P channel mosfet in series

[jom]

Hi,


can i connect a p channel mosfet in series with both the source and gate tied?

how does the biasing of the circuit happen ? the circuit is shown in the attachment.



thank you

 

[jpanhalt]

Remember the adage that just because two things can be connected doesn't mean they should be.

The best way to get productive answers is for you to describe what you want to accomplish.

For example, are you trying to make some sort of push/pull driver?

John

 

[jom]

ctually i had to disconnect the load when the battery goes low which is monitored by a comparator which sends a high signal to the CTL pin of 4412, but when this happens the mosfet F2 diode conducted so to overcome that i used F1 in series,

can i use it like this ?? or can i just change the soucre and drain of F2 ??i have attached the schematic.

 

[jom]

Remember the adage that just because two things can be connected doesn't mean they should be.

The best way to get productive answers is for you to describe what you want to accomplish.

For example, are you trying to make some sort of push/pull driver?

John

I m trying to build a circuit to cut the battery supply when the voltage is low.so initially i had only one mosfet F2 but the diode in it conducts even if gate is off , so the load draws current frm battery to over come that i connected the other mosfet...

or can i just reverse the drain and source of the mosfet ??

 

[jpanhalt]

If the intrinsic diode of the P-channel mosfet is conducting from the V+ to the load and ground when the mosfet is off, you have the mosfet connected backward. With P-channel, the source goes to the positive supply.

See these drawings: https://en.wikipedia.org/wiki/MOSFET

EDIT: Here's a better discussion for your purpose: https://www.electro-tech-online.com/custompdfs/2013/01/70611.pdf

John

 

[jom]

so just reversing the source and drain of the lower mosfet will solve the problem of intrinsic diode??

 

[jom]

thanks for that

 

[jom]

is there any problem if i connect the source to the battery ???

 

[jpanhalt]

is there any problem if i connect the source to the battery ???

Of course not, but it makes a difference which terminal of the battery you connect it to.

First, be absolutely sure you properly identify the source, drain and gate pins of your mosfet. If in doubt, read the datasheet.

Second, yes, with a PNP, the source is often connected to the positive terminal of the battery or power source.

Did you look at the two links I gave you? The second one, in particular, shows exactly that configuration.

John

 

[jom]

will the mosfet above ie F1 whether it will turn on ?? wen the mosfet F2 is on ???

 

[jom]

Of course not, but it makes a difference which terminal of the battery you connect it to.

First, be absolutely sure you properly identify the source, drain and gate pins of your mosfet. If in doubt, read the datasheet.

Second, yes, with a PNP, the source is often connected to the positive terminal of the battery or power source.

Did you look at the two links I gave you? The second one, in particular, shows exactly that configuration.

John

i did try using the online schematic editor and simulation using circuitlab.com ,when the battery supplied 9v and the gate s wer at 7v then only one mosfet on the left turned ON , the simulation result is shown on the left side of the attachment, the OUT node is zero even after the first mosfet is ON .In my design i use a p channel mosfet FDN306P, the symbol used in that editor is for P channel only.

Thanks

 

[jom]

as i lowered the gate voltage it conducted well ... here is what i did .

 

[jpanhalt]

After all of this discussion, why are you still trying to use that failed design? Try using a single P-channel mosfet in the proper orientation.

John

 

[jom]

After all of this discussion, why are you still trying to use that failed design? Try using a single P-channel mosfet in the proper orientation.

John

if i use a single p mosfet it would still conduct from drain to source if i connected drain to battery ,but if i connect battery to the source the direction of diode will be reversed and it will not conduct but when there is a voltage from the main supply the diode will be forward biased and will conduct and flow to battery ...am i correct ?

 

[jpanhalt]

My Post #13 was written in response to your Post#11. I was writing while you were posting #12.

Simply put, if you look at the circuit in #12, why do you have M11 and its diode? (I am assuming you cannot find the proper symbol for this mosfet, so D7 and D8 were added to represent the intrinsic diodes.)

John

 

[jpanhalt]

I get the feeling that there are some things in your design that are not shown in the schematic, like another power supply.

If you want to stop that hypothetical supply from "back charging" the battery when it is in the circuit, which appears to be why you have M11 and its intrinsic diode, then you can just eliminate M11 and keep a diode as you have shown. So long as that other supply's voltage is greater than the battery's voltage, current will not flow back into the battery. That, of course, begs the question of why you need either mosfet.

It is always confusing to try to discuss a circuit that is not shown. Schematics are a great bridge to solving language problems. Please post the whole schematic related to the battery and other power supply, if one exists.

John

 

[jom]

sorry i will post the schematics, and yes i wanted to use that diode to show the intrinsic diode of fet but that editor's symbol does not show the diode so i used ..but i could simulate it..the voltage drop across the diode will be more if i use a diode in place of the lower mosfet .i have attached my schematic.

 

[jpanhalt]

So, you appear to want a dual supply (Vbatt and Vin). Why not replace everything in the red outline with a diode. Thus the battery would go to diode anode (+), and the cathode ("bar" side) would go to the cathode of the diode from Vin. In other words, the diode from Vin (D11) does the same thing as this proposed new diode from the battery would do.

When Vin > Vbatt, then Vin will supply power to the rest of the circuit. When Vbatt > Vin or Vin is absent, then Vbatt will supply the power.

John

 

[jom]

So, you appear to want a dual supply (Vbatt and Vin). Why not replace everything in the red outline with a diode. Thus the battery would go to diode anode (+), and the cathode ("bar" side) would go to the cathode of the diode from Vin. In other words, the diode from Vin (D11) does the same thing as this proposed new diode from the battery would do.

When Vin > Vbatt, then Vin will supply power to the rest of the circuit. When Vbatt > Vin or Vin is absent, then Vbatt will supply the power.

John

View attachment 69977

i have added those circuits to cut the battery supply when it is low, that is in the absence of the main supply vin,to prevent the complete drain out of battery if the mains is not connected .

 

[jpanhalt]

OK, but you still only need one mosfet to shut down the battery when voltage is to low. The LTC4412 datasheet shows how that should be arranged.

Where are you getting the reference voltage for the comparator? A comparator and logic-level mosfet would be pretty simple.

Have you considered the LTC1998, which is a single package designed for this purpose? I am sure there are other devices that do the same. That part number was just the first to pop up.

John