Can a buck converter provide a lower voltage based on a target differential - instead of an absolute target?

Question

I have an application where I need to provide a lower voltage from a starting voltage. In other words - I need a buck converter. My question is - if I have a known required differential but a potentially variable input voltage how close to my target difference can I get?

Example: let's say I need to have two voltages with a difference of 8VDC. So, starting at a nominal 30VDC, I would need a converter that produces 22VDC.

So the question - can a single converter circuit produce that same difference but for a different starting voltage? Certainly within a certain specified range/tolerance. So in my case...let's say anything from 20VDC - 40VDC for the starting voltage. Can the same circuit that products 22VDC from 30VDC also provide 32VDC from 40VDC?

Answer 1

You can make a buck regulator do this if you design it yourself. Basically you can regard the positive rail as 0 volts and the (former) 0 volt rail as the negative voltage supply for the buck regulator. The buck regulator would then produce negative 8 volts relative to the "new" 0 volts.

What you need to find is a true negative rail buck regulator. I'm sure there are a few but, if you used an isolating DC-to-DC converter that produces an isolated 8 volt DC output, you can tie the positive output to the incoming power rail that feeds it. This gives you many more options: -

Answer 2

Conceptually (I've never built this, although the question has come up before), this could be a done with modifications to a standard synchronous buck converter.

First, rather than have a fixed voltage reference, such as a resistor to the input and a zener diode to GND, reverse these two components, with the zener value of 8 V. Now, the reference input to the voltage regulation loop always is 8 V less than the input. The problem here is that the voltage reference usually is inside the controller chip.

Next, connect the output directly to the feedback input to the controller. Now the output will track the input minus 8 V.

Or something like that.

Most buck regulator controller chips have an internal reference. If you can't find one that needs an external reference, you might have to grow the basic buck circuit from scratch.

Answer 3

Can a buck converter provide a lower voltage based on a target differential - instead of an absolute target?

Yes.

There are a number of ways to modify a buck converter to track the difference between the source voltage (\$V_{source}\$) and the output voltage of the buck converter, (which we shall call \$V_{bucked}\$), rather than tracking the output voltage itself. Note, we don't call the output of the buck converter \$V_{out}\$, because in a moment, we will show a circuit used to drive the feedback pin of a buck converter IC, and \$V_{bucked}\$ is the input to that circuit, and it would be confusing to call the input to a circuit \$V_{out}\$!

Somewhere in most IC controlled buck converters is a voltage divider between \$V_{bucked}\$ and ground, with the divided voltage going to a feedback pin, which in our schematic is labeled \$V_{fb}\$ thus:

^{simulate this circuit – Schematic created using CircuitLab}

As is readily apparent, if the feedback pin neither sinks nor sources current, then:

$$V_{fb} = \frac{R2}{R1+R2}V_{bucked}$$

What can be done is to replace the above feedback network with this:

^{simulate this circuit}

We shall show that in this case,

$$V_{fb} \approx \frac{R4}{R3}(V_{source}-V_{bucked})$$

We will assume that the current feeding the bases of the transistors is small, and the current to the feedback pin is small also.

We will also assume, because the bases of the two transistors are tied together, that the the voltages at the emitters of the two transistors are approximately equal.

With these assumptions, we see that the current through R3 is equal to the current through R4.

$$I_{R3} \approx I_{R4}$$

$$\frac{V_{source}-V_{bucked}}{R3} \approx \frac{V_{fb}}{R4}$$

$$\frac{R4}{R3}(V_{source}-V_{bucked}) \approx V_{fb}$$

Q.E.D.

So, if the buck controller IC attempts to keep \$V_{fb}\$ fixed, it must needs keep \$V_{source}-V_{bucked}\$ fixed.

\$R_{bias}\$ is not critical, but the same value as R4 is a good choice.

Note that you may run into compliance issues if either \$V_{source}\$ or \$V_{bucked}\$ is too small. That is, the current mirror circuit may cease to operate as a current mirror.

Also note that there will be some offset due to the fact that the two transistor will undoubtedly not be perfectly matched.

The same maths can be performed by an op-amp configured as a differential amp. That solution also takes 5 components, an op-amp and 4 resistors. Again with an op-amp, you will run into compliance issues if insufficient headroom is available for the particular op-amp chosen.