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'[EE]: MOSFET as a digital switch'
2003\02\10@142238 by Jai Dhar

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Hello all again,

I am nearing the end of my project (that as some of you may know, includes a
power supply that I happened to get a nice zing from) - and I have to make a
decision for the types of switches I am going to be using. Now obviously, I
could use a good ol' set of mechanical switches to turn the 5V/12V output on
and off etc... but this seems rather boring. So I thought since I am using an
f877 in there, I might as well try and do it digital - see if I can get it
working. So my question is that are MOSFET's the best solution for using a
switch controlled by logic-level outputs? Of course, I would use MOSFET
drivers.. but I have never used them before, so I'm not sure of the
advantages/disadvantages. I am aware that I would need to select mosfets
depending on the source/drain voltage (I think: which would range anywhere
from 5V to +25) , but is there anything else I need to take into
consideration? I have searched google for simple schematics using mosfets for
switching, but haven't found many - maybe because they aren't the best
solution to this type of problem. Anyway, any help is appreciated.

Jai



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2003\02\10@150211 by Rajiv Thakur

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Hi,

Sharing some of my experience.

I have seen mosfets mainly being used in Inverters on the DCside of the
transformer that converts 12 v battery Dc to 240 V AC (square output).. The
reason being they can handle high DC amperage (10 amps) and are good
switches as they are basically used to generate square waves in transformer
DC side based on IC sqaure waves fed through the controller/IC. Can be usedi
in parallel to reduce the current rating

But they can burn off with fire  if amperage is exceeded and require
wheeling diode circuitary for protection.

Rajiv Thakur


From: Jai Dhar <.....jdharKILLspamspam@spam@ENGMAIL.UWATERLOO.CA>
To: <PICLISTspamKILLspamMITVMA.MIT.EDU>
Sent: Tuesday, February 11, 2003 12:51 AM
Subject: [EE]: MOSFET as a digital switch


> Hello all again,
>
>  I am nearing the end of my project (that as some of you may know,
includes a
> power supply that I happened to get a nice zing from) - and I have to make
a
> decision for the types of switches I am going to be using. Now obviously,
I
> could use a good ol' set of mechanical switches to turn the 5V/12V output
on
> and off etc... but this seems rather boring. So I thought since I am using
an
> f877 in there, I might as well try and do it digital - see if I can get it
> working. So my question is that are MOSFET's the best solution for using a
> switch controlled by logic-level outputs? Of course, I would use MOSFET
> drivers.. but I have never used them before, so I'm not sure of the
> advantages/disadvantages. I am aware that I would need to select mosfets
> depending on the source/drain voltage (I think: which would range anywhere
> from 5V to +25) , but is there anything else I need to take into
> consideration? I have searched google for simple schematics using mosfets
for
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2003\02\10@161904 by John Snider

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On Mon, 10 Feb 2003 14:21:33 -0500, Jai Dhar <snip>
> working. So my question is that are MOSFET's the best solution for using
> a
> switch controlled by logic-level outputs? Of course, I would use MOSFET
> drivers.. but I have never used them before, <snip>
---------------------------------------
I use IRLZ Mosfets (ie IRLZ24) - logic level switched HexFets - switching
12 volts, using just a 100 ohm resistor from the port pin to the gate
(don't need drivers). They have a very low Rds (0.1 ohms),  and are rated
at 60volt / 17 amps.
(http://rocky.digikey.com/scripts/ProductInfo.dll?Site=CA&V=21&M=IRLZ24 )
About $1 CDN at Digikey.

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2003\02\11@080820 by i.tronics

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Hello John,

> I use IRLZ Mosfets (ie IRLZ24) - logic level switched HexFets - switching
> 12 volts, using just a 100 ohm resistor from the port pin to the gate
> (don't need drivers). They have a very low Rds (0.1 ohms),  and are rated
> at 60volt / 17 amps.

    I've been using IRF530's to switch 24V, "just a 100ohm resistor
from the port pin to the gate" also and it's working very well.

    Someone knows if this could lead to problems after some time?

    Best regards,

    Brusque

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2003\02\11@082309 by Spehro Pefhany

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At 11:06 AM 2/11/2003 -0300, you wrote:
>Hello John,
>
>>I use IRLZ Mosfets (ie IRLZ24) - logic level switched HexFets - switching
>>12 volts, using just a 100 ohm resistor from the port pin to the gate
>>(don't need drivers). They have a very low Rds (0.1 ohms),  and are rated
>>at 60volt / 17 amps.
>
>     I've been using IRF530's to switch 24V, "just a 100ohm resistor
>from the port pin to the gate" also and it's working very well.
>
>     Someone knows if this could lead to problems after some time?

You should be using a logic-level characterized part. This type is NOT
guaranteed to have very low resistance at 5V Vgs (it is guaranteed at
10V Vgs).  At 4V Vgs, the drain current could be as low as 250uA and still
meet specs.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
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2003\02\11@092933 by Jai Dhar

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Quoting Spehro Pefhany <KILLspamspeffKILLspamspamINTERLOG.COM>:

> At 11:06 AM 2/11/2003 -0300, you wrote:
> >Hello John,
> >
> >>I use IRLZ Mosfets (ie IRLZ24) - logic level switched HexFets - switching
> >>12 volts, using just a 100 ohm resistor from the port pin to the gate
> >>(don't need drivers). They have a very low Rds (0.1 ohms),  and are rated
> >>at 60volt / 17 amps.
> >
> >     I've been using IRF530's to switch 24V, "just a 100ohm resistor
> >from the port pin to the gate" also and it's working very well.
> >
> >     Someone knows if this could lead to problems after some time?
>
> You should be using a logic-level characterized part. This type is NOT
> guaranteed to have very low resistance at 5V Vgs (it is guaranteed at
> 10V Vgs).  At 4V Vgs, the drain current could be as low as 250uA and still
> meet specs.

