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This is a basic lithium battery protection circuit, but looking at the dual mos-fet part of the circuit, It doesn't make sense to me. It's a 8205A dual mos-fet, with its drain connected together and each of its source connected to the negative of the input and output. Normally, the drain of a mosfet is connected to the positive. in this case both are connected to the negative if they are turned-on. I don't understand! what voltage will be the drain with respect to the ground?

  • \$\begingroup\$ Ive seen this circuit a few places. Until now I have never see a 8205 mosfet reference in this circuit example. The 8205 has two mosfets but internally the drains are connected. That can not be used to build the circuit above. Or am I missing something? \$\endgroup\$
    – netskink
    Commented Feb 4, 2021 at 20:10

4 Answers 4


Here is some reference reading about MOSFETs.

When an "enhancement-mode" MOSFET gate charge exceeds \$VGS_{TH}\$, the channel from D to S goes into a low impedance or "closed" state. From the device's perspective, current can go through this channel either way - it doesn't matter - as long as \$VGS_{TH}\$ is met, current can flow.

Note that M2 cannot disconnect the battery during normal operation due to it's internal parasitic body diode, and this is the reason why M1 is required in this orientation (to block "forward" flow when off.)

In normal operation, both 8205A MOSFETS have their \$VGS_{TH}\$ met (usually far exceeded to "really turn them on hard"), so each device imposes their minimum \$RDS_{ON}\$ resistance (about 0.03 Ohms for this device.) So the voltage at the combined drains will be a resistor-divider's-worth above ground, say perhaps 0.05v, which of course will vary with current consumption. This is assuming you measure in a "forward" direction from the BATT- terminal to the drains.

Unlike BJT devices, there are no parasitic voltage drops (aside for \$VGS_{TH}\$) in MOSFETS, and the gate is not operated via current but voltage. The steady-state gate current \$I_{GSS}\$ is very small (100nA) as opposed to mA's for BJT's. However a MOSFET gate can have significant capacitance (\$C_{ISS}=600pf\$) especially for power devices. 600pF isn't a lot; some power MOSFETS are 10x that or more, but this makes it difficult when trying to switch them on and off very quickly.

  • \$\begingroup\$ sorry for my dullness. I still don't understand it. I have read but not comprehend that wiki page many times. according to it, When VGS > Vth and VDS ≥ ( VGS – Vth ), the fet is in Saturation or active mode; and since Vds is supposed to be 0 zero, how can it be bigger or equal to Vgs-vth? more doubtful than before, that you mentioned this internal parasitic body diode. since both side are negative, how do I know which side the current goes?? and a diode should have a forward voltage? how can there be enough forward voltage for that diode, both ends are ground, zero voltage?? \$\endgroup\$
    – user83582
    Commented Nov 30, 2015 at 9:00
  • 1
    \$\begingroup\$ @Atmega328 i think you confusion is that the active and saturation regions for a mosfet and bjt are opposite. Another thing to understand is that when Vgs rises Vds falls. With an increasing Vgs the mosfet turns on and as it turns on Vds decreases. As long as Vgs is not too high then Vgs-Vth is < Vds and the mosfet is in the saturation region. When Vgs gets high enough Vds will drop below Vgs-Vth and the mosfet will go into the linear region. \$\endgroup\$
    – vini_i
    Commented Feb 18, 2017 at 16:27

Think of the power MOSFET as a switch. As long as \$ V_{gate} \$ is more positive than \$ V_{source} \$, it'll turn on and as long as \$ V_{gate-source} \$ is much greater than \$ V_{threshold} \$ (\$ V_{gs} >> V_{th} \$) then the MOSFET will be fully on, i.e. it basically acts like a really low value resistor (tens of milliohms is a pretty common drain-source resistance for power FETs).

Power MOSFETs have body diodes present within their structure, this is a by-product of how they bare designed and it not always a desired feature (this being one of those times). To ensure that the battery can be completely isolated (i.e. power cannot flow in either direction) you need two back-to-back MOSFETs as any single power MOSFET can only block power flow one way while its body diode will allow power to flow in the reverse direction irrespective of whether the MOSFET is On or OFF.

Now although MOSFETs have these diodes that conduct in the reverse direction whether you want them to or not, the actual MOSFET itself doesn't care which way power is flowing and will happily behave like a low value resistor regardless of which way the power is flowing, the ideal MOSFET is a symmetric structure (the drain and source look exactly the same, and it is possible to make a MOSFET that has no body diode).

Another important point is that MOSFET terminology is a bit reversed compared to that of BJTs. With MOSFETs the "Saturation Region" is when the MOSFET is acting as a constant current source as the channel (the current carrying part) is... well, saturated in that it can't carry any more current (not without a higher gate voltage). While the "Active Region" is when the MOSFET is acting like a switch (it's like a relay, think "active" like "the relay is active"). The active (and cut-off or just plain "OFF") region is where almost all Power MOSFETs spend most their time. In the MOSFET active region, as long as the gate voltage is much higher than the threshold voltage, then you'd be hard pressed to tell a MOSFET apart from a piece of wire. The drain voltage is only important when either

  1. the MOSFET's about to breakdown because Vdsmax has been exceeded or

  2. the MOSFET's in the saturation region (where the MOSFET behaves like a constant current source) and you see a large voltage drop across Vds.

