Datasheet Link Here <==

This is the"typical application" from the datasheet "typical application" from the datasheet

Short description of my understanding of this ICs operation My understanding of this IC is that it takes two same voltage power supplies and combines their current output. If one supply's V starts to droop, the IC senses that change from Vin1 (or2) to OUT1 (or2), and drops the voltage of the other supply using the MOSFET to continue voltage matching.

I'm attempting to combine 2 outputs from two USB battery packs.

My question is mostly about the current sense resistors, I believe that they are supposed to create a drop in the voltage in order to know how much current is passing, but why is this necessary? When one of the battery's voltage starts to drop, the Vout sense pin should see it and drop the other battery's voltage in kind. No resistance needed? Why does it even need to know how much current is being sourced if all it has t do is match voltage?

Also, compared to the 2mOhm resistance of those resistors, the resistance of the MOSFET and wire connection is much higher. The resistors are also placed after the Vout sense pins on the IC, so how do they do anything?

The goal is to create ~24W and put it through a DC-DC converter to charge my laptop. If there are any obvious problems with this idea, it'd be much appreciated if you point it out! I'm about to buy all the components.


You've misunderstood how the circuit works. It doesn't care about absolute voltage to the load. You have to keep in mind that the two sense resistors are tied together at the load, so the voltage into the the chip differs only according to the currents through the sense resistors. What the IC does is try to keep the two currents totaling some predetermined level. So if one supply hits its limit and can provide no more current, the voltage across the two resistors is adjusted by increasing the current through the other side, drawing more current from the other supply.

  • \$\begingroup\$ Thank you! I was under the impression voltage drops would only occur after the resistor and the output pin is before. How does this detect anything? Does this mean that when one side droops the other side pushes through both resistors to the other side, where it is detected by the output pin? \$\endgroup\$ – Aaron Apr 29 '17 at 22:23
  • \$\begingroup\$ @AaronSchulte - No. The output pins are not outputs - they are inputs to the chip. See the chip block diagram in the data sheet. The two currents are controlled by the 2 MOSFETS, so the chip outputs are the two gate drives. The internal logic tries to equalize the two currents (zero voltage between the two output pins.) See page 8 of the data sheet. \$\endgroup\$ – WhatRoughBeast Apr 29 '17 at 23:22
  • \$\begingroup\$ Yeah I knew they were inputs, but they are sampling input from before the resistors. So how do the resistors do anything? Thanks for taking the time to explain. \$\endgroup\$ – Aaron Apr 30 '17 at 0:21
  • \$\begingroup\$ @AaronSchulte - The two resistors have the same resistance, and they both feed the load. So, when the current through both resistors is the same, the voltage on each will be the same, and the difference of the two voltages will be zero. \$\endgroup\$ – WhatRoughBeast Apr 30 '17 at 3:11
  • \$\begingroup\$ ooooohh that made it finally click. duh \$\endgroup\$ – Aaron Apr 30 '17 at 3:43

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