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Output Voltage Selector

collatz

Apr 11, 2016
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Hi!

I have two P channel MOSFETs, one to a 6 volt source and one to a 4 volt source. I want to be able to select which voltage is supplied to the load.

However, in my circuit, I am pushing current from the 6 volt supply back through the 4 volt supplies controlling MOSFET, even when the MOSFET is fully off.

In other words, on my board, if I have just the first MOSFET I can turn on and off the 6 volt supply to the load.
If I have just the second MOSFET I can turn on and off the 4 volt supply to the load.

If both MOSFETs are installed simultaneously, the 6 volt supply starts working extra hard (apparently sending current to the 4 volt supply) even though the 4 volt MOSFET is OFF.

I don't know what I'm doing wrong!!!

Please help me!
 

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duke37

Jan 9, 2011
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Mosfets contain an inherent diode which will conduct when the fet is used in the 'reverse' connection so the higher power supply will supply power to the lower supply.
The way to stop this is to put a diode in series with the lower supply but this will give a voltage drop of about 0.6V, leaving you with 3.4V to play with.
 

Alec_t

Jul 7, 2015
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Try using a respective pair of back-to-back PFETS for each supply, like this (only one pair shown):-
PFETswitch.PNG

Edit:
One of the FEts shown should be flipped left to right so that they are back-to-back.
 
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AnalogKid

Jun 10, 2015
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Alec, you want both sources tied together in the middle. What you show has the two internal diodes in series, but one has to be reversed with respect to the other.

ak
 

Alec_t

Jul 7, 2015
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Oops! Well spotted. Forgot to flip the Spice symbol.
 

collatz

Apr 11, 2016
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Thanks everyone!

I've made the circuit change. Rather than using two opposing pairs I only did so on the lower voltage circuit. So when the 6 volt supply is on it goes to the load and can't go back through to the 4 volt supply. When the 4 volt supply MOSFET pair is on, 4 volt goes to load, but can't go through the single 6 volt mosfet.

So far so good.

BUT now, in the event that both MOSFETs are simultaneously active (which unfortunately I cannot guarantee that such a case will not occur, but wasn't sure that it would matter anyway), it seems to destroy the MOSFET pair, any idea what could cause this??

My mosfet drivers use a 7.8 volt boost converter, so the MOSFETs' gates are driven from 0v (on) to 7.8v (off), could the two MOSFETs in series plus the fact that when the 6 volt source and 4 volt source are both being "outputted" be changing what the MOSFET sees as it's gate to source voltage? The max, according to the datasheet is +/- 8 volts. I don't see how I could be exceeding this. Typically, in my circuit (since the current can peak at 20 amps) the MOSFETs catch on fire when they fail, in this case it seems like it just burned out (no longer turns on...)
 

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duke37

Jan 9, 2011
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Q2 and Q3 are in series. There is no need for this, one fet should be adequate to switch 4V. When one transistor is on, so is the other so all you are doing is killing two transistors rather than one.

To stop reverse current, you need one n channel and one p channel fet in series with appropriate switching.
Look up H bridge.
 

AnalogKid

Jun 10, 2015
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Q2 and Q3 are in series. There is no need for this,
Yes, there is.

The two FETs are connected such that their internal parasitic diodes are back-to-back. This prevents reverse current flow when the FETs are off. Because it uses two identical transistors and only one drive signal, it is considerably easier than implementing a full or half bridge circuit. It also allows one end of the load to be connected directly to a power rail. Often this is a requirement.

ak
 

duke37

Jan 9, 2011
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Yes, there is.

The two FETs are connected such that their internal parasitic diodes are back-to-back. This prevents reverse current flow when the FETs are off. Because it uses two identical transistors and only one drive signal, it is considerably easier than implementing a full or half bridge circuit. It also allows one end of the load to be connected directly to a power rail. Often this is a requirement.

ak
Ah, I didn't spot that one transistor was up-side down.:(
But the drive needs to enhanced?
To turn them both on, the gate needs to be positive with respect to the source. What drive is needed when the source and drain are interchanged?

It would help if the power supplies were labelled with their polarity.
 

Alec_t

Jul 7, 2015
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Also, the OP wants to be able to select which source connects to the load, so an active switch is required.
 

collatz

Apr 11, 2016
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Hey guys! You are all so helpful! I definitely wished I could have just used a diode, this thing just keeps getting more complicated. I am prototyping these revisions on an already made circuit board. All parts are surface mount QFN...and 0402 resistors and caps... I've been lucky to connect new wires to vias, as I can't solder under the QFN chips! The whole circuit board measures 450 mm^2 / 0.697 square inches.

Anyway, so since I don't think there is any way for me to programmatically ensure that both the 6 volt and 4 volt select lines are not simultaneously active, I've added a NOT gate and NAND gate (EDIT!! I mean NOR gate) to apply the rule to the "4 volt select" MOSFET driver, which should work, but since board space is a real concern, I don't know if I'll be able to add in CMOS / TTL type gate chips... I might have better luck with adding a few transistors to make the "rule". It might actually be smaller. I don't know...

I really wish I could implement the rule (no 4 volt if 6 volt is triggered) in the microcontroller firmware, but the 4 volt line is controlled by an output compare timer, and I don't think I can disabled and reconfigure back and forth fast enough to not cause glitching.

Does anyone have any tricks?? =)
 

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Alec_t

Jul 7, 2015
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If you haven't got room for those logic gates, perhaps use a BJT to clamp the 4V drive enabling signal, like this?
6Von4Voff.PNG
 
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