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Does a diode in the Vcc line disable clamp diodes to Vcc

Started by Bill September 14, 2012
I have a case where I want to put about 20 microamps into an 
AVR at 12v.  AVRs have clamp diodes to both ground and Vcc on
their digital inputs.   It seems to me that a diode inline with
the Vcc supply would disable the internal clamp diodes of the 
chip.   Am I correct on this or is the internal circuitry 
fancier than a simple clamp diode.

Thanks,
   Bill
On Fri, 14 Sep 2012 16:42:18 -0700, Bill wrote:

> I have a case where I want to put about 20 microamps into an AVR at 12v. > AVRs have clamp diodes to both ground and Vcc on their digital inputs. > It seems to me that a diode inline with the Vcc supply would disable > the internal clamp diodes of the chip. Am I correct on this or is the > internal circuitry fancier than a simple clamp diode.
What you are saying doesn't make sense -- do you have a schematic posted somewhere that you can point to? If you take a source that can only supply 20 microamps and connect it to the power pin of a chip that takes more, the supply won't rise to 12V. If you do succeed in raising the VCC pin of your AVR to 12V, you'll destroy the chip. If you put a diode in the VCC line so that it's forward conducting, you won't change the behavior of the chip's clamp diodes -- those only change what happens to the individual pins with respect to VCC (VDD, actually) and VSS. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Bill wrote:

> I have a case where I want to put about 20 microamps into an > AVR at 12v. AVRs have clamp diodes to both ground and Vcc on > their digital inputs. It seems to me that a diode inline with > the Vcc supply would disable the internal clamp diodes of the > chip. Am I correct on this or is the internal circuitry > fancier than a simple clamp diode. > > Thanks, > Bill
The input diodes you speak of are connected to the VSS and VDD internally and are there for a couple of reasons. 1. is to prevent over drive of either low or high of course. 2. Since the internals are CMOS, most likely, it's not a good idea to allow inputs to exceed the +/- rails, otherwise, you'll get a latch up and a misbehaved chip if input signals for even a moment, exceeding a diode drop above the rail voltage. You need not to worry about the supply diode you have in line to the (+) supply of the chip. It's all about making sure the inputs do not exceed the +/- of the rails at the chip. The voltage at the (+) rail of the chip, for example, will cause current to flow in the upper diode if it reaches a diode drop (~ 0.6) above the (+) rail, at which point the input signal will flow through the diode to the (+) rail instead, against the supply. The clamping force will still be provided. It's possible they're using sk diodes so it could even be closer to .3 volts when (If) is started. You really want to avoid hitting the inputs hard enough to activate these diodes. It's not good practice and it will most likely extend the skew. If you need to clip an input signal to compress it to a square wave for an adequate trigger, do that before with an external circuit. Jamie
On Fri, 14 Sep 2012 20:23:19 -0400, Jamie
<jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:

>Bill wrote: > >> I have a case where I want to put about 20 microamps into an >> AVR at 12v. AVRs have clamp diodes to both ground and Vcc on >> their digital inputs. It seems to me that a diode inline with >> the Vcc supply would disable the internal clamp diodes of the >> chip. Am I correct on this or is the internal circuitry >> fancier than a simple clamp diode. >> >> Thanks, >> Bill > > The input diodes you speak of are connected to the VSS and VDD >internally and are there for a couple of reasons. > > 1. is to prevent over drive of either low or high of course. > > 2. Since the internals are CMOS, most likely, it's not a good idea to >allow inputs to exceed the +/- rails, otherwise, you'll get a latch up >and a misbehaved chip if input signals for even a moment, exceeding a >diode drop above the rail voltage. > > You need not to worry about the supply diode you have in line to the >(+) supply of the chip. It's all about making sure the inputs do not >exceed the +/- of the rails at the chip. The voltage at the (+) rail of >the chip, for example, will cause current to flow in the upper diode if >it reaches a diode drop (~ 0.6) above the (+) rail, at which point the >input signal will flow through the diode to the (+) rail instead, >against the supply. The clamping force will still be provided. >
Some chips, like 3.3v things that are 5-volt tolerant, have a clamp diode for the negative swing direction but more like a zener in the positive swing direction. They tend to clamp at +7 or some such. -- John Larkin Highland Technology, Inc jlarkin at highlandtechnology dot com http://www.highlandtechnology.com Precision electronic instrumentation Picosecond-resolution Digital Delay and Pulse generators Custom laser drivers and controllers Photonics and fiberoptic TTL data links VME thermocouple, LVDT, synchro acquisition and simulation
On Fri, 14 Sep 2012 16:42:18 -0700 (PDT), the renowned Bill
<billosh@gmail.com> wrote:

>I have a case where I want to put about 20 microamps into an >AVR at 12v. AVRs have clamp diodes to both ground and Vcc on >their digital inputs. It seems to me that a diode inline with >the Vcc supply would disable the internal clamp diodes of the >chip. Am I correct on this or is the internal circuitry >fancier than a simple clamp diode. > >Thanks, > Bill
It's perhaps a bit fancier, but yes this will likely "disable" the clamp diode to Vdd. Meaning that if you pull an input up to 12V, you might think that only leakage current would flow back into the Vdd supply. Unfortunately, the diode is still in there and will conduct the 12V (less a diode drop) to the Vdd of the internals of the chip, and it will probably die a painful death. As someone else said, this will likely cause latch-up, the Vdd supply at the chip Vdd pin will come crashing down to a volt or so as the parasitic SCR turns on, and then your series diode will conduct in the forward direction (a whole bunch of current). Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Spehro Pefhany wrote:

> On Fri, 14 Sep 2012 16:42:18 -0700 (PDT), the renowned Bill > <billosh@gmail.com> wrote: > > >>I have a case where I want to put about 20 microamps into an >>AVR at 12v. AVRs have clamp diodes to both ground and Vcc on >>their digital inputs. It seems to me that a diode inline with >>the Vcc supply would disable the internal clamp diodes of the >>chip. Am I correct on this or is the internal circuitry >>fancier than a simple clamp diode. >> >>Thanks, >> Bill
> > > It's perhaps a bit fancier, but yes this will likely "disable" the > clamp diode to Vdd.
I fail to see how that would happen ? If the Vcc(Vdd) is getting supplied via this external diode from the supply, the chip is only going to see that VDD there. It will still act as the limiting factor for the upper side of the internal clamp diode. If anything, if the input signal is getting a reference from the supply that maybe even .7 higher, it'll just cause the diode to clamp sooner.. Even if the input some how manages to raise the Vdd, due to internal current passing through the clamp to the Vdd, it'll just elevate the Vdd. The most that can happen is a slight elevation of Vdd but since latch up involves input being above Vdd, I still fail to see how it can happen? I may not express myself with 101 "Harvard English", but I think I get my point across, most of the time. Maybe I am blind, due to all the magic smoke that has escaped me over the years.
> > Meaning that if you pull an input up to 12V, you might think that only > leakage current would flow back into the Vdd supply. Unfortunately, > the diode is still in there and will conduct the 12V (less a diode > drop) to the Vdd of the internals of the chip, and it will probably > die a painful death. > > As someone else said, this will likely cause latch-up, the Vdd supply > at the chip Vdd pin will come crashing down to a volt or so as the > parasitic SCR turns on, and then your series diode will conduct in the > forward direction (a whole bunch of current). > > > > Best regards, > Spehro Pefhany
"Once I thought I was wrong, then I found that every one else was mistaken" Jamie
fOn Fri, 14 Sep 2012 22:25:08 -0400, the renowned Jamie
<jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:

