I had a feeling you were going to say that.
Is this because you don't understand how the circuit works?
If this is the case you might find it hard to build and it'd be a pretty pointless exercise since you won't have learnt anything.
However, I will help you to understand it.
First you need to understand how the LM2576 works when used as an ordinary constant voltage switching regulator. If you look at the data-sheet (use Google) you'll find that pin 4 (FDB, Feedback) is connected to the middle a potential divider which is connected to across the output of the regulator. To keep the output voltage constant LM2576 ensures that the voltage across the lower resistor in the potential divider is equal to the internal voltage reference of 1.237V; it does this by using an internal error amplifier which adjusts the switching duty cycle accordingly. When the voltage on the lower resistor starts to drop below 1.237V, it increases the duty cycle and when the voltage on the lower resistor starts to rise above 1.237V, it increases the duty cycle; this is know as PWM (Pulsed Width Modulation) regulation control.
The formula for calculating the output voltage comes from the potential divider equation. I'm not going to post the actual formula here, it can be found on the data-sheet for the LM2576. However, I'll give an example so you can understand it. 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.
Got it? If not then you might want to research more on potential dividers and then how linear regulators work (LM317, LM350 etc.). You can't understand something like this until you've grasped basic concepts like this.
Now let's look at the circuit in the article. Rsc is a current sense resistor, ohm's law says that the voltage across it is directly proportional to the current flowing through it. If you used a 1
hm: resistor it will have 1V across it when 1A flows through it, if you used a 500m
hm: resistor it would have 500mV across it when 1A flows through it.
So how does that help us build a constant current source?
I know, we could find a way to fix the voltage across Rsc so the current through it will remain the same! This is exactly what this circuit does.
In this circuit the duty cycle is adjusted so the voltage drop across Rsc, and therefore the current though Rsc, is kept constant regardless of the load on the regulator's output.
Now how do we do that?
Remember our LM2576 alway adjusts the duty cycle so the voltage on the FDB pin always remains at 1.237V. All this circuit does is measures the voltage across Rsc, multiplies it by a certain value and feeds it into the FDB pin.
What if we if we used 1
hm: for Rsc and multiplied the voltage across it by 2 then feed the output into the FDB pin?
The LM2576 would keep the voltage on the FDB pin at 1.237V which is twice the voltage across Rsc (remember we multiplied it by 2). Therefore the voltage across Rsc would be 1.237/2 = 0.6185V and I = V/R so the output current would be 0.6185A or 618.5mA.
Alright let's look at the circuit again to see how it measures the voltage across Rsc and multiplies it by a certain value.
IC2a forms a differential amplifier with a gain of one (Google it if you don't understand) which measures the voltage across Rsc. IC1b forms a non-inverting amplifier (again, Google it) which multiplies the output from IC1a by a scaling factor equal to R7/R6+1 in this case it's 5.15.
Now look at it again, we measure the voltage across Rsc, multiply it by 5.15 then feed it into the FDB pin. The LM2576 will try to keep the voltage on the FDB pin at 1.237V which is 5.15 times greater than the voltage across the resistor. Therefore the voltage across the resistor will always be kept at 1.237/5.15 = 0.24V, the value of Rsc is 0.12
hm: so if the ohm's law says if the voltage across it is 0.24 the current thought it = V/R = 0.24/0.12 = 2A.
To summarise this there are two ways of changing the current: you can either change Rsc or change the gain of IC2b.
Here is a list of formulae for calculating the current:
[latex]I_{OUT} = \frac{V_{REF}}{VR_{SC}}[/latex]
[latex]VR_{SC} =R_{SC} \times Av[/latex]
[latex]Av = 1+\frac{R_7}{R_6}[/latex]
Av is the gain of the non-inverting amplifier.
Vref is the reference voltage of the LM2576.
VRsc is the voltage across the sense resistor.
Putting all of the above into one formula:
[latex]I_{OUT} = \frac{V_{REF}}{R_{SC}(1+\frac{R_7}{R_6})}[/latex]
For optimum performance, R5 needs to be the equlavent value of R6 and R7 in parallel.
[latex]R_5 = \frac{R_5 \times R_6}{R5+R_6}[/latex]
Setting the gain of the amplifier and voltage across the current sense resistor is a bit of a compromise. For optimum performance you want the voltage across the sense resistor to be as low as possible but for the circuit to remain stable the amplifier's gain shouldn't be too high. in the example given, 0.24
hm: dissipates only 480mW at 2A and the amplifier will be stable with a gain of only 5.15.