Using a buck controller (LM2576) in buck-boost inverting topology - Page 1
Forum section: electronics / beginners

author: andreasvitikan
author_rank: Newbie
timestamp: July 08, 2018, 07:21:46 pm
content: My question is regarding using a buck controller such as the LM2576 but in an inverting topology (the inductor and catch diode "swap" places) to produce a regulated negative voltage that can be bigger or smaller than the input voltage (in absolute value, of course). My questions are this?1. Can I use the output of this configuration "in reverse" from the way it's supposed to be? I mean normally it outputs 0V and -12V and the -12V line sinks current while the "virtual ground" = 0V sources current. Can I use it to power a 12V load with + connected to 0V and - to -12V? In other words if I use the "ground terminal" as the positive and the output terminal "As the negative" (which actually is a negative voltage) can I use it as a normal monopolar positive supply?2. Can I make the output in the inverting topology adjustable? The way I figured it, just using the same configuration (voltage divider: coil output/ground to 50k 10turn pot followed by 1.21k resistance to negative output, and the feedback pin connected to the midpoint of the voltage divider) to make the inverting buck-boost output voltage adjustable?3. Is it wrong to think of it in the following terms: Redefine ground/0 V to be the negative output. Then the output simply becomes +5/12/whatever in adjustable configuration volts and the input voltage is the input+output voltage? Is it wrong to think about it like this? Is it only useful in theory but does not carry over in reality to think about this circuit like this?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 08, 2018, 09:59:21 pm
content: 1. Yes, if you don't need a common ground on the load.2. No, it's a fixed regulator. Well, there's always a way around that, but it's easier with the ADJ version which has V(FB) = V(GND) + 2.5V or whatever it is. In this configuration, V(FB) to VOUT (with respect to ground) is a voltage follower. Typically, you'll have a voltage divider to set the output voltage (the fixed voltage versions have this divider internally), and you can add a third resistor to the FB node to tweak the output from a control voltage.3. That's right. Voltage is relative. The downside is, the regulator needs to switch the sum of voltages, and you don't get a common ground. You might find it's better to use a Zeta topology, or a boost regulator in SEPIC configuration, or either kind with an isolated (forward/flyback) topology (even if you aren't using the isolation).Tim

author: David Hess
author_rank: Super Contributor
timestamp: July 09, 2018, 12:57:38 am
content: Quote from: andreasvitikan on July 08, 2018, 07:21:46 pmSure, just recognize that the output is not isolated from the input. If you "ground" the -12 volt output, then the input gets pushed up by 12 volts. If you do this, then you have to watch out that the "+12V" supply current is greater than the "Input+12V" supply current because the inverting converter can only sink current. At least I think that is right; these things give me a headache unless I am current with them.QuoteJust adjust the feedback divider.QuoteI think I covered this in 1 above.Linear Technology wrote an application note covering various configurations for buck regulators including this one:http://www.analog.com/media/en/technical-documentation/application-notes/an44fa.pdf

author: andreasvitikan
author_rank: Newbie
timestamp: July 09, 2018, 08:03:01 am
content: Thanks alot for the answers and the Linear Tech application note, that helped out. Yes, with the voltage divider I didn't express myself right I think, I meant something like this (the TI LM2576 datasheet, page 14) with the wiper connected to feedback, top to output (which is "GND" in the inverting topology) and bottom to the negative voltage output. This didn't quite look right at first because it meant feeding the "feedback" pin a negative voltage (Since it's a voltage divider between 0V and something negative).And one final question, if I power 2 different 2576 from the same input, both in the inverting buck-boost topology, the grounds will be the same and the negative outputs will be different? So if I were to use them as positive supplies the grounds would not be equal to each other (because they're redefined different negative voltages)?

author: David Hess
author_rank: Super Contributor
timestamp: July 09, 2018, 03:44:48 pm
content: Quote from: andreasvitikan on July 09, 2018, 08:03:01 amYes.QuoteKeeping in mind that the "input ground" and "output ground" are not shared when you are treating the output as positive. With two outputs, you have to make sure that the current drawn from the lower "positive" output is less than the current drawn between the outputs. I doubt this will work.

author: andreasvitikan
author_rank: Newbie
timestamp: July 09, 2018, 03:48:25 pm
content: I thought it would be a bad idea, I just wasn't sure how bad, thanks for clearing that up!