2
\$\begingroup\$

In a buck boost inverting converter is there any difference exchanging the position of the diode from the positive to the negative rail keeping the same current rotational direction ?

enter image description here

\$\endgroup\$
  • 4
    \$\begingroup\$ Even though it's a common and well known circuit, it would make your question more clear to include a schematic of exactly the version you want to ask about. \$\endgroup\$ – The Photon Apr 15 at 19:16
  • 1
    \$\begingroup\$ An image always helps @ThePhoton \$\endgroup\$ – George Kourtis Apr 15 at 19:43
  • \$\begingroup\$ They are not buck boost converters. \$\endgroup\$ – Andy aka Apr 15 at 21:27
  • 1
    \$\begingroup\$ Wikipedia shows as buck-boost inverting converter the following image en.wikipedia.org/wiki/File:Buckboost_operating.svg \$\endgroup\$ – George Kourtis Apr 15 at 21:41
  • 1
    \$\begingroup\$ @Andyaka aka the first one is. \$\endgroup\$ – Jasen Apr 15 at 22:34
1
\$\begingroup\$

In your second circuit, the switch is directly connected between the input and output, and this means that the signal on the switched node is directly connected to the output, meaning that the output is not referenced to ground, so that circuit is probably broadcasting RFI all over the place, which will not only make you unpopular with the neighbors, but it will also hurt efficiency.

A normal buck-boost, which is your first diagram if "+" is considered to be ground, shares the grounded end of the inductor between input and output and takes the inverted rail from the switched node via the diode.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ It does not change anything if you exchange the battery poles at the entrance, the only reason I designed it like that is that I use an N mosfet instead of a P mosfet. The polarity of electricity does not change's the working of the circuit. \$\endgroup\$ – George Kourtis Apr 15 at 22:39
  • \$\begingroup\$ yes. just call it ground, for the purpose oc considering the buck-boost converter topology. \$\endgroup\$ – Jasen Apr 15 at 22:44
  • 1
    \$\begingroup\$ I don't understand the difference, could you explain it ? \$\endgroup\$ – George Kourtis Apr 15 at 22:53
  • 1
    \$\begingroup\$ immagine an oscilloscope connected between the output terminals and ground. \$\endgroup\$ – Jasen Apr 16 at 19:53
  • 1
    \$\begingroup\$ Then examining the upper diagram the upper output is always ground while the lower output is positive compared to ground due to the capacitor that keeps charging and discharging. \$\endgroup\$ – George Kourtis Apr 17 at 11:07
1
\$\begingroup\$

The difference is in the mechanism by which the energy is transferred. In the top diagram, the charged-up inductor pushes its energy through the diode (at the bottom position) into the capacitor's bottom plate, while the top rail remains anchored to the rest of the circuit, establishing how the voltage potential relates to the rest of the circuit.

In the bottom diagram, the charged-up inductor actually "sucks" current out of the diode (at the top position) out of the capacitor's top plate, while the bottom rail remains anchored to the rest of the circuit, establishing how the voltage potential relates to the rest of the circuit.

It's confusing because, in some ways, the circuits are the same, and in some ways, the circuits are different. And you may find even more (subtle) ways in which they are different (like the interference example given by a commenter).

The ways that they are about equivalent is they will both give about the same power output (equivalently, about the same voltage into the same load). So if you're powering a set of LED's, or a portable soldering iron, for instance, you should be able to use either circuit.

But the voltages at the output will be different relative to the rest of the circuit.

It may be useful to think in terms of a shop vac attached to a very air-tight cardboard box. A shop-vac has two ports, a suction port, and a blower port, and depending on what you need, you attach to the appropriate port. Either way you transfer energy. But if you connect the blower port to the box, it will deform outwards as it experiences an air pressure above atmospheric pressure. If you instead connect the suction port, the box will crumple as it experiences negative pressure in relation to atmospheric.

Assuming you don't lose any air, and a perfect shop-vac, you will keep transferring the same amount of energy to the cardboard box until it either explodes or implodes. But you used the same energy pump to do it.


EDIT:

For the following inverted topology Joule-Thief-like circuit:

inverted topology Joule-Thief-like schematic showing both diode positions

Here are the LTSpice simulation numbers:

numerical figures showing output rail voltages the only real difference

The red boxes show the only significant difference between the two topologies -- the voltages of the bottom and top output rails. Hope this helps you to "see" it.

Finally, here are the two traces for the two output rails (as once again suggested by Jasen):

both output rail voltages, graphed by LTSpice, relative to ground

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.