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 1hm: resistor it will have 1V across it when 1A flows through it, if you used a 500mhm: 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 1hm: 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 0.12hm: 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:







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:


For optimum performance, R5 needs to be the equlavent value of R6 and R7 in parallel.


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, 0.24hm: dissipates only 480mW at 2A and the amplifier will be stable with a gain of only 5.15.

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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. 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

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 !!!!

 

Perhaps I'm simple, but couldn't you install a large heatsink with a powerful fan to cool this beast? Also, if I remember correctly, Peltier's are slow to move the heat energy, so you may toast the processor because heat is not displaced fast enough. Reducing it's current to less than 25% of it's requirement will probably make the problem mentioned even worse.
There are water-cooled systems available for situations like this. Check with your local computer chains. Water has six times the displacement capacity of air. I heard of a guy who didn't install this properly and came home one night to find his PC leaking profusely. Fortunately the PC's power was off at the time.

How much luck have you had with this Peltier so far?

Good luck with this!
kenjj

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Thank you for the suggestion. I am well aware of H2O cooling systems and I have build custom ones in the past. I already tested my peltier at 3A of current through a power resistor and it works very well. I am not going to use large heat sinks because that’s just extra heat in my case (maybe its ok for 1 peltier but what if I decide to cool my video card, North Bridge, RAM, and CPU all at the same time ? than I will have a total of 4 peltiers and my power resistors will be wasting almost 40Watts of heat!). 40 Watts of HEAT is a LOT inside any computer case and I will have to install like another dozen fans just to keep the air in the case cool.

 

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.

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0.22hm: would be fine but would result in a power dissipation of 1.98W (check the resistor is rated to 2W). Alternatively you could parallel two to give 0.11hm: to reduce the power dissipation to 0.99W but you'd obviously need to double the gain.

That would be fine but finding those resistor values could be difficult, although I suppose get them by connecting resistors in series and parrallel.

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.

You're missing the point.

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.

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I contacted Electro sonic and they told me that they have resistor values as low as 0.05Ohm! Perhaps I will use something like 0.1 Ohm for Rsc since it only cost around 2 dollars vs. 8 dollars for 0.05 Ohm. Anyhow, thank you for all your help and suggestions!

Sincerely,

Andrew Frost

 

$2 for 0.1hm: and $8 for 0.05hm:? If you really wanted 0.05hm: units it's cheaper to use two 0.1hm: resistors in parallel!

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Sorry, I missed the change to switch-mode! - but as he's struggling understanding a simple linear current source, the more complicated switch-mode type probably isn't too good an idea?.

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Dunno, with modern IC's switch modes are easier to understand than simple linear supplies. You just have to value the analog components more carefully, such as the primary inductor and capactive filtering elements.

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What is peltier? A heater?
Just a though. Why not just create heat from power transistor? If the transistor has 12Volt at Vce and you limit the emitor current for only 3A, it will provide 36 watts of heat. But might be maximum temperature only 125 Celcius.

 

A peltier is a cooler, not a heater. Why would he want to heat his overclocked processor up anymore?

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Actually it's both, and neither, it's a heat transfer device, it transfers heat from one side to the other - so to cool a CPU you use it to transfer heat from the CPU to outside the case.

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Alright then, a peltier is a heat pump.

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Will such a low value ( 0.05 Ohm ) be ok ? I mean it will only have a potential of 150mV. Will this be ok for the regulator ?

Thanks again!