So assuming that I used a logic-level hexfet for example, what are some
important parameters that I should be looking at - just the on resistance? And
will these parts work for AC also? From what I understand of MOSFET's, I don't
see why they won't... but everything is a bit more complicated in reality
compared to theory :-)



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2003\02\11@114351 by Adi Linden

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> You should be using a logic-level characterized part. This type is NOT
> guaranteed to have very low resistance at 5V Vgs (it is guaranteed at
> 10V Vgs).  At 4V Vgs, the drain current could be as low as 250uA and still
> meet specs.

Are there logic level MOSFET's in TO-92 or a similar small thru hole case
style?

Adi

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2003\02\11@115321 by Jai Dhar

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Most MOSFET's come in TO-220's, don't they?

Quoting Adi Linden <spamBeGoneadilspamBeGonespamADIS.ON.CA>:

{Quote hidden}

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2003\02\11@115619 by Mike Harrison

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Yes - e.g. Zetez ZVN/ZVP series. There are better and cheaper ones in SM though.
On Tue, 11 Feb 2003 10:48:42 -0600, you wrote:

>> You should be using a logic-level characterized part. This type is NOT
>> guaranteed to have very low resistance at 5V Vgs (it is guaranteed at
>> 10V Vgs).  At 4V Vgs, the drain current could be as low as 250uA and still
>> meet specs.
>
>Are there logic level MOSFET's in TO-92 or a similar small thru hole case
>style?
>
>Adi

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2003\02\11@143710 by Dwayne Reid

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At 11:06 AM 2/11/03 -0300, TakeThisOuTc.i.tronicsEraseMEspamspam_OUTterra.com.br wrote:
>Hello John,
>
>>I use IRLZ Mosfets (ie IRLZ24) - logic level switched HexFets - switching
>>12 volts, using just a 100 ohm resistor from the port pin to the gate
>>(don't need drivers). They have a very low Rds (0.1 ohms),  and are rated
>>at 60volt / 17 amps.
>
>     I've been using IRF530's to switch 24V, "just a 100ohm resistor
>from the port pin to the gate" also and it's working very well.
>
>     Someone knows if this could lead to problems after some time?

What is your load?  If the FET is enhanced adequately for your target drain
current - no problem.  The manufacturer's data sheets are your friend:
study the curve that shows Rds vs Vgs.  You want to make sure that a
different batch of FETs won't cause you problems just because they are
somewhat different than the FETs you are using now.

Personally - I'd use the 'L' version of the FET.

dwayne

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2003\02\11@143713 by Dwayne Reid

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At 10:48 AM 2/11/03 -0600, Adi Linden wrote:

>Are there logic level MOSFET's in TO-92 or a similar small thru hole case
>style?

<http://www.vishay.com/mosfets/>  Look for the VN series: vn2222, vn2106, etc.

dwayne

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Celebrating 18 years of Engineering Innovation (1984 - 2002)
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2003\02\11@162105 by Larry Bradley

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International Rectifier has some in a 4-pin DIP package, for example, the
IRLD024 - about $0.60 from Digikey - I have some on order.

RDSon of 0.1 ohms at 5 volts , 60V, 2.5 amp rating

Larry

At 10:48 AM 2/11/2003 -0600, you wrote:
{Quote hidden}

Larry Bradley
Orleans (Ottawa), Ontario, CANADA

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2003\02\11@163531 by i.tronics

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Hello Dwayne,

>>     I've been using IRF530's to switch 24V, "just a 100ohm resistor
>> from the port pin to the gate" also and it's working very well.

> What is your load?

    it's a solenoid that works at about 2A. When at maximum usage
(rarelly), it's turned on for 25ms ten times a second. When at normal
usage it's fired once each 2 seconds.

> If the FET is enhanced adequately for your target drain
> current - no problem.  The manufacturer's data sheets are your friend
> study the curve that shows Rds vs Vgs.

    That's exactly what I did. IRF's datasheet curves for the IRF530N
says about Id=20A for Vgs=5V. For the IRF530 numbers are not so
optimistic with Id=4A for Vgs=5V. There's no Rds vs Vgs curve, but I
assume it's no problem as I've read very low voltage across the MOSFET
when ON.

    Anyway, there's no sign of heating or other problems on the
IRF530's I'm using.

>You want to make sure that a
> different batch of FETs won't cause you problems just because they are
> somewhat different than the FETs you are using now.

    This should be no problem, because every unit is extensivelly
tested before leaving factory. What worries me is the possibility of
(say) in some months units returning because the solenoid isn't working
anymore or are not as powerfull as before.

> Personally - I'd use the 'L' version of the FET.

    You mean IRF530L? I can't find it on http://www.irf.com.

    Thank you,

    Brusque

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2003\02\11@205608 by William Chops Westfield

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>> Are there logic level MOSFET's in TO-92 or a similar
>> small thru hole case style?
>
> RDSon of 0.1 ohms at 5 volts , 60V, 2.5 amp rating

Doesn't "logic level" usually mean "well characterized at Vgs significantly
LOWER than 5V"?  In particular, you want the data for TTL "on" voltage spec
of 3.something, don't you?  (or is "real" TTL so rare these days that it no
longer matters?)

BillW

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2003\02\11@233818 by Spehro Pefhany

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At 05:54 PM 2/11/2003 -0800, you wrote:
> >> Are there logic level MOSFET's in TO-92 or a similar
> >> small thru hole case style?
> >
> > RDSon of 0.1 ohms at 5 volts , 60V, 2.5 amp rating
>
>Doesn't "logic level" usually mean "well characterized at Vgs significantly
>LOWER than 5V"?  In particular, you want the data for TTL "on" voltage spec
>of 3.something, don't you?  (or is "real" TTL so rare these days that it no
>longer matters?)

