I am not that familiar with electronics, it's just a hobby.

I am reviewing the TP4056 and DW01A ICs that are used for battery charging.

Actually there is nothing special about these ICs but I was wondering how the red marked area works.

I know the two DW01A pins prevent overcharging and discharging, but how does it work? I searched for an explanation but nothing found.


  • \$\begingroup\$ merci infiniment pour cette courageuse explication d une manière très claire \$\endgroup\$ – robert goriot Dec 12 '20 at 8:18

enter image description here

Monitoring the battery voltage
The DW01A monitors the battery voltage using its power supply pins (pin 5 and 6 of DW01A ) as shown above. The comparators shown on the left side trigger in case of overcharge (battery voltage too high) or overdischarge (battery voltage too low).

The battery voltage is not measured directly. There is an RC filter (R5 & C2) that filters out spikes in the battery voltage due to e.g. inrush currents of the load.

Monitoring the battery current
When the battery is being discharged, the battery's current will flow B+ to OUT+ through the load, entering OUT-, flowing through the dual n-channel mosfet FS8205A back to B- as shown with the red arrowed line.
When the battery is being charged, the charge current flows as shown with the blue arrowed line. (I omitted the load current, note the load current is not measured while charging.)

enter image description here

The FS8205A has a drain-source on-resistance \$ R_{DS(ON)} \$. The current through this resistance will cause a voltage drop with respect to GND (pin 6 of DW01A), which is applied via R6 to pin 2 of the DW01A(1). The comparators on the right side of the DW01A all have a small voltage source on the + input, so they can measure positive and negative currents through the FS8205A.
The charger detector, short circuit detector and overcurrent detector will trigger based on the voltage drop across the FS8205A. So, the \$ R_{DS(ON)} \$ of the FS8205A determines at which current this will happen.

(1) Theoretically, there will be no current flowing through R6, and therefore no voltage drop across R6, because the inputs of the internal comparators is theoretically infinite. In practise, there will be a (negligible) leakage current. The purpose of R6 is to protect pin 2 of DW01A against ESD.

  • \$\begingroup\$ Thank you very much for your detailed explanation, it helped me very much! \$\endgroup\$ – J. Knabenschuh Jul 3 '19 at 9:52
  • \$\begingroup\$ Perfect thank you \$\endgroup\$ – AndrewT Aug 17 '20 at 3:09

over pin 2 and 6 the DW01A sense the votlage drop over the FS8205, if the voltage drop is to high it will detect over current or short circuit and so on.

However it's the voltage drop over the MOSFETs that sets the trip point! and the voltage drop over the mosfet varies due to that the RDS(on), varies depending on different operations modes like change in temperature, due to higher load for example.

So you choose different MOSFET depending on the trip point of the battery protection circuits fixed points, that usually is fixed.

  • \$\begingroup\$ Nice addition to mention the dependency of \$ R_{DS(ON)} \$ on amount of current, temperature, etc. I decided to not mentioning it in order to reduce the wall of text. \$\endgroup\$ – Huisman Jul 3 '19 at 10:20

Thanks Huisman et al. for thoroughly explaining the workings of the DW01A + FS8205 schematic.

Having arrived here with the same question as the OP in mind, I think I can give a more precise answer towards the nature of his original question. Because, how can this Dual N-Channel FS8205 work with only its gates and sources connected to the outside world? I was confused by this as well, because it's not the familiar Half H-Bridge P-Channel+N-Channel MOSFET combination you usually see.

It turns out that the symbolic representation of the FS8205 in the OPs schematic (which you will find all over the internet), is incomplete. The actual symbolic representation of the FS8205 (taken from the datasheet) is:


As you can see, the FS8205 internally has two parallel discharge diodes. It is these two discharge diodes which, depending on direction of the current, will bypass either one or the other MOSFET, making the shared drains connected to the outside world.

  • \$\begingroup\$ The diodes are not necessary for the purpose being discussed. Each FET can be turned on by application of the correct voltage between gate and source. Each FET DS connection is then a resistive low-Ohms path and there is thus a conductiove path from S1 to S2 \$\endgroup\$ – Russell McMahon Apr 20 at 23:35

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