Series MOSFETs possible floating source

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psnee1

Joined Sep 28, 2010
6
Hi

I have a Low-side driver application where we switch a solenoid on/off with a low-side N-channel mosfet. For failure redundancy we're actually using two N-channel mosfets in series. Someone pointed out to me that when both mosfets are commanded off, both their gates are grounded, and the bottom mosfet's source is grounded, but the upper mosfet source (bottom mosfet drain) is floating. They mentioned this could actually damage the top mosfet, because that floating node could float up over 20V (max Vgs rating). The load these are switching is tied to 28V.

Is this a real threat? I tried to attach a simplified schematic.

circuit.png
 

ScottWang

Joined Aug 23, 2012
7,397
About the Vgs problem that you just in parallel with a zener diode as 12V~15V and in series with a resistor as 1K~10K from input, but if you only input 0V and 10V then there is no Vgs problem, but you need to concern about the Vds rating voltage up to Vds = 28V * 1.5 = 42V, at least Vds = 28V*1.2 = 33.6V, the L1 needs to in parallel with a diode for discharge the back emf.
 

shortbus

Joined Sep 30, 2009
10,045
For failure redundancy we're actually using two N-channel mosfets in series.
While I'm no expert, that doesn't make sense to me. Wouldn't it be better, as far as redundancy goes, to put the mosfets in parallel? In series if one fails the other is just there, it can turn on but it can't conduct. Or am I missing something?
 

ScottWang

Joined Aug 23, 2012
7,397
While I'm no expert, that doesn't make sense to me. Wouldn't it be better, as far as redundancy goes, to put the mosfets in parallel? In series if one fails the other is just there, it can turn on but it can't conduct. Or am I missing something?
The circuit was changed from two NPN bjts and that is a AND gate function.
 

crutschow

Joined Mar 14, 2008
34,280
if you only input 0V and 10V then there is no Vgs problem
Yes, there can be.
If M1 has slightly more off leakage than M2, then M1's source can go to near 28V which will exceed its Vgs(max) rating.
Wouldn't it be better, as far as redundancy goes, to put the mosfets in parallel? In series if one fails the other is just there, it can turn on but it can't conduct.
Depends upon if the likely failure is a short or an open.
To cover both possibilities you need four MOSFETs in series-parallel.
 

ScottWang

Joined Aug 23, 2012
7,397
Yes, there can be.
If M1 has slightly more off leakage than M2, then M1's source can go to near 28V which will exceed its Vgs(max) rating.
Do you mean that when the M1=0V and the Ids1 leakage current causes Vds close to 0V, so Vgs1=0, and Vs1=28V?

If do so then do you have any information to support this point, because this is a big issue when we use the MOSFET still need to worry about the leakage current may causes Vds = 0V.
 

Bordodynov

Joined May 20, 2015
3,177
I will give an explanation of my schemes. The left side is the original one. I took transistors with a breakdown voltage of 20V. As it turned out with identical leaks, the worst case is when transistors turn off. The right part is a scheme in which all problems are solved.
But the same scheme with a leak of upper transistors is 100 times greater than the lower ones.
Look please, I brought all the necessary voltage.
Draft762.png
 

ebp

Joined Feb 8, 2018
2,332
"The right part is a scheme in which all problems are solved."

If M4 fails short-circuit, which is presumably what the TS is concerned with, the load will be turned on. Except for distributing the voltage between M3 and M4, which is hardly something that is necessary given the huge choice of FETs on the market, M3 does nothing that a piece of wire can't do better. M3 must be actively switched to make it useful in terms of redundancy.
 

Bordodynov

Joined May 20, 2015
3,177
"The right part is a scheme in which all problems are solved."

If M4 fails short-circuit, which is presumably what the TS is concerned with, the load will be turned on. Except for distributing the voltage between M3 and M4, which is hardly something that is necessary given the huge choice of FETs on the market, M3 does nothing that a piece of wire can't do better. M3 must be actively switched to make it useful in terms of redundancy.
See
Draft764.png
 

ebp

Joined Feb 8, 2018
2,332
That doesn't help. If M4 fails short circuit the gate of M3 can still go to 12 volts which is more than enough to turn it on. M4 failing short-circuit is essentially indistinguishable from it simply being ON, so you can't do anything relative to M4's drain with regard to pulling M3's gate down or M3 will turn off if M4 is ON.
 

crutschow

Joined Mar 14, 2008
34,280
Do you mean that when the M1=0V and the Ids1 leakage current causes Vds close to 0V, so Vgs1=0, and Vs1=28V?
Yes.
do you have any information to support this point
What information do you need?
Leakage current is a standard phenomenon and it does vary between parts.
Can't you just connect a 1-10 Meg resistor between the top FET source and ground, to prevent floating?
That should work, depending upon the leakage value.
 

ScottWang

Joined Aug 23, 2012
7,397
What information do you need?
Leakage current is a standard phenomenon and it does vary between parts.
Just to prove that the leakage current could causes Vds close to 0V, any information or data from datasheet or application note ...

Like at the other day, when I talk to someone about your point and then I have to take some information about this.
 

ebp

Joined Feb 8, 2018
2,332
For a circuit such as this, if it is critically important that the load not be energized due to a fault, it is usually best to start right at the beginning with the control signal. If the control is coming from outputs from a microcontroller, one way of reducing the likelihood of spurious turn-on is to use two output pins, one that must be asserted HIGH and the other asserted LOW. This is not absolute guarantee, but it does mean that no matter what happens at power up or power down or during reset, the probability of both signals being at their assertion level is reduced. To extend the fault tolerance to the power path would then require the two FETs to be driven with independent circuits.

From the wording of his original post I've been assuming that the TS's big concern is that one FET might short, not that there is need for higher voltage withstanding capability.
 

Bordodynov

Joined May 20, 2015
3,177
That doesn't help. If M4 fails short circuit the gate of M3 can still go to 12 volts which is more than enough to turn it on. M4 failing short-circuit is essentially indistinguishable from it simply being ON, so you can't do anything relative to M4's drain with regard to pulling M3's gate down or M3 will turn off if M4 is ON.
I agree that a short circuit of any transistor will lead to an accident. But the reliability of the circuit is determined by the margin of the operating voltages. But if you are so stubborn, then replace each transistor with a pair of transistors that are connected in series. What do you say about this scheme:
Draft765.png
 

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