On 2013-06-26, Joerg <invalid@invalid.invalid> wrote:> Spehro Pefhany wrote: >> >> I guess they assume you'll never exceed the absolute maximum voltage >> limits of +5.8/-0.3, so no significant current will ever flow. >> > > Ok, but in real life that's almost unprotectable. The way I usually do > it, there is a diode against the 5V rail, followed by a resistor, then > the port pin. If the pulse from hell comes along that diode can briefly > lean in so badly that Vf goes slightly above 2V, meaning over 7V total. > Bigger diodes aren't an option because that messes up the signal, on > account of their large capacitance.How about this: series resistor, shunt diode to 3.3V , port pin -- ⚂⚃ 100% natural --- news://freenews.netfront.net/ - complaints: news@netfront.net ---
8051F020 series, 5V tolerant input schematic?
Started by ●June 25, 2013
Reply by ●June 28, 20132013-06-28
Reply by ●June 28, 20132013-06-28
On 29 Jun 2013 00:03:04 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:>On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >> Spehro Pefhany wrote: >>> >>> I guess they assume you'll never exceed the absolute maximum voltage >>> limits of +5.8/-0.3, so no significant current will ever flow. >>> >> >> Ok, but in real life that's almost unprotectable. The way I usually do >> it, there is a diode against the 5V rail, followed by a resistor, then >> the port pin. If the pulse from hell comes along that diode can briefly >> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >> Bigger diodes aren't an option because that messes up the signal, on >> account of their large capacitance. > >How about this: series resistor, shunt diode to 3.3V , port pinThere seems to be considerable confusion about what a "5V-tolerant" input looks like, schematic-wise. Would everyone please post what they think it looks like? ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
Reply by ●June 28, 20132013-06-28
Jasen Betts wrote:> On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >> Spehro Pefhany wrote: >>> I guess they assume you'll never exceed the absolute maximum voltage >>> limits of +5.8/-0.3, so no significant current will ever flow. >>> >> Ok, but in real life that's almost unprotectable. The way I usually do >> it, there is a diode against the 5V rail, followed by a resistor, then >> the port pin. If the pulse from hell comes along that diode can briefly >> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >> Bigger diodes aren't an option because that messes up the signal, on >> account of their large capacitance. > > How about this: series resistor, shunt diode to 3.3V , port pin >That doesn't protect very well because the parasitic substrate diode will take a substantial portion of the hit. The only thing that works is: R1 -> shunt diodes -> R2 -> port pin. R2 can be small because there will only be 1-2V max across it and it only has to make sure the substrate diode gets no more than a few mA. But ... too many parts in this case. -- Regards, Joerg http://www.analogconsultants.com/
Reply by ●June 28, 20132013-06-28
Joerg wrote:> Jasen Betts wrote: >> On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >>> Spehro Pefhany wrote: >>>> I guess they assume you'll never exceed the absolute maximum voltage >>>> limits of +5.8/-0.3, so no significant current will ever flow. >>>> >>> Ok, but in real life that's almost unprotectable. The way I usually do >>> it, there is a diode against the 5V rail, followed by a resistor, then >>> the port pin. If the pulse from hell comes along that diode can briefly >>> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >>> Bigger diodes aren't an option because that messes up the signal, on >>> account of their large capacitance. >> How about this: series resistor, shunt diode to 3.3V , port pin >> > > That doesn't protect very well because the parasitic substrate diode > will take a substantial portion of the hit. The only thing that works is: > > R1 -> shunt diodes -> R2 -> port pin. > > R2 can be small because there will only be 1-2V max across it and it > only has to make sure the substrate diode gets no more than a few mA. > > But ... too many parts in this case. >Oops, sorry, your are right. At a 5V-tolerant pin it would work. Problem is that we only have 5V in the area where the protectors are. -- Regards, Joerg http://www.analogconsultants.com/
Reply by ●June 28, 20132013-06-28
