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Options for triggering a mosfet

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ffddoo

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Hi!

I'm designing an usb circuit using a FT232. The circuit gets it's power from the 5V USB power and uses 5V and 3.3V (through a 5V to 3.3V LDO). When the USB is in idle I have to swith power off, so I think the best option is a P channel MOSFET.

I've a signal PWREN from the FT232, it is HIGH when the usb is suspended and LOW when the USB is ok (power enable).
If all the power signals in my system were 5V I would use a P channel mosfet to swith it off, the source of the mosfet to the 5V power USB, the PWREN to the gate of the mosfet and the drain to the rest of the circuit.

The problem is that PWREN is a 3.3V signal so It will switch off the mosfet ever.

I'm trying to figure out different ways of triggering the mosfet but I don't find good enough any of them.

Any ideas please?

Thanks in advance
 

Have you not looked at the FTDI application notes?
I am sure there is an application note that will help.

https://www.ftdichip.com/Support/Documents/AppNotes/DG232_20.pdf


yes I did. But none of them fits my system. The most similar is the one in figure 6.0. My FTI and my micro are powered at 3.3V as in the figure, but the mosfet in my sistem must be placed before the LDO because I'm powering 5V chips too, then the source of the mosfet is at 5V, but PWREN is only 3.3V when high, not enough to switch the mosfet off.
 

I'm trying to figure out different ways of triggering the mosfet but I don't find good enough any of them.

Any ideas please?

It seems you like to switch off the circuit 5V supply and the 3.3V (which will be off automatically when 5V is off).

You can try this simple solution.
It needs a red LED with a nominal current of 20mA (most popular).
PWREN will be connected to the cathode of the LED.
The LED anode will be connected to the PMOS gate.
Between the gate and the USB 5V supply, you add a 1K resistor.

Now, if PWREN is high the forward voltage of the red LED will be about 1.7V while passing a very small current (about 10uA). In this case, Vsg will be about 0.1V which will likely turn the PMOS off.

When PWREN is low Vsg= 5-1.8 = 3.2 V which lets the PMOS be on.

This is not a bright solution but with the right LED... it works ;-)

Good luck,

Kerim
 
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    ffddoo

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It seems you like to switch off the circuit 5V supply and the 3.3V (which will be off automatically when 5V is off).

You can try this simple solution.
It needs a red LED with a nominal current of 20mA (most popular).
PWREN will be connected to the cathode of the LED.
The LED anode will be connected to the PMOS gate.
Between the gate and the USB 5V supply, you add a 1K resistor.

Now, if PWREN is high the forward voltage of the red LED will be about 1.7V while passing a very small current (about 10uA). In this case, Vsg will be about 0.1V which will likely turn the PMOS off.

When PWREN is low Vsg= 5-1.8 = 3.2 V which lets the PMOS be on.

This is not a bright solution but with the right LED... it works ;-)

Good luck,

Kerim

I have been thinking in something similar: using a 1,5-2V zenner reverse polarized (instead of the led) but I was not sure if It was going to work because PWREN is not open drain and when PWREN is high, current is suppouse to be sourced by the pin, not sinked.
 

What is important is that PWREN voltage will not exceed 3.3V since the LED will drop 1.7V (some may drop a bit more) even at a very low current.
You are right PWREN is supposed to source but if the load doesn't need any current, PWREN won't source any current too. For example if the pin is left open how much it will source? zero current ;) . The important idea is that we don't let the voltage on it exceeds its maximum (3.3V in our case). and even if it is exceeded, the current that tries to enter it should be very small (we have about 300mV margin above 3.3V) so a few micro ampere won't harm.

Edited:
If you don't have a red LED you can replace it with 3 small diodes as 1N4148 (or equivalent) connected in series.

Edited:
To verify the idea, assemble the PMOS, 1K between its gate and source and the 3 small diodes. Then manually connect the cathode of the last diode to ground and check that the PMOS is fully on. Then insert 470K between the cathode (also of the last diode) and ground and check that the PMOS is off. I think you can do the second test without the 470K, instead you replace it by the two probes of a voltmeter (positive probe to cathode and the negative probe to ground). The meter will likely read a voltage about 3.3 V (obviously the exact value depends on the internal resistance of the voltmeter).
 
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    ffddoo

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I got you a better solution, it uses a N mosfet bidirectional level converter.
You can use any mosfet with a low gate threshold

control_5v_mosfet_with_3v3.gif

Alex
 
It looks very smart. If I understand it well: PWREN=0 makes VGS=3.3V so NMOS is on, otherwise PREN=3.3V makes VGS=0 so NMOS is off.
 

Yes, when PWREN is high (3v3) the Nmosfet Vgs will be 0v so it turns off and R7 works as a pull up and provides 5v to the gate of the Pmosfet and it turns off.
When PWREN is 0v then the Vgs of the N mosfet becomes 3v3 and conducts, the current so the drain becomes almost 0v and the P mosfet turns on

Alex
 

I see... so there is something between the gate of the NMOS and PWREN pin, not shown on the pic... right? I thought that PWREN is directly connected to the gate of Q9 (that is Vpwren = Vgs).
 

The gate is connected to the 3v3 power supply (as shown), the source should be connected to the PWREN

Alex
 

It looks very smart. If I understand it well: PWREN=0 makes VGS=3.3V so NMOS is on, otherwise PREN=3.3V makes VGS=0 so NMOS is off.

Isn't the source of Q9 connected to ground?
Isn't the gate of Q9 connected to PWREN pin?
So how the voltage of the gate and PWREN are not equal... ?!!!
If I am wrong than ... I think I have to stop working in electronics :???:

---------- Post added at 21:25 ---------- Previous post was at 21:22 ----------

The gate is connected to the 3v3 power supply (as shown), the source should be connected to the PWREN

OK... now I get it though I have to think of it more ;-)
Hmmm... It seems I am getting old... really :)
 
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The source of Q9 connects to the PWREN which provides a voltage of either 0 or 3v3.
The gate of Q9 is connected to the 3v3 power supply
when the source has SPIEN=0 then Vgs becomes 3v3(gate)-0v(source)
when the source has SPIEN=3v3 then Vgs becomes 3v3(gate)-3v3v(source)
Are you searching for excuses to stop working in electronics...;-)

Alex

---------- Post added at 21:31 ---------- Previous post was at 21:28 ----------

The operation of the N mosfet bidirectional level converter is based on
http://ics.nxp.com/support/documents/interface/pdf/an97055.pdf

and a more simple version
**broken link removed**

Alex
 

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