In Section 2.5 of Battery Power Management for Portable Devices, by Yevgen Barsukov, the author explains the Dynamic Power Path Management (DPPM) found in battery chargers. Figure below describes the basic DPPM circuit found in linear battery chargers.

enter image description here

From what I've read, when the system current \$I_{SYS}\$ grows, so that \$I_{SYS}+I_{CHG}> I_{ADP_{MAX}}\$, i.e., the adapter current grows above it's maximum value, the voltage \$V_{OUT}\$ starts to decrease, till it reaches some prefixed value \$V_{DPPM}\$. The DPPM circuit than starts to descrease the charging current \$I_{CHG}\$, so that \$I_{SYS}+I_{CHG}=I_{ADP_{MAX}}\$ and the voltage \$V_{OUT}\$ is kept at \$V_{DPPM}\$.

enter image description here

My question is: Why the voltage \$V_{OUT}\$ drops when the current \$I_{ADP}\$ exceeds the adapter's limit current \$I_{ADP_{MAX}}\$?


2 Answers 2


Your diagram shows that USB is a possible power source for the DPPM circuit.

For something like USB, the host/charger is obligated to follow whichever parts of the USB specification it claims to comply with. If the charger is following the USB battery charging specification V1.2 then the output current and voltage must comply with "Figure 4-1 CDP Required Operating Range". In this figure the output voltage is obligated to drop once the output current gets high enough.

enter image description here

Anything attached to the charger must be designed to handle the fact that the input voltage will drop as the load current increases.

Texas Instruments document SLUA400A contains a good discussion on the DPPM with respect to the BQ2512X family of chargers.

In the document they explain that all the chips in the design have some minimum volage that they need in order to run. Below that voltage they may fail to perform their function, or even malfunction.

As the battery draws more charging current the input voltage will tend to drop. Therefore, once the input voltage drops past a certain threshold the BQ2512X reduces the battery charge current in order to prevent it from droping further. This guarantees that all the chips will have enough supply voltage to keep running.

The next logical question is "Why does the charger voltage need to drop". There are a few reasons.

  1. As far as I know it's impossible to design a charger that will output infinite current. From a physics standpoint this is going to happen at some point when you put enough load on the charger.

  2. For safety purposes. Somebody somewhere is going to short out a USB cable or plug something broken into it, and it's a lot safer if the charger output current is limited. The usual way to limit output current is by dropping voltage once the current limit is exceeded.

Specifications like USB just specify how much power the adapter is required to deliver and what the current limit should be.

  • \$\begingroup\$ But why the output voltage drops. Is it because as the current increases, the voltage drop in the MOSFET Q1 increases? \$\endgroup\$ Commented Jun 30, 2022 at 3:11
  • \$\begingroup\$ dV/dt of the battery = Ic/C in effective capacitance C in kiloFarads and the charger is in CC mode not CV \$\endgroup\$ Commented Jun 30, 2022 at 5:05
  • \$\begingroup\$ @LuizGustavoMartins In your diagram Q1 is a MOSFET. The difference between input and output voltage does drop across the MOSFET. The charger will adjust the MOSFET gate voltage to get the output voltage it wants (linear region of operation). The DPPM circuit intentionally reduces the magnitude of VGS to cause voltage to drop across the MOSFET. But even if it were to turn the MOSFET all the way on, the MOSFET still has some resistance. So, voltage would still drop proportional to current in that case (V = I * R). \$\endgroup\$
    – user4574
    Commented Jul 1, 2022 at 2:15

If the battery charger cannot meet load demand, then the battery supplies the extra current as the voltage decays.

I do not know why they don't sense battery current separate from charge current where the sum is load current. If the battery stays on charge continuously above 4V it can erode the lifespan. In this case, I would use a timer to truncate the charge time and keep the voltage at 3.9 to 4 V or 80% SoC until the excess capacity rises above 50%. But this is a tradeoff between short-term gain and significant long-term loss of life cycles.


  • \$\begingroup\$ Ok, but why the output voltage is decaying? \$\endgroup\$ Commented Jun 30, 2022 at 1:44
  • \$\begingroup\$ Because battery charge is being depleted by the excess load \$\endgroup\$ Commented Jun 30, 2022 at 3:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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