On Fri, 28 Jun 2013 17:53:15 -0700, Joerg <invalid@invalid.invalid> wrote:>Joerg wrote: >> Jasen Betts wrote: >>> On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >>>> Spehro Pefhany wrote: >>>>> I guess they assume you'll never exceed the absolute maximum voltage >>>>> limits of +5.8/-0.3, so no significant current will ever flow. >>>>> >>>> Ok, but in real life that's almost unprotectable. The way I usually do >>>> it, there is a diode against the 5V rail, followed by a resistor, then >>>> the port pin. If the pulse from hell comes along that diode can briefly >>>> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >>>> Bigger diodes aren't an option because that messes up the signal, on >>>> account of their large capacitance. >>> How about this: series resistor, shunt diode to 3.3V , port pin >>> >> >> That doesn't protect very well because the parasitic substrate diode >> will take a substantial portion of the hit. The only thing that works is: >> >> R1 -> shunt diodes -> R2 -> port pin. >> >> R2 can be small because there will only be 1-2V max across it and it >> only has to make sure the substrate diode gets no more than a few mA. >> >> But ... too many parts in this case. >> > >Oops, sorry, your are right. At a 5V-tolerant pin it would work. Problem >is that we only have 5V in the area where the protectors are.If you only have 5V around driving a 5V-tolerant pin... what are you fretting over? ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
Reply by ●June 29, 20132013-06-29
Jim Thompson wrote:> On Fri, 28 Jun 2013 17:53:15 -0700, Joerg <invalid@invalid.invalid> > wrote: > >> Joerg wrote: >>> Jasen Betts wrote: >>>> On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >>>>> Spehro Pefhany wrote: >>>>>> I guess they assume you'll never exceed the absolute maximum voltage >>>>>> limits of +5.8/-0.3, so no significant current will ever flow. >>>>>> >>>>> Ok, but in real life that's almost unprotectable. The way I usually do >>>>> it, there is a diode against the 5V rail, followed by a resistor, then >>>>> the port pin. If the pulse from hell comes along that diode can briefly >>>>> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >>>>> Bigger diodes aren't an option because that messes up the signal, on >>>>> account of their large capacitance. >>>> How about this: series resistor, shunt diode to 3.3V , port pin >>>> >>> That doesn't protect very well because the parasitic substrate diode >>> will take a substantial portion of the hit. The only thing that works is: >>> >>> R1 -> shunt diodes -> R2 -> port pin. >>> >>> R2 can be small because there will only be 1-2V max across it and it >>> only has to make sure the substrate diode gets no more than a few mA. >>> >>> But ... too many parts in this case. >>> >> Oops, sorry, your are right. At a 5V-tolerant pin it would work. Problem >> is that we only have 5V in the area where the protectors are. > > If you only have 5V around driving a 5V-tolerant pin... what are you > fretting over? >The concern are surge events that make these lines carry nasty transients, which in turn can lock up or damage the uC. -- Regards, Joerg http://www.analogconsultants.com/
Reply by ●June 29, 20132013-06-29
On Sat, 29 Jun 2013 08:00:01 -0700, Joerg <invalid@invalid.invalid> wrote:>Jim Thompson wrote: >> On Fri, 28 Jun 2013 17:53:15 -0700, Joerg <invalid@invalid.invalid> >> wrote: >> >>> Joerg wrote: >>>> Jasen Betts wrote: >>>>> On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >>>>>> Spehro Pefhany wrote: >>>>>>> I guess they assume you'll never exceed the absolute maximum voltage >>>>>>> limits of +5.8/-0.3, so no significant current will ever flow. >>>>>>> >>>>>> Ok, but in real life that's almost unprotectable. The way I usually do >>>>>> it, there is a diode against the 5V rail, followed by a resistor, then >>>>>> the port pin. If the pulse from hell comes along that diode can briefly >>>>>> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >>>>>> Bigger diodes aren't an option because that messes up the signal, on >>>>>> account of their large capacitance. >>>>> How about this: series resistor, shunt diode to 3.3V , port pin >>>>> >>>> That doesn't protect very well because the parasitic substrate diode >>>> will take a substantial portion of the hit. The only thing that works is: >>>> >>>> R1 -> shunt diodes -> R2 -> port pin. >>>> >>>> R2 can be small because there will only be 1-2V max across it and it >>>> only has to make sure the substrate diode gets no more than a few mA. >>>> >>>> But ... too many parts in this case. >>>> >>> Oops, sorry, your are right. At a 5V-tolerant pin it would work. Problem >>> is that we only have 5V in the area where the protectors are. >> >> If you only have 5V around driving a 5V-tolerant pin... what are you >> fretting over? >> > >The concern are surge events that make these lines carry nasty >transients, which in turn can lock up or damage the uC.So you have a "dirty" 5V supply? ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
Reply by ●June 29, 20132013-06-29
