Hello everyone, I am currently designing a bidirectional high-side switch using a P-channel MOSFET for DIY BMS, and I came across a circuit tutorial based on an article from Texas Instruments (page 4). Here is the circuit from the tutorial:

My first response when looking at this circuit was, "Will the MOSFET's be floating?" because as far as I know, to create a P-channel high-side MOSFET switch, a pull-up resistor is needed as shown in the diagram below to prevent floating. Is the placement of the pull-up resistor in the first diagram correct?
i know that the placement of pullup resistor in these 2 diagram are the same, between gate and source. Because my understanding so far has been that the position of the pull-up resistor is between the gate and Vdd just like diagram below and i am open to criticism if my understanding are not correct

 

My first response when looking at this circuit was, "Will the MOSFET's be floating?"

No. Your definition of floating is too restrictive.
To be defined as floating, a node must have no connection to any other node so that its voltage is unknown, and that's not the case here.
When the control MOSFET of the top MOSFET circuit is off, the gates of the top MOSFETs are connected to their sources through the 10k resistor, which makes their gate voltages equal to the source voltage and keeps them both off.
That voltage may vary with respect to ground as determined by the voltages at A and B, but that does not make them "floating".

 

I think that this may be what You had in mind .........
.
.
.

.

 

I think that this may be what You had in mind .........
.
.
.
View attachment 299097
.

Since the MOSFETs in his circuit are rated for 50V, and there is no mention of AC voltages, I don't see a good reason for your suggestion to change the circuit, as it will switch low DC voltages just fine.
Also it seems unrelated to his question.

 

It depends on what the Tread-Starter means by "floating".
Sometimes "floating" and "isolated" are used interchangeably.
.
300-volts, not "50"
.

.
And, the ratings of the FETs are completely irrelevant to the general Circuit design suggested,
any N-Chan FET can be substituted for the ones in the picture.

The difference is the fact that there is complete isolation between
the Control-Voltage and any Voltages that may be present across the Switching FETs.

The Tread-Starters suggested Circuit does not have any "isolation".
This may be perfectly fine for some specific applications,
but usually, complete isolation is a requirement, or at least a bonus,
and being able to use N-Channel FETs,
rather than the P-Channel FETs required by the Thread-Starters Circuit,
is an additional added bonus.

The only drawback to the Circuit that I suggested is that it has
poor performance at very-high Switching-Frequencies,
but that can be fixed if necessary.
.
.
.

 

The Tread-Starters suggested Circuit does not have any "isolation".

And no mention was made of wanting that.
I think your convolving "floating" with "isolation" is not pertinent to the TS's question here, even if the added complication to achieve that may be a "bonus".

 

To clarify the meaning of "floating" as I intended, it refers to the condition where the MOSFET can turn on by itself without any voltage applied to its gate (for example, when it is touched by hand). I have tried to assemble the MOSFET circuit based on the first diagram, and it works well, solving my problem.
However, I am interested in the circuit shared by Mr.LowQCab here because this circuit uses an N-channel MOSFET which has a lower Rdson value than a P-channel. Does this circuit use an optocoupler to drive the MOSFET? The TI article also mentioned about N-Channel MOSFET application, but it using charge pump to drive the gate as shown below. Can optocoupler eliminate the use of charge pump?

 

The FDA215 is not an "Opto-Coupler" in the usual sense,
it is a "Photo-Voltaic" FET-Driver,
which creates it's own Voltage, but at a very low Current,
and since the FET Gates behave very similar to a Capacitor,
almost zero-Current is required to turn the FETs "On".

The Output-Voltage of a single FDA215 is around ~8-Volts,
this may not be enough Voltage to completely "turn-On" some FETs.
In that case, the Outputs of 2 FDA215s can be connected in series to provide around ~16-Volts,
at the expense of having to drive 2-LEDs instead of 1.
This should be more than adequate Gate-Voltage for complete turn-on of any common FET.
The 2-LEDs may also be connected in series if You have enough Driving-Voltage for
2 LED Forward-Voltage-Drops, which may be ~4 to ~5-Volts, at ~20 to ~30mA of Current.

You must check the Spec-Sheet of the FETs that You would like to use to
see what the Rds/on will be at a particular Gate-Voltage.
There will be a Graph that shows this relationship.

The lower the Rds/on, the less Heat will be generated.
Using 4 FETs in this configuration will reduce the Heat even further.
But keep in mind that the more Gate-Capacitance that the FDA215 must charge-up,
the slower the Turn-On-Time of the FET will be,
so High-Frequency-Switching is not practical with this setup.
.
.
.

 

The FDA215 is not an "Opto-Coupler" in the usual sense,
it is a "Photo-Voltaic" FET-Driver,
which creates it's own Voltage, but at a very low Current,
and since the FET Gates behave very similar to a Capacitor,
almost zero-Current is required to turn the FETs "On".

Thank you very much for your answer, it's very helpful to sate my curiosity. A photovoltaic driver sounds like the component I need now, but it's not available in my country, and I want to reduce expenses by using components sold here. Can using an optocoupler with an external voltage source be a solution?

 

A useful answer would require a detailed explanation of what your project is intended to do.
.
.
.