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
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.
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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.
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The input diodes you speak of are connected to the VSS and VDD internally and are there for a couple of reasons.
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
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
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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.
"Once I thought I was wrong, then I found that every one else was mistaken"
Jamie
Vdd o +5 | V External diode - | +19.3 Vdd pin of chip o-------------------+---------------------+ | | | | | --- - | --- bypass ^ ||-+ | +20V | ||-> === o--------------+--------------|-||-+ GND | | | | | +---------o - | | ^ | ||-+ | | ||It will still
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
** It does, if you read it properly.
** The OP is talking about the "digital inputs ".
** But you remove the current sinking effect of PSU source impedance from the Vcc pin.
... Phil
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
All inputs except for the reset pin.
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.
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Stick a 100k resistor across the vcc and gnd of the AVR. That will provide a path so the clamp diodes to work. You want to check exactly what is being shoved into the input and choose the resistor from there.
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No problem.... the internal diodes will take 20uA easy.
Is this for a battery voltage detect system ?
I have not tested the AVR, but other chips I have tested don't behave quite like a small diode. There is quite a lot of ohmic resistance in there IIRC, at least on the "diode" to Vdd. According to what IIRC JT said, the exact ESD-protection circuitry is generally proprietary.
That said, usually you can think of it as a pair of diodes, with the caveat that running current through those diodes can make strange things happen, from slight errors in an ADC to outright destruction of the chip at very high peak current. Another slightly more accurate (IME) model is a diode to Vdd with a resistor in series, and (more or less) a diode to Vss.
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
That's the issue right there. Once the diode in the AVR supply line is reverse biased, the overvoltage, through the (forward biased) clamping diode will push the Vcc line up to whatever the AVR and other Vcc load impedances will let it rise to.
Perhaps a TL431 downstream of the supply diode could clamp the bus fast enough.
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Thanks for the answers.
I had modeled this in LTSPICE and had found that the clamp didn't protect it if there was a diode in the Vcc line.
I blew it. I had included the impedance of the signal source but I forgot to include the internal load of the circuit of the circuit in the IC. I have now put a
500 ohm resistor in to represent this load and the diode clamps are doing their job in the LTSPICE model.Good Luck, Bill
They call that a oops. Normally if you have a bypass cap there and only present a transient, it should model ok. Like I said, should.
I don't use Ltspice or spice in general all that much, I've done everything with a calculator, paper, bread boards and test jigs. It wasn't up to about 2 or so years ago I even tried Ltspice.
Years ago I tried using emulators, but the computers were so slow and software so bad that I was able to get my answer faster by hand.
WHen my little one when to trade school, is when I saw the use of workbench, I really didn't think much of that. The graphical view of the circuit was more like an arcade game, didn't really turn me on.
Now days, I walk around at work with my TI in my shirt pocket and a small laptop that is easy to carry. lately the bench hasn't gotten much exercise.
Jamie
What do you mean by emulators? Not generally but specifically a product, can't imagine the use for something like that, like the resistor box joke thing.
=20 Well not really. !
1.) use a Schottky diode if you can and a circuit to limit Vdd, like a transistor beefed up TL431. =20 2.) it is not really a diode but a more complex circuit that acts similar to a diode and is very fast (must be, as ESD is very fast).?-)
imagine the use for something like that, like the resistor box joke thing.
My son used a program while in trade school which was called "Work Bench", it actually showed you a bread board, jumper wires, real view of a DMM and explosions and smoke as things didn't turn out right.
Maybe I have the product name incorrect but that is what I remember seeing it called on the screen..
But when it came time to build a AM&FM receiver kit, he presented it to me after he had assembled it with a few minor problems. While I helped him correct those problems, I also did a little modification on it to bump up gain and quiet the receiver.
He got to enter that in a parents get together for public view and it ended up in a glass case at school for a while, along with a couple other kids, which also seem to have had some help that were in the business.
I really don't think the instructor caught on, maybe he did and didn't say any however, he may have noticed those transistor types I changed if he really looked a littler closer, along with some added by passes etc. :)
Jamie
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