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I'm trying to perform communication with a Motorola MPC5606B microcontroller using a cable that I already had previously. Using this cable will make things easier in my lab because everyone has one the same. But I have a concern: there is a diode 1N4007 in line with Vcc.

schematic

simulate this circuit – Schematic created using CircuitLab

So, because of the diode on the VCC line, the I/O voltage can be greater than that specified on the datasheet. The datasheet specification is given in the figure below.

So, my concern is: if the voltage on the I/O pin is 0.3V greater than Vcc, but below 6V, can it damage the I/O pin of the microcontroller?

extract from datasheet showing max voltage on I/O pins as VDD+0.3V

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    \$\begingroup\$ there is a diode 1N4007 in the line of Vcc What does it mean exactly? Draw circuit diagram. \$\endgroup\$ – Anonymous Jan 24 '18 at 12:30
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    \$\begingroup\$ It is outside of the SOA/absolute maximum ratings, what does the datasheet say about that? \$\endgroup\$ – PlasmaHH Jan 24 '18 at 12:30
  • \$\begingroup\$ A question why are you putting a 1n4007 diode there? it's Vf is 1.1V with the current configuration you will fry the chip unless insted of 5V you use a 6.1 Vcc. And I will delete my previus commet due to not seeing the diode.Still recomend on using a transciever: SN74LVCC3245ADW. \$\endgroup\$ – Dimitri Jan 24 '18 at 13:10
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The simple answer is yes, it can. The specifications clearly show what the absolute rating is, and you are violating it, since there may be cases where VDD is 4.3V but your input is 5V. This means that the manufacturer is not guaranteeing that the part will function correctly.

The real question is, will it probably work? The answer is maybe. If there's not much current going into the pin (i.e. whatever you're connecting to the input has a high output impedance), it's probably fine. All your IO pins will have internal diodes connected to VDD and GND. These are what will clamp the input voltages, and these are what will break if you put too much current through them. They are usually rated for very small amounts of current. It looks like, in your case, they may be rated for 10ma, so if you're putting 1ma through them, that might be alright.

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    \$\begingroup\$ If you have to rely on the other side being a relatively high impedance output, you are in a world of trouble. Put some sort of level translation on your board (a simple resistor can be enough to do the trick (limiting the current to a safe value)). \$\endgroup\$ – Arsenal Jan 24 '18 at 13:06
  • \$\begingroup\$ If he really has 5V and not 3.9V (due to 1N4009) than there shouldn't be any problem. It "Would" work, but in the limit. \$\endgroup\$ – Dimitri Jan 24 '18 at 13:07
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    \$\begingroup\$ What I mean is that if he really has 5V in Vdd and low current in I/O ports, hw shouldn't have any problem with 5+0.3V . It's not so weird to have fluctuations. However from the schematic i'm not sure that there is 4V in Vdd. Usually the manufacturer leaves a margin for I/O but in this case the Vdd is lower than 5V due to 1N4007. I'm not sure if I explain myself, sorry if I don't. \$\endgroup\$ – Dimitri Jan 24 '18 at 14:06
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    \$\begingroup\$ @Dimitri The diode is there because its part of the cable that this person wants to use in their project. The question they're asking is whether or not it's ok to do so... \$\endgroup\$ – BeB00 Jan 24 '18 at 14:26
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    \$\begingroup\$ ok. now i get it. Thanks. In this case he really should use a transciever 5V to 3V3 and try working close to Voh which is 0.8Vdd in his case. \$\endgroup\$ – Dimitri Jan 24 '18 at 14:30
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Despite the fact that you are violating an absolute maximum rating on the devices in this case you may actually be ok.

Here is the reasoning..

The max input voltage is so you do not blow the protection diode and input stage of the device by trying to over-voltage Vcc.

schematic

simulate this circuit – Schematic created using CircuitLab

However, in this instance, since Vcc is supplied through a diode, you actually have this...

schematic

simulate this circuit

As such, the current that will pass through the protection diode will not be fighting with the power supply.

However, the IO pin will be trying to power the rest of the circuit so, depending on what else is connected to that supply, it could still be potentially far too much current for the pin to withstand. Further, that rail will have noise injected by the input signal, which, on anything but a basic circuit, will be problematic.

Let's assume the latter is not an issue. The question then becomes can whatever is driving that signal line supply enough current to keep the pin voltage up at that level.

Remember \$Voh\$ is dependent on the current drawn from an output pin. Generally what happens is the protection diode pulls down the output pin. You can assist it in that mission by adding an extra load resistor from the input pin to ground to "soak-up" some of the driver's \$I_{OH}\$. But all you are doing here is passing the stress over to the other device. A series resistor can also be added to limit the current through the pin to a level that prevents damage.

However: These as just stop-gap measures and are bad design practices. You really ought to be fixing the design to use the appropriate rail voltages at both ends. Even if you get away with this on input signals you also need to be concerned about the logic levels of any outgoing signals.

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Many data sheets are needlessly under-specified. Generally, the design of most parts will, at no cost, allow three guaranteed ways to prevent stress or damage.

  1. Unless the voltage on a pin is driven above or below a certain level, the device will not source nor sink a harmful or stressful amount of current, no matter how much current the external device would be capable of supplying.

  2. Unless a certain amount of current is fed into or drawn out of a pin, it will be impossible for the voltage on that pin to reach a harmful or stressful level.

  3. If a voltage on a pin starts at a level that would neither cause the device to source nor sink current, a certain amount of charge would need to be added or removed for the voltage on that pin to reach a harmful or stressful level.

Unfortunately, for whatever reason, data sheets only quantify the first. I can understand that even if today's manufactured parts would require that 10mA be driven onto a pin before the voltage could reach a stressful level, the manufacturer may want to allow for the possibility that parts manufactured in future might not be able to withstand that much. On the other hand, there are many situations where having a guarantee that 10uA would be harmless would be much better than having no guarantee at all, and if today's parts can handle 10mA it would seem unlikely that future parts would be incapable of handling 1/1000 of that.

With regard to your situation, it's likely that driving one I/O pin to 5 volts while the device is trying to drive some other pins high would result in current flowing from the externally-supplied pin to the pins you are trying to drive. If the total load on the pins you are trying to drive would be sufficiently low, your design would be unlikely to unduly stress the part. Unfortunately, it's hard to know what "sufficiently low" means in this context.

Further, I have observed chips which behaved as though each pair of inputs was clamped to VDD using a PNP transistor. Driving one input above VDD while the other was being externally pulled down would cause some current to flow into the positive rail, but would cause more current to flow into the other input. I don't know that this was particularly stressful for the device, but it did cause the other input to read high when it should have been low, thus yielding incorrect operation. If the current pushed into the pin above VDD had been low enough this wouldn't have been an issue, but again data sheets give no useful guidance.

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