose get them by connecting resistors in series and parrallel.So I calculated the following:
Gain of Ic2b = 1.87 because 1.237V \ 1.87 = 0.66V across Rsc
And 0.66V \ 0.22Ohm = 3A
Thus R6 needs to be 115K Ohm because (1 + 100k\115k) = 1.87G
R5 should therefore be 53.5K ohm
Don't worry about Vout, the whole point of this circuit is to regulate the current, not the voltage. I was providing a bit of background by explaining how the LM2576 works when it's configured as a constant voltage regulator. My intention was to provide a bit of background which would make it easier to understand. I'm sorry if it confused you but reading the datasheet and understanding how the LM2576 works as a voltage regulator will help a lot.I am not very confident about the Vout.
Can you please explain
"Suppose you make both the resistors the same in the potential divider, so it will divide the output voltage by 2.The voltage across the bottom resistor will be 1.237V which is half the output voltage (remember the divider always divides the output by 2) so the output voltage will be double the reference or 2.474V. If you configure the potential divider so it divides the output voltage be 3 then the output voltage will be three times the reference voltage of 1.237V or 3.711V."
What resistors are you talking about in "Suppose you make both the resistors the same in the potential divider, so it will divide the output voltage by 2." ?
I found the formula in the datasheet:
Vout - Vref(1+R2\R1)
but not sure how to implement it into this circuit because I don't know what resistors serve as voltage devider in this circuit.
Thank you Hero999 !!!!
You're missing the point.Nigel Goodwin said:You appear to be missing the point?, using the IC regulator dissipates exactly the same heat inside your case - it's no different to using a resistor. Use a simple (LARGE) resistor, and mount it outside the case on a suitable heatsink. Or use an IC regulator, and mount that on a similar sized external heatsink.
If you use a switching regulator it dissipates much less power than a linear regulator; with switcher it's possible for the regulator to only burn a couple of Watts.
Last edited: