I have a circuit to drive a simple resistive load, and it works quite well. On the high side is 2.5V from a regulator, and the low side is 0V (taken as reference to a battery's negative terminal which supplies the regulator).

I have realized that it would be desirable in my application to have the high side of the load be 0V, and the low side to be -2.5V instead. To do this, I have considered implementing the TPS63700 (link below).

What I can't get my head around, is the fact that the regulator would need to sink current rather than source it in this situation (about 60mA in my case). I would imagine it can, since i can't understand what situation wouldn't require it unless there was an even lower voltage in the circuit to source to. I've gone through the datasheet but the definition it uses is 'output current', and i don't know if that means current at the output or current that the device puts out.

Could someone explain where my thinking is wrong, and give a brief explanation of why the part (or other inverting regulators) would or wouldn't work for this application? Also, how does the current flow if the input side of regulator is connected to a battery (terminals 0V and ~4V), and the output side (-2.5V) is connected to the battery's neg terminal (0V) by a series resistor?

TPS63700 datasheet: http://www.ti.com/lit/ds/symlink/tps63700.pdf



1 Answer 1


Consider the topology of an inverting buck converter:


simulate this circuit – Schematic created using CircuitLab

The regulator output does not sink current. The coil does!

When the MOSFET inside the regulator is conducting, there's a current running clockwise in the left loop. Then the regulator opens the MOSFET, but the current through the coil cannot be turned off in an instant. That's why there's now a counterclockwise current in the right loop.

  • \$\begingroup\$ After studying the topology of buck converters alongside your explanation i think i understand it now. Thanks. \$\endgroup\$
    – Joe
    Mar 12, 2018 at 5:40

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