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PMOS device reversed

Q

QQ

Jan 1, 1970
0
Hi all,

Is there any problem with using an on-chip (integrated) pmos transistor
upside-down..ie current flows from drain to source instead of source to
drain. I know that there will be leakage current from drain to source
in the off state and I can live with that in this application. But will
there be any other problem? Is there a possibility of permanent damage
to the device?

I greatly appreciate any advice.

Thanks
QQ
 
W

Winfield Hill

Jan 1, 1970
0
QQ wrote...
Is there any problem with using an on-chip (integrated) pmos transistor
upside-down..ie current flows from drain to source instead of source to
drain. I know that there will be leakage current from drain to source
in the off state and I can live with that in this application. But will
there be any other problem? Is there a possibility of permanent damage
to the device?

No problem, we do it all the time. Keep in mind, if the reverse
ON voltage exceeds a diode drop, the MOSFET's intrinisic substrate
diode will start conducting, in parallel with the ON FET. This is
fine, except the diode usually has a reverse-recovery-time issue
that can affect fast H-bridge switching, or example.
 
Q

QQ

Jan 1, 1970
0
Thank you for your reply sir. (I am a fan of A-of-E!)

I know that the sync fet of a buck regulator is sometimes expected to
reverse conduct during the dead time so I know that discretes can do
that. I just wasnt too sure if it was safe to do so with a pmos device
which is sharing the die with other components as in the case of an ic.

Thanks
QQ
 
P

Pooh Bear

Jan 1, 1970
0
QQ said:
Thank you for your reply sir. (I am a fan of A-of-E!)

I know that the sync fet of a buck regulator is sometimes expected to
reverse conduct during the dead time so I know that discretes can do
that. I just wasnt too sure if it was safe to do so with a pmos device
which is sharing the die with other components as in the case of an ic.

Hmmmm . An IC ? You want to check if that reverse voltage is going to cause
any other problems. It might turn on parasitic diodes in an IC with
unpredictable results.


Graham
 
Q

QQ

Jan 1, 1970
0
Thanks Graham, I was concerned about the same but I am not sure if it
will or will not turn on any parasitic diodes given that all devices
are sharing a common p-substrate. Any specific idea on what parasitics
may be prone to turn on?

Thanks
QQ
 
W

Winfield Hill

Jan 1, 1970
0
QQ wrote...
Thanks Graham, I was concerned about the same but I am not sure if it
will or will not turn on any parasitic diodes given that all devices
are sharing a common p-substrate. Any specific idea on what parasitics
may be prone to turn on?

Yes, I agree it could be a problem. If you carefully monitor the
chip's supply current with and without the FET's parasitic diode
conducting, that may provide a clue.
 
J

Jim Thompson

Jan 1, 1970
0
Thanks Graham, I was concerned about the same but I am not sure if it
will or will not turn on any parasitic diodes given that all devices
are sharing a common p-substrate. Any specific idea on what parasitics
may be prone to turn on?

Thanks
QQ

The "common p-substrate" will be at the body potential of all NMOS
devices... desirably the most negative potential on-chip.

PMOS devices are built in an N-well, thus forward biasing the body
diode will activate a vertical PNP.

If enough current flows to regenerate, this is called LATCH-UP.

PMOS devices that need to handle bi-directional currents usually have
their well (body) tied to +VDD.

...Jim Thompson
 
Q

QQ

Jan 1, 1970
0
Hi Jim,

Thanks for your reply. Through my reading up on the net I was coming to
the same conclusion but the information is quite hard to find. Would it
help if this device was a "high voltage" pmos with a N buried layer?
Would the N buried layer effectively cut off the vertical pnp
transistor? (I read this somewhere..)
PMOS devices that need to handle bi-directional currents usually have
their well (body) tied to +VDD.

I dont understand. How will this help to cut off the vertical pnp?. My
source is supposed to be connected to the highest potential already. Is
it sufficient to connect body to source?

Will this problem occur in the case on NMOS device as well?

Thanks
qq
 
J

Jim Thompson

Jan 1, 1970
0
Hi Jim,

Thanks for your reply. Through my reading up on the net I was coming to
the same conclusion but the information is quite hard to find. Would it
help if this device was a "high voltage" pmos with a N buried layer?
Would the N buried layer effectively cut off the vertical pnp
transistor? (I read this somewhere..)


I dont understand. How will this help to cut off the vertical pnp?. My
source is supposed to be connected to the highest potential already. Is
it sufficient to connect body to source?

Will this problem occur in the case on NMOS device as well?

Thanks
qq

You originally indicated a desire to tie body to drain (rather than
source), which would provide the possibility of forward bias.

However, if you were really asking, "Are source and drain
interchangeable?"

The answer is yes (generally, some HV devices have different source
and drain structures), PROVIDED body is always most positive.

Draw some pictures to envision how the junctions are arranged.


...Jim Thompson
 
Q

QQ

Jan 1, 1970
0
Hi Jim

Okay I drew a picture and I see now that if the body is always
connected to the higher of the two potentials then the base of the
vertical pnp transistor is always at the highest possible potential. So
the base is at a voltage higher than the emmitter at all times thus
turning the vertical device off.

No I am not asking this..your original interpretation is correct. The
"source" end (this device is assymetric) is connected to the body and
the voltage at the source can go to a voltage equal to the highest
available on chip voltage. So tying the body to a voltage higher than
either source or drain is not an option.

Does having an N buried layer help in any way, to cut off the vertical
pnp? This is a HV device which has an n-buried layer (whose function I
dont know). The low voltage devices in the same process do not have
buried layers.

Thanks
qq
 
J

Jim Thompson

Jan 1, 1970
0
Hi Jim
[snip]

No I am not asking this..your original interpretation is correct. The
"source" end (this device is assymetric) is connected to the body and
the voltage at the source can go to a voltage equal to the highest
available on chip voltage. So tying the body to a voltage higher than
either source or drain is not an option.

No separate body connection?
Does having an N buried layer help in any way, to cut off the vertical
pnp? This is a HV device which has an n-buried layer (whose function I
dont know). The low voltage devices in the same process do not have
buried layers.

Thanks
qq

N+ buried layer _will_ cut the vertical beta, but I wouldn't count on
it for safety.

Can you describe your application in more detail?

What situation forces you into the potential risk of a forward-biased
body diode?

...Jim Thompson
 
Q

QQ

Jan 1, 1970
0
N+ buried layer _will_ cut the vertical beta, but I wouldn't count on
Could you please explain why does it cut the pnp? What risk still
exists?

I would like to but cannot because of confidentiality. Hope you will
understand.

Thanks
qq
 
Q

QQ

Jan 1, 1970
0
No separate body connection?

Sorry missed this...there is a separate body connection but there is no
available voltage that is higher than the maximum possible voltages at
source or drain, to which to tie it.

Thanks
qq
 
J

Jim Thompson

Jan 1, 1970
0
Could you please explain why does it cut the pnp?

Richer doped base _generally_ loses beta, but not always... depends on
the other doping levels.
What risk still
exists?

Latch-up. If your connection was an absolute necessity you could add
an N-well ring around the device tied to +VDD.

However I smell that you are using some kind of array where you don't
have the latitude afforded by a custom layout??
I would like to but cannot because of confidentiality. Hope you will
understand.

Thanks
qq

Well! You could hire me and I could sign an NDA ;-)

...Jim Thompson
 
J

Jim Thompson

Jan 1, 1970
0
Sorry missed this...there is a separate body connection but there is no
available voltage that is higher than the maximum possible voltages at
source or drain, to which to tie it.

Thanks
qq

I have used diodes from source AND drain tied to body:

--->|------|<---[D}

Whichever end is acting as the TRUE source determines the body bias.

(Since the body leakage is very low, this scheme will prevent forward
conduction of the body diode.)

...Jim Thompson
 
Q

QQ

Jan 1, 1970
0
Thank you for the advice. Right now the design is still in simulation
stages so I cannot try this.

Thanks
qq
 
Q

QQ

Jan 1, 1970
0
Latch-up. If your connection was an absolute necessity you could add
I think there may be some freedom with the process. Why will an N-well
ring tied to +vdd help? Is vdd meant to be the highest voltage on chip?

I can only afford free advice at the moment... :(

Thanks
qq
 
J

Jim Thompson

Jan 1, 1970
0
I think there may be some freedom with the process. Why will an N-well
ring tied to +vdd help? Is vdd meant to be the highest voltage on chip?
Yes.


I can only afford free advice at the moment... :(

Thanks
qq

Lack of information makes it difficult to give cogent advice.


...Jim Thompson
 
Q

QQ

Jan 1, 1970
0
Jim, thanks for your advice so far. This has been very useful to me. I
guess I will have to do some more research before deciding what to do.

Thanks
qq
 
R

Robert Baer

Jan 1, 1970
0
Winfield said:
QQ wrote...



Yes, I agree it could be a problem. If you carefully monitor the
chip's supply current with and without the FET's parasitic diode
conducting, that may provide a clue.
Yes; be aware that a lateral NPN device can exist, where the
substrate is the base.
Also note that a transistor is still active with zero base-collector
voltage, and still operational even with small forward B-C bias.
And do not let the lousy beta (0.001 in some cases) fool you; just
remember "total loop gain".
 
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