> If the Vcc(Vdd) is getting supplied via this external diode from the >supply, the chip is only going to see that VDD there.
Vdd o +5 | V External diode - | +19.3 Vdd pin of chip o-------------------+---------------------+ | | | | | --- - | --- bypass ^ ||-+ | +20V | ||-> === o--------------+--------------|-||-+ GND | | | | | +---------o - | | ^ | ||-+ | | ||<- | |-||-+ | | === === GND GND The external diode isolates the Vdd pin of the chip from the external supply.
>It will still >act as the limiting factor for the upper side of the internal clamp >diode. If anything, if the input signal is getting a reference from the >supply that maybe even .7 higher, it'll just cause the diode to clamp >sooner.. > > Even if the input some how manages to raise the Vdd, due to internal >current passing through the clamp to the Vdd, it'll just elevate the >Vdd. The most that can happen is a slight elevation of Vdd but since >latch up involves input being above Vdd, I still fail to see how it can >happen?
Why slight? A micro, depending on mode, might be drawing only leakage current, so the voltage would rise until something breaks down. With such a small current (assuming it _is_ limited- it's not clear whether the current is limited or the OP wishes to draw a maximum current of 20uA or maybe there is another meaning altogether), it might not damage the chip, in fact, but it certainly won't allow the input to go much above the external supply safely since the internal Vdd of the chip sees the input voltage less a diode drop. An easy way to do this, as I think JL mentioned, is to use a CD4049/50 buffer, which has an abs max input voltage of 18V. But if the 12V signal is coming from the outside world and 20uA is allowable current draw, there are lots of ways of handling it more safely such as dividing it down or using a BJT. Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
"Tim Wescott"
> Bill wrote: > >> I have a case where I want to put about 20 microamps into an AVR at 12v. >> AVRs have clamp diodes to both ground and Vcc on their digital inputs. >> It seems to me that a diode inline with the Vcc supply would disable >> the internal clamp diodes of the chip. Am I correct on this or is the >> internal circuitry fancier than a simple clamp diode. > > What you are saying doesn't make sense --
** It does, if you read it properly.
> If you take a source that can only supply 20 microamps and connect it to > the power pin of a chip that takes more, the supply won't rise to 12V. > If you do succeed in raising the VCC pin of your AVR to 12V, you'll > destroy the chip.
** The OP is talking about the "digital inputs ".
> If you put a diode in the VCC line so that it's forward conducting, you > won't change the behavior of the chip's clamp diodes -
** But you remove the current sinking effect of PSU source impedance from the Vcc pin. ... Phil
Spehro Pefhany wrote:

> fOn Fri, 14 Sep 2012 22:25:08 -0400, the renowned Jamie > <jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote: > > >> If the Vcc(Vdd) is getting supplied via this external diode from the >>supply, the chip is only going to see that VDD there. > > > Vdd o +5 > | > V External diode > - > | +19.3 > Vdd pin of chip o-------------------+---------------------+ > | | | > | | --- > - | --- bypass > ^ ||-+ | > +20V | ||-> === > o--------------+--------------|-||-+ GND > | | | > | | +---------o > - | | > ^ | ||-+ > | | ||<- > | |-||-+ > | | > === === > GND GND > > The external diode isolates the Vdd pin of the chip from the external > supply. > > > > >>It will still >>act as the limiting factor for the upper side of the internal clamp >>diode. If anything, if the input signal is getting a reference from the >>supply that maybe even .7 higher, it'll just cause the diode to clamp >>sooner.. >> >> Even if the input some how manages to raise the Vdd, due to internal >>current passing through the clamp to the Vdd, it'll just elevate the >>Vdd. The most that can happen is a slight elevation of Vdd but since >>latch up involves input being above Vdd, I still fail to see how it can >>happen? > > > Why slight? A micro, depending on mode, might be drawing only leakage > current, so the voltage would rise until something breaks down. > > With such a small current (assuming it _is_ limited- it's not clear > whether the current is limited or the OP wishes to draw a maximum > current of 20uA or maybe there is another meaning altogether), it > might not damage the chip, in fact, but it certainly won't allow the > input to go much above the external supply safely since the internal > Vdd of the chip sees the input voltage less a diode drop. > > An easy way to do this, as I think JL mentioned, is to use a CD4049/50 > buffer, which has an abs max input voltage of 18V. > > But if the 12V signal is coming from the outside world and 20uA is > allowable current draw, there are lots of ways of handling it more > safely such as dividing it down or using a BJT. > > > > Best regards, > Spehro Pefhany
Hmm, it seems that our descriptions of what is and what isn't, is getting mixed in translation. If I read you correctly, you now agree with me, at first it seem to be the opposite? Oh well, no matter anyway. I think most understand the logic. On to the next... Jamie
On 2012-09-14, Bill <billosh@gmail.com> wrote:
> I have a case where I want to put about 20 microamps into an > AVR at 12v. AVRs have clamp diodes to both ground and Vcc on > their digital inputs.
All inputs except for the reset pin.
> It seems to me that a diode inline with > the Vcc supply would disable the internal clamp diodes of the > chip. Am I correct on this or is the internal circuitry > fancier than a simple clamp diode.
No. it's that simple, The diode will continue to function. depending on the model of AVR and what it's doing 20 microamps may be more than the current the AVR is consuming in which case you'll loose supply voltage regulation for the chip. -- &#9858;&#9859; 100% natural --- Posted via news://freenews.netfront.net/ - Complaints to news@netfront.net ---