The current carrying channel of a MOSFET is symmetric so you could break one big MOSFET into lots of smaller MOSFETs in series all sharing the same gate and it'd behave the same as one big MOSFET (like how breaking a magnet gets you two smaller magnets), but because of the changing voltage along the channel, the "sources" of the tiny MOSFETs closer to the main Drain terminal are at a higher voltage than those further back. Raising the source voltage while keeping the gate voltage constant is the same as dropping the gate voltage while keeping the source constant, and seeing as a smaller \$ V_{gs} \$ leads to a reduced current carrying capacity, that's why the Drain-Source voltage is important too when operating a MOSFET in the Saturation Region.

Now that we've gotten some background info out of the way we can get to answering your question. As a few others have mentioned, the drains of that dual MOSFET will be pretty close to the same potential as the negative battery terminal. This isn't a problem, the MOSFET doesn't care about the absolute voltage (how would it know what it was anyway?), when "fully ON" it only cares about the relative difference between its gate and its source. So even if the source and drain were both at -387V and the gate was at -387+5V (-382V) then it's turn on quite happily and would allow current to flow in either direction with very low loss (makes for both efficient charging AND discharging). The reason we use two back-to-back MOSFETs in this situation is because we want to make sure that we can completely block power from flowing in either direction when necessary (which is a bit tricky with those pesky body diodes). Because both source terminals of the two back-to-back MOSFETs are connected together, it's really easy to put a positive bias on the gates as we can tie both gates together and drive them as one. And seeing as MOSFETs don't care which way power's flowing when they're fully ON, this arrangement basically makes something which looks just like a relay.

This turned into a much longer ramble than I expected and I may have missed a few things or used a few B-grade analogies. Hope this at least clears up how MOSFETs work and why someone would even want to put two back-to-back like in that battery protection circuit.


The MOSFET drains are connected so that the intrinsic diodes are in opposition to each other. If a MOSFET were reversed, you would lose the ability to protect against either over-discharge or over-charge. Here's the example where M1 is reversed (over-discharge protection fails).


simulate this circuit – Schematic created using CircuitLab

In this case, the protection circuit would detect the battery reaching its cutoff voltage, and would bring OD to ground, thus turning off M1. However, M1's intrinsic diode would still be conducting! Even if both MOSFETs were turned off, there is still a current path and the battery would continue discharging well past its safe discharge voltage. If you flipped M2, a similar situation exists where the protection circuit loses its ability to protect against over-charging.

To answer your question about what the drain voltage will be relative to ground (the battery's negative terminal), there are a few cases:

1) Both MOSFETs are on
In this case, the drain voltage will be a few millivolts above ground if discharging and a few millivolts below ground if charging, as determined by Ohm's Law and the \$RDS_{ON}\$ of M1.

2) Over-discharge protection has kicked in (M1 is off)
In this case, M2 is still conducting, so the drain voltage will be BATT-, plus a few millivolts (\$-I_{Battery}*RDS_{ON}\$ of M2) if the voltage across the BATT+ and BATT- terminals is greater than the protected battery's voltage plus the diode voltage of M1 (the battery is being charged).

3) Over-charge protection has kicked in (M2 is off)
In this case, M1 is still conducting, so the drain voltage will be ground, plus a few millivolts as determined by \$I_{Battery}*RDS_{ON}\$ of M1.

4) Both MOSFETs are off
In this case, the exact voltage for the drain pin will depend on BATT- and the leakage characteristics of the diodes, but will be within one diode voltage of ground or BATT-. Any more in either direction and one of the diodes would start conducting. For example, if BATT- were +5V, the drain voltage could ~4.5V, but not 4.0V, as M2's diode would begin conducting.

  • \$\begingroup\$ The circuit given by Alex to simulate appears to be wrong as it connects drain to source, while the device (8250 A) under discussion has both drains shorted. In this we may get wrong results. Suggested to correct the circuit for simulation. Thanks, BV Ramesh. \$\endgroup\$
    – BV Ramesh
    Commented Jan 16, 2018 at 5:54
  • \$\begingroup\$ That was sort of the point, to show the mosfets connected drain-to-source with their body diodes shown to demonstrate why they aren't connected that way in the actual circuit \$\endgroup\$
    – Alex
    Commented Jan 16, 2018 at 12:27

You need to read the datasheet of DW01-P: https://cdn.sparkfun.com/assets/learn_tutorials/2/5/1/DW01-P_DataSheet_V10.pdf You will understand that OD is the cut-off of discharge, OC the cut-off of charge.

You can also read the two questions I asked about the same circuitry: Over current protection for a 1-cell battery and Remove over-current protection of battery protection circuit


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