Inverting buck boost feedback question.

[mb]

Hi everyone.

I want to assemble the circuit of an inverting buck boost regulator which I came across from an old NS App note AN1118 ( see fig1), but I want to use an adjustable version of the LM2595, but am unsure of the correct feedback I should use. Fig2 is what I think it should be, but fig3 is one I came across for another one of TI's regulators that is very similar to my circuit. Can anyone help on this?

Thanks
Martin.

[T3sl4co1l]

The chip is performing a differential measurement, namely,
Verr = V(FB) - V(GND) - 1.25V
So whatever arrangement has FB at 1.25V above GND will do.  Or, whatever the ADJ voltage is -- check the datasheet.

Note that it's pulling itself, er, down by the bootstraps, as the case may be.  So from a "ground" (0V) perspective, FB will be negative too, just slightly less so.

Tim

[mb]

Tim, if I understand your reply correctly, my cct has FB below 0V (negative), which is not ok

Verr = V(FB) - V(GND) - 1.25V

Are you  refering to fig1, the cct in the App note? OR my cct?

Vref=1.23V, is this the ADJ voltage you are referring to?

So, that would mean fig3 is the way to go?

Martin.

[metacollin]

Ground is arbitrary.  The circuit topology is identical regardless of the polarity of the output vs. input.  The feedback is unchanged, it is done the exact same way as any other resistor devider. 

Look at the circuit but pretend vnegative is ground, ground is output voltage.  See? It's not actually different.  The regulator doesn't know or care that it's making a negative voltage because ground is whatever potential it's ground pin is at, which, if you connect it to the output to ground, , will be regulated to be whatever voltage difference needed in the feedback divided to hit vref- usually 1.25 volts or so.  The extra positive output is of no concequince, it's just another inductor that happens to be isolated so you can do whatever you want with it.  I this case, they are just using it in the opposite polarity, but you could also put it in series and double the negative voltage or whatever else. 

As for the feed back, it's a resistor divider to whatever you want.  If it's across the output to the ground terminal, it will regulate so that voltage difference produced 1.25V at the divider connection.  it's just in series with the output, so -12V and 12V is no different than 12V and 24V. So you can do either. 

Always remember, there is no such thing as ground or positive or negative voltage. Voltage is relative.  You chose a potential as your reference, and that is your ground and all voltages are simply above or below that potential.  What you chose as your reference potential/ground is up to you, no matter what you pick, it won't actually alter anything except how label your potentials. 

It's just like how you're always standing on the ground no matter how high above sea level you are.  I live up in the mountains and it's about 8000 feet above sea level - but when I fall, it's still only a few feet to the ground .

 Unless the fall is off a cliff hehe.

[metacollin]

Sorry for typos, on cellphone

[mb]

Thanks for clearing that up metacollin.

[mb]

The LM2595 data sheet gives the method for programming the output voltage as:
Vout=Vref(1+R2/R1), Vref=1.23V

If R1 is known, then: R2=R1(Vout/Vref-1)

Will this calculation also work if I use the cct in fig3 as the feedback method.

The TPS54xx data sheet lists the calculation as:
R1=R2[(Vout-0.8V)/0.8V]

If I use the method shown in fig3 as the feedback for the LM2595 cct, can I use the calculation as listed in the TPS datasheet and replace the value of Vref from 0.8V to 1.23V?

Also SLVA369A.pdf lists the output voltage calculation as R1=R2(Vopos – Voneg/Vref-1), these two are not the same, I assume they both will work just fine, but why the difference and which is preferred?

Can you tell I’m a switcher newbie?

Thanks
Martin.