No longer matters. Logic level is usually characterized around 5V -
tolerance(4.5V perhaps), though I'm seeing some parts
now characterized for 3V-ish (2.7V/2.8) Vgs, but they usually have low
breakdown (eg. IRF7470) and will do better Rds(on) if driven with 4.5V.

Non-logic-level MOSFETs are usually characterized at 10V Vgs, which
requires a bit of extra circuitry to drive, especially if you want to
drive it fast.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
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2003\02\12@075542 by Olin Lathrop

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> Doesn't "logic level" usually mean "well characterized at Vgs
significantly
> LOWER than 5V"?

No, it usually means characterized at 5V.

> In particular, you want the data for TTL "on" voltage spec
> of 3.something, don't you?  (or is "real" TTL so rare these days that it
no
> longer matters?)

"Logic level" doesn't imply the FET has the same 1 and 0 thresholds as
logic (which varies a lot anyway), but that it can go from off to on with
a gate swing from 0 to the logic supply level.  In other words, it is
meant to work with the output of logic, not look like a logic input.


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2003\02\12@123350 by Russell McMahon

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> > Doesn't "logic level" usually mean "well characterized at Vgs
> > significantly LOWER than 5V"?

The term lacks some precision as I believe it was more a marketing term
which has caught on.

To satisfy most definitions, Zetex have some very nice small FETs with a 1.1
volt or so threshold (minimum turn on voltage). Most logic families will
drive these :-). Alas these may be available in SMT packages only. Typically
20v x 5A partsin pkg that can be lost if you sneeze. (very technical :-)).
I used one of these in a prototype self oscillating 1 cell to white LED
conveter recently with superb results. (2 bipolars plus FET/ more details in
due couurse).

Selection guide at

       http://www.zetex.com/3.0/b2-2.asp        <= ***** SUPERB devices
*****
       (even Roman will like the specs !   :-) )

MSOP-8, SO8, SOT23, SOT23-6 ...

They don't give Vth in table but all are specd at 2.5v and 4.5v

As an example the utterly superb ZXMN2A02X8 (where do they get these part
numbers from) claims

   Vth = 0.7v <- *********** Turns on with 0.7v drive  ***********
   20V
   7.6A
   1.8w
   Rdson 0.04r at Vgs = 2.5v / 0.02r at 4.5v
       (note while Vth is 0.7v it is still being enhanced more at 4.5v than
2.5v)
   MSOP8

Note that the Vds max mustn't be exceeded (sniff)
After hand soldering one of these in a non-PCB prototype layout it's a shame
to lose it :-(

Apart from price and size this would have to be one of the most useful
transistors available for amateur load driver use. Up to 7 amps with any
drive source in a truly tiny package.

Superb for dc to dc converters- but do watch Vds max !!!

There are lower rated parts available in even small packages.


       Russell McMahon

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2003\02\12@124631 by Spehro Pefhany

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At 06:33 AM 2/13/2003 +1300, you wrote:


>     Vth = 0.7v <- *********** Turns on with 0.7v drive  ***********

Do keep in mind that "turns on" in this case means that you get
250 microamps of drain current- not bad, it looks like a few K ohms, but
not in the ampere range.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
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2003\02\12@163410 by Ian McLean

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Yes.  Look at the 2n7000.  It has a very small base current turn on, and can
be driven from logic levels.  I have often used this as the first stage in a
cascaded mosfet switch.  The 2n7000 drives a larger Mosfet, like an MTP3055E
for driving high currents.  I have a schematic if you are interested.

Rgs
Ian

{Original Message removed}

2003\02\13@121058 by Micro Eng

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All this talk of FET's is bringing back some fuzzy memories of it when in
class.  Thats what happens when you just do digital design for years.

These are bidirectional devices right? Such that it doesnt matter if the
load is attached to the drain or source? Or does it matter?






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2003\02\13@121626 by Jai Dhar

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From what I understand from MY class (heh), you simply need the polarity
correct for current flow. I don't think they are truly 'digital' devices
(except with the logic level ones). But then again, this is my speculation -
I'm not an expert on this stuff.

Quoting Micro Eng <RemoveMEmicro_engspam_OUTspamKILLspamHOTMAIL.COM>:

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2003\02\13@123909 by Ned Konz

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On Thursday 13 February 2003 09:09 am, you wrote:
> All this talk of FET's is bringing back some fuzzy memories of it
> when in class.  Thats what happens when you just do digital design
> for years.
>
> These are bidirectional devices right? Such that it doesnt matter
> if the load is attached to the drain or source? Or does it matter?

Yes, because there is a "body diode" between D and S.

On a N-channel device, the body diode will conduct (even if the
transistor is OFF) when the S becomes more than a little positive
than the D.

P channel devices are reversed from that.

So when there's a real bidirectional current flow requirement, people
tend to use two FETs with their diodes back to back.

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2003\02\13@130541 by Micro Eng

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duh....of course!!!  Sometimes I feel like my light bulb is just a little
dim.


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2003\02\13@130748 by Chris Loiacono

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Of course any MOSFET can be used as a high or low side switch, but some
devices can be used bidirectionally, others not so well. Now let's see if I
can remember which is which...

>
> These are bidirectional devices right? Such that it doesnt
> matter if the
> load is attached to the drain or source? Or does it matter?

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2003\02\13@131207 by Jai Dhar

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Quoting Chris Loiacono <chrisSTOPspamspamspam_OUTMAIL2ASI.COM>:

> Of course any MOSFET can be used as a high or low side switch, but some
> devices can be used bidirectionally, others not so well. Now let's see if I
> can remember which is which...

What is the definition of a high/low switch... informally...

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2003\02\13@132442 by Spehro Pefhany

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At 01:10 PM 2/13/2003 -0500, you wrote:
>Quoting Chris Loiacono <spamBeGonechrisSTOPspamspamEraseMEMAIL2ASI.COM>:
>
> > Of course any MOSFET can be used as a high or low side switch, but some
> > devices can be used bidirectionally, others not so well. Now let's see if I
> > can remember which is which...
>
>What is the definition of a high/low switch... informally...

A high side switch switches the supply rail so that the load can be grounded.

The is especially desirable in automotive applications where one side of
load is typically grounded. You need a pair (one high, one low) to make a
half-bridge.

It's trivial (use a P-channel MOSFET) when the voltage is Vdd, but for
higher voltages (12V or 300V or 600V), you run into problems - for example,
there are no common P-channel MOSFETs rated for >500VDC AFAIK. Also IGBTs,
which come into their own at high voltages are only available in "N" type.
To use N-channel MOSFETs (say) as a high side switch you need a drive voltage
that is above the +ve rail.

An H-bridge is two half bridges, a 3-phase bridge is 3 half-bridges.

               A            B                C


            High side                      +V
         +V                                 |  Half bridge
          |   _/                            o
           -o/  o--                         '\  High
                  |                           \
                 .-.       +V               o  \
                 | |        |               |
            Load | |       .-.              .
                 '-'   Load| |              +---------+
                  |        | |              '     Out
                  |        '-'              |
                 ===        |               o
                 GND        |               '\   Low
                            o                 \
                            '\              o  \
                 Low side     \             |
                            o  \            |
                            |              ===
                            |              GND
                           ===
                           GND


(view with fixed pitch font only)

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2003\02\13@134555 by Chris Loiacono

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Theoretically, unlike the JFET or IGFET, MOSFET's, whether N or P channel
can be driven with either + or - Vg. That would imply bidirectional
application. Also theoretically, it should be safer to do so with
enhancement mode devices (that's the description I was looking for earlier).

I have designed also for Bipolar NPNs then slipped in N MOSFETs, since
another nice advantage is that you don't have to worry about gate leakage.
This has been a nice way to play with LED display brightness when needed,
due to the low Rds MOSFETs of today.

I always try to use N channel & P channels appropriately, but have seen all
N channel bridges and am itching to give this a try....Today it seems to be
a matter of how much complexity you want to make your gate
circuitry....which seems to be a matter of how hardened or protected it
needs to be. There are so many switching device types available that once
these decisions are made, it's fairly easy to read data sheets a make a
choice.

Chris

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2003\02\13@143004 by Russell McMahon

face
flavicon
face
> Theoretically, unlike the JFET or IGFET, MOSFET's, whether N or P channel
> can be driven with either + or - Vg. That would imply bidirectional
> application.

Not quite, I'm afraid.
An enhancement mode MOSFET (by far the most normal type) (never actually
seen a depletion mode MOSFET) is on when -

- For an N channel, the Gate is more positive than the Source.
- For a P channel, the Gate is more negative than the Source

by a suitable voltage level (see spec sheet) in each case.

In both cases the FET will then conduct resistively and bidirecionally
between Drain and Source.

However,, as others have noted, there is in each case a reverse "body diode"
across Drain & Source which is an inherent part of the FET and cannot be
eliminated. This will conduct when the FET is reverse biases Drain to Source
even if the FET is off.

As noted by others, this can be overcome by operating two FETS in series
drain1 in,  source1 to source2, drain 2 out.
Join gate1 to gate2. Drain1 & Drain2 are now bidirectional inputs. Drive
gate1&2 positive relative to source1&2 to turn on (for an N channel). The
gate-source supply needs to effectively float relative to the input/output
signals. This is a useful tool when necessary (I have used it when
desperate) but can be difficult to drive due to the floating gate signal.



       Russell McMahon

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2003\02\13@144446 by Chris Loiacono

flavicon
face
Actually, check your facts.. I don't ever want to argue on this forum, but I
just pulled an old text from the shelf, and it clearly explains how IN
THEORY, MOSFET's, both N & P channel will work with either positive or
negative gate bias. I have tested this years ago and found it to be true in
SOME cases, not so in others. What you mention is the normal application,
and some devices will not work both ways. Then, of course there is the
interal diode to consider...Well, Russ, I guess you didn't write all the
books...

But I did say that I refuse to argue just wanted to clear this up.

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2003\02\13@170626 by Russell McMahon

face
flavicon
face
> I just pulled an old text from the shelf, and it clearly explains how IN
> THEORY, MOSFET's, both N & P channel will work with either positive or
> negative gate bias.

No arguments here :-)

I mentioned enhancement and depletion mode FETs and that the overwhelmingly
most common form is enhancement mode. This means that they are OFF until
ENHANCED on. A depletion mode FET is on until DEPLETED off. This will all be
covered in the text books. If we are dealing with real world devices then we
have to run with what they make.

Also note that IF you could reasonably make an N Channel MOSFET which
operated with either positive or negative gate bias relative to the source
and IF it was about as good as existing real N Channel FETS then it would
seeep away almost all P Channel FET applications (as it does what they do as
well) and probably replace most N FET designs as well.  I'd buy them :-).

> Well, Russ, I guess you didn't write all the books...

Indeed. But I read all the data sheets that I can get hold of :-)

What people make and sell is the ultimate arbiter (alas) of what you have to
do to make them work. Current MOSFETS are sold as 2 quadrant devices at
present (ie gate with one polarity, drain with either) and the 2nd quadrant
is not overly useful in most cases due to the body diode.



           Russell McMahon

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2003\02\13@174845 by Sean Breheny

face picon face
Any ideas why almost all mosfets have the source and substrate (body)
internally attached? There is no fundamental reason why this needs to be
(mosfets on ICs often do not do this), and it would allow you to
(optionally) eliminate the body diode. The fet would be totally
symmetrical with respect to source and drain, as long as you kept the
body terminal tied to the lowest potential in the circuit (highest for
Pmos). Since many fets come in to-220 packages, they could easily make
the tab the body (or keep the tab as the drain and make the drain pin the
body).

Sean


On Fri, 14 Feb 2003, Russell McMahon wrote:

{Quote hidden}

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2003\02\13@195456 by Ian McLean

flavicon
face
This is quite correct.  Which way you drive them depends on the design.  You
can drive them either from the negative side (open drain) or from the
positive side (open source) - which way will depend on your switching and
driving requirements.

{Original Message removed}

2003\02\13@200629 by Ian McLean

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part 1 2950 bytes content-type:text/plain; (decoded 7bit)

Seeing as I have started a topic of some interest - I think for everyones
sake and the sake of not getting any more requests personally for the
circuit schematic, I'll just post it to the list.  Any comments are
welcomed.

This circuit is good for 12 amps.  It is negative side driven (open drain).
I drive it at 12V but it could go quite a bit higher.  The input can be
driven directly from a PIC port pin.

Their is a little impedance oscillation in motors to contend with (hence the
diode).  The cap (100n in this circuit) will need to be sized for the
circuit as this is what is controlled the rising and falling edges of the
output.  100n does for most applications - do not raise it much above this.

Rgs
Ian.

{Original Message removed}
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part 3 2 bytes
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2003\02\13@234015 by Tal

flavicon
face
The reversed body diode can be 'eliminated' as follows:

1. Take two mosfets (either both N or both P, N typically have lower
RDSon)

2. Connect the two sources together

3. Connect the two gates together

4. Connect a resistor (e.g. 30K) between the sources and the gates
(important)

5. Use the two drains as the two ends of the switch

6. Drive the gates with appropriate voltage (when designing the voltage
range, examine the voltage on the sources and forget to consider the
reversed diode, in this case they actually help to set the bias
voltage).

In OFF state, both mosfets will be off and one of the reversed diodes
will be off which will block the current.

In ON state, both mosfets will be on and the affective RDSon is double
the RDSon of each of the mosfets.

We have used it in several circuits and it worked well for us.

Tal


> {Original Message removed}

2003\02\14@080031 by Olin Lathrop

face picon face
> Seeing as I have started a topic of some interest - I think for
everyones
> sake and the sake of not getting any more requests personally for the
> circuit schematic, I'll just post it to the list.  Any comments are
> welcomed.
>
> This circuit is good for 12 amps.  It is negative side driven (open
drain).
> I drive it at 12V but it could go quite a bit higher.  The input can be
> driven directly from a PIC port pin.
>
> Their is a little impedance oscillation in motors to contend with (hence
the
> diode).  The cap (100n in this circuit) will need to be sized for the
> circuit as this is what is controlled the rising and falling edges of
the
> output.  100n does for most applications - do not raise it much above
this.

Some comments on your schematic:

1  -  Did you really need to draw R2 going up to ground, with the ground
sideways?  This would have been easier to draw and easier to understand if
it was drawn down to the existing ground line.  That makes it more obvious
that it is a pulldown on the gate of Q1.

2  -  While this should work fine, Q1 could be a dirt cheap NPN
transistor.  Instead of a pulldown resistor, all you need is a series
resistor from the PIC pin to the base.

3  -  Note that the output is switched on if the input to this circuit is
left open.  Perhaps that is what you intended, but the "safe" thing is
usually the other way around.


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2003\02\14@135934 by David Minkler

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Russell McMahon wrote:

<snip>

> ... and the 2nd quadrant is not overly useful in most cases due to the body diode.

"most" is right.  Very useful however, if you are doing synchronous
rectification.

Dave

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2003\02\14@161403 by Russell McMahon

face
flavicon
face
> > ... and the 2nd quadrant is not overly useful in most cases due to the
body diode.

> "most" is right.  Very useful however, if you are doing synchronous
> rectification.

Yes. A MOSFET is often the best performing rectifier you can get if you are
prepared to provide the drive signals. Often enough this is not too hard at
all. Even a power Schottky will have around 0.4 volts drop at currents over
an amp whereas a MOSFET forward drop is essentially resistive and dependent
on how small an Rdson FET you are walling to pay for. As an example the
rather nice and well priced IRFP250 (which I use as an example because I use
it as my "workhorse" FET in this class) has an Rdson in REAL applications of
about 0.15 ohm *. At 0.4v drop that's 0.4/.15 = 2.66A so below that it's
superior to a Schottky in a synchronous rectifier. That's a fraction of what
this FET is rated at of course. There are many FETs with FAR lower Rdson at
low voltage.

* Do read FET data sheets VERY carefully when looking at claimed Rdson. The
claimed values may be gained under low duty pulse conditions where the die
has lots of time to cool between pulses. Looking at the typical Vds versus
Ids curves under high case temperature is a better guide to what you will
probably get in the real world..

       Russell McMahon

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2003\02\14@174419 by David Minkler

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Russell McMahon wrote:
> ... At 0.4v drop that's 0.4/.15 = 2.66A so below that it's superior to a
> Schottky in a synchronous rectifier. That's a fraction of what
> this FET is rated at of course. There are many FETs with FAR lower Rdson at
> low voltage.

And, it'll never get worse than the body diode!

Dave

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2003\02\14@184452 by David Harris

picon face
Hi-
Is anyone aware of any products that do not have the intrinsic diode?
David

David Minkler wrote:

{Quote hidden}

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2003\02\14@195018 by Spehro Pefhany

picon face
At 03:42 PM 2/14/2003 -0800, you wrote:
>Hi-
>Is anyone aware of any products that do not have the intrinsic diode?
>David

http://www.micrel.com/_PDF/mic2544.pdf

Best regards,

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2003\02\14@195634 by Russell McMahon

face
flavicon
face
> >>... and the 2nd quadrant is not overly useful in most cases due to the
body diode.

> >"most" is right.  Very useful however, if you are doing synchronous
> >rectification.

> Is anyone aware of any products that do not have the intrinsic diode?

Not in the form of a MOSFET as it is known to us, as the body diode is an
unavoidable part of the basic structure of this type of semiconductor. Sean
suggested you could potentially (no pun intended) separate substrate layer
from FET souirce and connect the substrate to the most negative point in the
circuit. THEN the body diode would not conduct under reverse bias of
drain -source. This owuld require, at least, a 4 terminal device and the
fact that (AFAIK) nobody does it suggests there are other problems.

`        RM

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2003\02\14@204843 by Russell McMahon

face
flavicon
face
> >Is anyone aware of any products that do not have the intrinsic diode?

> http://www.micrel.com/_PDF/mic2544.pdf

Well, its still got one BUT they have kindly separated it out from the FET
which gives you the flexibility people on this thread were looking for - god
for 1.5A but alas its effectively only a 5 volt part :-(

Interesting device - not in its intended role as a high side switch where
its pretty boring, but as a device which, as Sean suggests, has its
substrate brought out separately from the MOSFET source. Here the substrate
is (presumably) connected to ground and the body diode is thus never reverse
biased when the device is operated within its spec sheet ratings.

Unfortunately, because they don't expect people to be usually exploiting
this as a "diodeless" MOSFET, they don't put specs in the data sheet which
demonstrate that the body diode still exists. They DO note the
bidirectionality of the switch element but they put only an UPPER limit on
Vin and Vout and no lower limit. Naively this suggests and infinite negative
going Vin but in practice suggests that they don't mean you to take Vin or
Vout under ground (when the body diode would suddenly magically appear
:-) ).




           Russell McMahon

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'[EE]: MOSFET as a digital switch'
2003\03\09@144320 by Jai Dhar
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part 1 2789 bytes content-type:text/plain; charset=ISO-8859-1 (unknown type 8bit not decoded)

Sorry to resurrect this thread, but I finally got a hold of some IRLZ24's, and
decided to try out a little electronic switch with a fan. I basically have the
output of a FF connected to the gate of the mosfet through a resistor, which
then has a fan going across it. When the fF output is on, the gate see's +5V,
which is well above the threshold, but the source only gets about +3V. I tried
removing the FF and just applying +5V directly to the gate, and same results.
But when I apply +12V to the gate, the fan turns on fully and source gets
+12V. Am I doing something wrong? (I have attached schematic)

Jai

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2003\03\09@145530 by Spehro Pefhany

picon face
At 02:42 PM 3/9/2003 -0500, you wrote:
>Sorry to resurrect this thread, but I finally got a hold of some IRLZ24's, and
>decided to try out a little electronic switch with a fan. I basically have the
>output of a FF connected to the gate of the mosfet through a resistor, which
>then has a fan going across it. When the fF output is on, the gate see's +5V,
>which is well above the threshold, but the source only gets about +3V. I tried
>removing the FF and just applying +5V directly to the gate, and same results.
>But when I apply +12V to the gate, the fan turns on fully and source gets
>+12V. Am I doing something wrong? (I have attached schematic)

Connect the source of the MOSFET to ground, and connect the fan between 12V
and the drain.

Best regards,

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2003\03\09@145939 by Ned Konz

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face
On Sunday 09 March 2003 11:42 am, Jai Dhar wrote:
> Sorry to resurrect this thread, but I finally got a hold of some
> IRLZ24's, and decided to try out a little electronic switch with a
> fan. I basically have the output of a FF connected to the gate of
> the mosfet through a resistor, which then has a fan going across
> it. When the fF output is on, the gate see's +5V, which is well
> above the threshold, but the source only gets about +3V. I tried
> removing the FF and just applying +5V directly to the gate, and
> same results. But when I apply +12V to the gate, the fan turns on
> fully and source gets +12V. Am I doing something wrong? (I have
> attached schematic)

Yes, you've connected the fan to the wrong place.

The fan should go between +12 and the drain of the transistor; the
source should go to ground.

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2003\03\09@151017 by Jai Dhar

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Works like a charm :-) Thanks a lot Spehro.
Quoting Spehro Pefhany <@spam@speffspam_OUTspam.....INTERLOG.COM>:

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2003\03\09@151358 by jim barchuk

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Hi Jai!

On Sun, 9 Mar 2003, Jai Dhar wrote:

> Sorry to resurrect this thread, but I finally got a hold of some IRLZ24's, and
> decided to try out a little electronic switch with a fan. I basically have the
> output of a FF connected to the gate of the mosfet through a resistor, which
> then has a fan going across it. When the fF output is on, the gate see's +5V,
> which is well above the threshold, but the source only gets about +3V. I tried
> removing the FF and just applying +5V directly to the gate, and same results.
> But when I apply +12V to the gate, the fan turns on fully and source gets
> +12V. Am I doing something wrong? (I have attached schematic)

Here're a couple of informative links.

In all these inductive/resistive examples see the diode across load.
Strongly recommended to protect the controlling device.

PIC temp sensors for PC fan control:
http://liquid-nexus.net/fanctrl/schematics.htm

Simple MOSFET circuits:
http://www.uoguelph.ca/~antoon/gadgets/mostest.htm

This one's a blast with tons of intros on almost any topic:
http://www.web-ee.com/primers/primer_bottom.htm

More theory on many semi topics:
http://www.americanmicrosemi.com/tutorials.htm

Have a :) day!

jb

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2003\03\09@181318 by Jai Dhar

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So with a working circuit such as this:

                        ------IRLZ24-------
+12V <-------| FAN |-----DRAIN  GATE  SOURCE---------- GND
                                 |
                                 |
                                 |
                                Logic input

would it be correct calling this a low-side switch? Is it possible to convert
it to a high side, since I'm hoping to use the IRLZ24's as a power switch for
my supply.. ie:

+12V <------ (IRLZ24) ----- LOAD ---- GND

I hope I haven't mixed up my high/low side terminology. Would I need a P-
channel for this application?



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2003\03\09@184714 by Herbert Graf

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> So with a working circuit such as this:
>
>                          ------IRLZ24-------
> +12V <-------| FAN |-----DRAIN  GATE  SOURCE---------- GND
>                                   |
>                                   |
>                                   |
>                                  Logic input
>
> would it be correct calling this a low-side switch?

       That would be correct.

> Is it
> possible to convert
> it to a high side, since I'm hoping to use the IRLZ24's as a
> power switch for
> my supply.. ie:
>
> +12V <------ (IRLZ24) ----- LOAD ---- GND

       Although "possible" I would STRONG recommend against it. The reason is in
this config you are using the MOSFET in source follower config. Why this is
bad is power dissipation, the MOSFET will drop a rather large amount of
voltage and probably burn up (if you are running at "rated" current). This
is due to the transistor not being fully "on" (hard in the triode region).

> I hope I haven't mixed up my high/low side terminology. Would I need a P-
> channel for this application?

       Correct, you need a p-ch to do this properly. TTYL

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2003\03\09@202542 by Jai Dhar

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Quoting Herbert Graf <.....mailinglistRemoveMEspamFARCITE.NET>:

{Quote hidden}

Can anyone recommend a good p-ch that can be driven by logic level's? (do they
even exist since they need a negative gate voltage to turn on?).

>
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2003\03\09@203957 by Herbert Graf

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> >         Correct, you need a p-ch to do this properly. TTYL
>
> Can anyone recommend a good p-ch that can be driven by logic
> level's? (do they
> even exist since they need a negative gate voltage to turn on?).

       Actually with P channels you often don't need "logic level" parts. To turn
"on" a p-ch transistor you bring it's gate to a voltage at least |Vtp| below
the source. If you are using it as a high side driver then the source is
connected to Vdd; by setting it's gate to GND you are getting a Vsg of Vdd,
in most cases Vdd is more than 5V and therefore you don't need a logic level
part. If OTOH you ARE using a Vdd of only 5V (or less) you may need a logic
level part. TTYL

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2003\03\10@112152 by Tal

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Any P channel will do. Drive it with an open collector and a pullup
resistor and you will have a good Vgs drive voltage of 12V (your high
side voltage). Make sure your mosfet can handle 12V Vgs (otherwise, add
another resistor for voltage divider).

Also, I am not sure what are the charactersistis of the fan you plan to
use. If this is an inductive load, you may need add a reverse diode
parallel to it. Anybody wants to comment on this ?

BTW, N chan can also do high side but the drive is a little bit more
complex (e.g. a charge pump to generate voltge above your VDD) so
typically you want to avoid it.

Tal

> {Original Message removed}

2003\03\10@202506 by Jai Dhar

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Quoting Tal <.....talSTOPspamspam@spam@ZAPTA.COM>:

> Any P channel will do. Drive it with an open collector and a pullup
> resistor and you will have a good Vgs drive voltage of 12V (your high
> side voltage). Make sure your mosfet can handle 12V Vgs (otherwise, add
> another resistor for voltage divider).
>

The fan was just an example. Basically, replace the fan with any load. I just
want to make a simple on/off switch for my power supply that's not mechanical
(so ugly!!). I am just learning that it matters whether your mosfet is on the
high side or low side, so all of this is new to me. I will order up some P-
channel samples from Fairchild and play with them, see if I can get this
working..

As a side note, how come BJT's aren't used for this kind of application?

> Also, I am not sure what are the charactersistis of the fan you plan to
> use. If this is an inductive load, you may need add a reverse diode
> parallel to it. Anybody wants to comment on this ?
>
> BTW, N chan can also do high side but the drive is a little bit more
> complex (e.g. a charge pump to generate voltge above your VDD) so
> typically you want to avoid it.
>
> Tal
>
> > {Original Message removed}

2003\03\10@212542 by Herbert Graf

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> As a side note, how come BJT's aren't used for this kind of application?

       BJTs are, and have been, used for this sort of thing for years. The are
actually some applications where a BJT would be a superior device. That
said, BJTs aren't used as often these days mainly because of base current.
With a MOSFET you don't need to supply any DC current to the device's gate
terminal (you do need current to discharge and charge the gate capacitance
but that isn't of much concern when switching very slowly). With a power BJT
you do. How much depends on how much current you are supplying to the load
and what the Beta of the BJT is. For higher currents it is possible to have
Betas as low as 10 (even lower). That means to control 1A, you need ~100mA
of current flowing into the base. As you can see that can be a big problem,
often you will need another transistor driving the power transistor!

       As an aside, there are some devices that take advantage of both worlds. For
example, IGBTs (insulated gate bipolar transistors) have a MOSFET
controlling a BJT. You get the benefit of no gate current and the benefits a
BJT gives you.

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2003\03\11@203509 by Ian McLean

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The diode is definately needed, otherwise the output tends to oscillate from
fan motor impedance.  Place the anode of the diode to the drain output (low
side output) of the P-chan mosfet.  The cathode goes to +12V which is also
the positive output to the fan.  The fan on low side drive is turned on and
off from the fan negative terminal, which is connected to the mosfet drain
pin.  I recommend at least an 1N4001 for this, but a slightly faster diode
is better (such as 1N5407 or 1N5408).

IME, you may also wish to place a small (100nF) mono cap between the source
and drain of the mosfet as this is needed to control the rise and fall time
of the output - wether this is needed or not will be largely determined by
what sort of frequency your driving the fans at.  If you are using PWM, then
the cap is definately needed for any usable PWM frequency, although, YMMV,
bepending on which Mosfet you are using.