Jim Thompson wrote:> On Sat, 29 Jun 2013 08:00:01 -0700, Joerg <invalid@invalid.invalid> > wrote: > >> Jim Thompson wrote: >>> On Fri, 28 Jun 2013 17:53:15 -0700, Joerg <invalid@invalid.invalid> >>> wrote: >>> >>>> Joerg wrote: >>>>> Jasen Betts wrote: >>>>>> On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >>>>>>> Spehro Pefhany wrote: >>>>>>>> I guess they assume you'll never exceed the absolute maximum voltage >>>>>>>> limits of +5.8/-0.3, so no significant current will ever flow. >>>>>>>> >>>>>>> Ok, but in real life that's almost unprotectable. The way I usually do >>>>>>> it, there is a diode against the 5V rail, followed by a resistor, then >>>>>>> the port pin. If the pulse from hell comes along that diode can briefly >>>>>>> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >>>>>>> Bigger diodes aren't an option because that messes up the signal, on >>>>>>> account of their large capacitance. >>>>>> How about this: series resistor, shunt diode to 3.3V , port pin >>>>>> >>>>> That doesn't protect very well because the parasitic substrate diode >>>>> will take a substantial portion of the hit. The only thing that works is: >>>>> >>>>> R1 -> shunt diodes -> R2 -> port pin. >>>>> >>>>> R2 can be small because there will only be 1-2V max across it and it >>>>> only has to make sure the substrate diode gets no more than a few mA. >>>>> >>>>> But ... too many parts in this case. >>>>> >>>> Oops, sorry, your are right. At a 5V-tolerant pin it would work. Problem >>>> is that we only have 5V in the area where the protectors are. >>> If you only have 5V around driving a 5V-tolerant pin... what are you >>> fretting over? >>> >> The concern are surge events that make these lines carry nasty >> transients, which in turn can lock up or damage the uC. > > So you have a "dirty" 5V supply? >No, rock-stable. But clamping against it will lead to up to 7V spike amplitude if it's a huge transient. Those small clamper arrays can easily reach 2V Vf or more for a few usec. -- Regards, Joerg http://www.analogconsultants.com/
Reply by ●June 29, 20132013-06-29
On Sat, 29 Jun 2013 08:13:15 -0700, Joerg <invalid@invalid.invalid> wrote:>Jim Thompson wrote: >> On Sat, 29 Jun 2013 08:00:01 -0700, Joerg <invalid@invalid.invalid> >> wrote: >> >>> Jim Thompson wrote: >>>> On Fri, 28 Jun 2013 17:53:15 -0700, Joerg <invalid@invalid.invalid> >>>> wrote: >>>> >>>>> Joerg wrote: >>>>>> Jasen Betts wrote: >>>>>>> On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >>>>>>>> Spehro Pefhany wrote: >>>>>>>>> I guess they assume you'll never exceed the absolute maximum voltage >>>>>>>>> limits of +5.8/-0.3, so no significant current will ever flow. >>>>>>>>> >>>>>>>> Ok, but in real life that's almost unprotectable. The way I usually do >>>>>>>> it, there is a diode against the 5V rail, followed by a resistor, then >>>>>>>> the port pin. If the pulse from hell comes along that diode can briefly >>>>>>>> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >>>>>>>> Bigger diodes aren't an option because that messes up the signal, on >>>>>>>> account of their large capacitance. >>>>>>> How about this: series resistor, shunt diode to 3.3V , port pin >>>>>>> >>>>>> That doesn't protect very well because the parasitic substrate diode >>>>>> will take a substantial portion of the hit. The only thing that works is: >>>>>> >>>>>> R1 -> shunt diodes -> R2 -> port pin. >>>>>> >>>>>> R2 can be small because there will only be 1-2V max across it and it >>>>>> only has to make sure the substrate diode gets no more than a few mA. >>>>>> >>>>>> But ... too many parts in this case. >>>>>> >>>>> Oops, sorry, your are right. At a 5V-tolerant pin it would work. Problem >>>>> is that we only have 5V in the area where the protectors are. >>>> If you only have 5V around driving a 5V-tolerant pin... what are you >>>> fretting over? >>>> >>> The concern are surge events that make these lines carry nasty >>> transients, which in turn can lock up or damage the uC. >> >> So you have a "dirty" 5V supply? >> > >No, rock-stable. But clamping against it will lead to up to 7V spike >amplitude if it's a huge transient. Those small clamper arrays can >easily reach 2V Vf or more for a few usec._Where_ are these "huge transients" coming from? ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
Reply by ●June 29, 20132013-06-29
On Fri, 28 Jun 2013 17:39:27 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon@On-My-Web-Site.com> wrote:>On 29 Jun 2013 00:03:04 GMT, Jasen Betts <jasen@xnet.co.nz> wrote: > >>On 2013-06-26, Joerg <invalid@invalid.invalid> wrote: >>> Spehro Pefhany wrote: >>>> >>>> I guess they assume you'll never exceed the absolute maximum voltage >>>> limits of +5.8/-0.3, so no significant current will ever flow. >>>> >>> >>> Ok, but in real life that's almost unprotectable. The way I usually do >>> it, there is a diode against the 5V rail, followed by a resistor, then >>> the port pin. If the pulse from hell comes along that diode can briefly >>> lean in so badly that Vf goes slightly above 2V, meaning over 7V total. >>> Bigger diodes aren't an option because that messes up the signal, on >>> account of their large capacitance. >> >>How about this: series resistor, shunt diode to 3.3V , port pin > >There seems to be considerable confusion about what a "5V-tolerant" >input looks like, schematic-wise. > >Would everyone please post what they think it looks like? > > ...Jim Thompsonhttp://ics.nxp.com/features/5v-io/ Absolute max "rating" is 7V, but "simple" inputs (no drain or source connected, just gates) can probably take 20V. ...Jim Thompson -- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