Rgs
Ian.

{Original Message removed}

2003\03\11@221125 by Jai Dhar

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Ok, I'm really trying to understand these darned MOSFET's here, so I managed
to find some good articles regarding them. I will summarize what I know to see
if it's true (for pMOS)

If the voltage drop from gate to source is greater than the threshold, the
MOSFET should turn on...

I took my IRF9530 (just a basic pmos with a Vgs threshold of -4V) to test this
out. Connecting source to +12 and gate to GND, the drain showed a drop of
+12V. So far, that seems right.. since Vgs = -12V, and the threshold = -4V.

BUT...

When I connected the gate to +5V, the drain still showed +12V. Why is this?
Vgs = 5 - 12 = -7V, which is still greater than the threshold, right?

Another thing is the signs here... I think that's what may be messin me up.
Are we talking about algabraically greater?? Because -7 isn't greater than -
4... nor is -12. So somewhere in my explanation (and understanding), I'm
mistaken. Can anyone clarify?


Quoting Tal <talEraseMEspam@spam@ZAPTA.COM>:

{Quote hidden}

> > {Original Message removed}

2003\03\12@063520 by Banjo Spam

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Jai,

You are on the right track. In some textbooks, the
absolute values are used so that the equations for
PMOS and NMOS are the "same".

Note that the historic application of the term
"greater" in this context is strictly true for NMOS.
In terms of learning this stuff from scratch, it will
probably be easier to understand the NMOS devices
first, because most casual literature (i.e. one that
mentions "Vgs" and "greater" without using absolute
values) will be strictly and algebraically correct.
Since the source of NMOS devices is generally
connected to the lowest potential, Vgs and Vds are
positive numbers and you won't have to "invert" the
numbers in your head.

One thing that does make it confusing is the
definition of "source" and "drain". On a discrete PMOS
3-terminal device (where only the source, drain, and
gate are available), the source should always be
connected to a higher potential than the drain (i.e.
one closer to positive infinity). When the gate
voltage is more than a threshhold away from the source
voltage, then the device is "on" and the drain will
approach the source voltage.

Again, you're on the right track.

Hope this helps,

You're doing the right thing by playing with one and
verifying your understanding!

Good luck!

--- Jai Dhar <RemoveMEjdharspamspamBeGoneENGMAIL.UWATERLOO.CA> wrote:
{Quote hidden}

> > > {Original Message removed}

2003\03\12@170251 by Byron A Jeff

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On Tue, Mar 11, 2003 at 10:09:29PM -0500, Jai Dhar wrote:
> Ok, I'm really trying to understand these darned MOSFET's here, so I managed
> to find some good articles regarding them. I will summarize what I know to see
> if it's true (for pMOS)
>
> If the voltage drop from gate to source is greater than the threshold, the
> MOSFET should turn on...
>
> I took my IRF9530 (just a basic pmos with a Vgs threshold of -4V) to test this
> out. Connecting source to +12 and gate to GND, the drain showed a drop of
> +12V. So far, that seems right.. since Vgs = -12V, and the threshold = -4V.

How did you measure the voltage on the drain? The best way I've found to test
this is to attach the drain to GND through a pulldown resistor. When the
MOSFET is off, then the drain will measure 0V, whereas if it's on it'll measure
somew voltage closer to the source.

BTW in this configuration the MOSFET is on and will pull the drain to +12V.

>
> BUT...
>
> When I connected the gate to +5V, the drain still showed +12V. Why is this?

Because the MOSFET is still on.

> Vgs = 5 - 12 = -7V, which is still greater than the threshold, right?

Right. Greater and less than can get you in trouble here. A threshold of
-4V means that the gate must be at least 4V below the source for the MOSFET
to turn on. Since it's 7V below, the MOSFET is on.

I'm understanding now that this doesn't meet your expectation. You expected
that the gate at 5V should turn the part off. It doesn't.

To do it right the gate should be pulled all the way up to 12V to turn off
the MOSFET giving a Vgs=0V which is not at least 4V below the Source.

A single NPN transistor with the emitter grounded, collector tied to the
gate, along with a pullup resistor between the source and gate/collector, and
the base tied to the PIC I/O through a series resistor, will do the job fine.

When the PIC I/O is 0V, the NPN is off, the gate moves to 12V, and the MOSFET
is off.

However when the PIC I/O is 5V, the NPN turns on, grounding the gate, And
the MOSFET turns full on.

Now a question: could a zener diode be used instead of the NPN? Consider
if the gate is pulled up via a pullup resistor. Of course the MOSFET is off.
Now connect a 5.7V zener between the gate and the PIC I/O pin. My though is
that the voltage of the gate will be 5.7V above the voltage of the PIC I/O
pin. So if the PIC pin is 0V, then the gate will be at 5.7V, and therefore
the MOSFET is on. However if the PIC pin is 5V, then the gate will be at 10.7V
and Vgs=-1.3V so the MOSFET will turn off.

Thoughts?
>
> Another thing is the signs here... I think that's what may be messin me up.
> Are we talking about algabraically greater?? Because -7 isn't greater than -
> 4... nor is -12. So somewhere in my explanation (and understanding), I'm
> mistaken. Can anyone clarify?

I Hope that I did so above. I always read Vgs=-4 as "The gate must be at least
4V below the source for the MOSFET to turn on."

Hope this helps,

BAJ

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2003\03\12@173334 by Jai Dhar

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Your NPN transistor idea is exactly what I thought of... AFTER I sent that
message. I hooked up a pot to the gate to see what level the MOSFET turned on
and off... low and behold, I found out that my definition of 'above the
threshold' was backwards. Either way, I understand it now, thank you for re-
affirming that what I did was a valid solution!



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