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I have an LED Driver IC that states a 5.5V tolerant Open Drain Output. It's Absolute Maximum section indicates 6.5V with a note saying "The input negative-voltage and output voltage ratings may be exceeded if the input and output current ratings are observed." The outputs can drive LEDs directly or be used as open-drain GPIO.

Typical Application:
enter image description here

My question is, how does a diode's forward voltage drop affect this? If I have a led strip segment of 3x 3.3Vf Leds + 100Ω Resistor in series and a 12V supply, shouldn't the voltage drops from the diodes bring the output within tolerance? Or not? Why?

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the current through the port is 20mA, half of the max recommended 40mA (Absolute Max is 50mA and a clamp current of ±20mA) per output and 200mA per port.

Edit: Found TI: Understanding and Interpreting Standard-Logic Data Sheets

Output Voltage, VO
This is the maximum voltage that can be applied safely to an output terminal, with respect to the ground of the device.

Helpful Hint:
If there are clamp diodes between the device outputs and the VCC supply (see Figure 20) for ESD protection or parasitic current paths in the output p-channel pullup transistor, the positive absolute maximum rating for the output voltage is specified as VCC + 0.5 V. This ensures that there will not be enough voltage applied between the output and VCC to forward bias the clamp diode and cause current to flow. You may exceed the negative rating if you ensure that you are not putting too much current through the ground-clamp diode. The maximum current that you may put through the ground-clamp diode is specified in the IOK absolute maximum rating.

As the IOk rating is 20mA, and the normal ON state current would be ~18mA, the normal OFF state with the drain left open should be a lot less, no? If the current through the ESD/Clamp Diode increases, the voltage drop across the leds and resistor should increase as well right?

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  • \$\begingroup\$ This question is meant to be general, not tied to a specific device, so I provided most of what should be the important information above. None the less, the device datasheet in question is the TI TCA6507 ti.com/lit/ds/symlink/tca6507.pdf \$\endgroup\$ – Passerby Nov 12 '14 at 6:45
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    \$\begingroup\$ The problem is when the output goes Hi-Z, since the forward voltage of the LEDs will drop to almost nothing. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 12 '14 at 6:48
  • \$\begingroup\$ Reading between the lines, I suspect the IC fabrication process TI used for this chip breaks down (permanent damage) if more than 6.5V appears anywhere across the chip for any length of time... there's always a process limit like this... If their chip could survive higher voltage, they'd want to advertise that. I don't know whether the LED leakage current would be enough to trigger this damage, but it might. \$\endgroup\$ – MarkU Nov 12 '14 at 6:53
  • \$\begingroup\$ Normally I'd suspect a latchup issue, but the missing PMOS makes me question that. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 12 '14 at 6:56
  • \$\begingroup\$ @IgnacioVazquez-Abrams see edit \$\endgroup\$ – Passerby Nov 12 '14 at 7:20
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I don't think the diodes would matter for that limit. When the output is open, the diode drops will be 0V and the pin will see the full 12V. Transistor voltage limits are there to prevent the transistor from being ripped apart by a strong electric field. It doesn't take much current to do that.

I'm not sure whether a clamp diode or a parallel resistor on the input would help, either. If it draws enough current to create significant diode drops, the LEDs will turn on. If it doesn't, the pin voltage will go up and exceed the limit. If the LED current exceeds the bias current for the driver IC, Vcc might start to rise unless your regulator can sink current. (Most won't.)

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I believe the circuit can be saved with a high value resistor (maybe 100K) from either end of R1 to the processor's VCC.

With the port pin on, the wasted power through this resistor is minimal; it has virtually no effect against the 2-20mA current flowing.

With the port pin off, any leakage through the diodes will be drawn from the 5V rail via this resistor, pulling the port pin above 5V by Ileakage * R. Set the value of R such that this value < 0.5V for the worst case leakage current across the temperature range (plus a safety margin). This analysis is left as an exercise : my 100k suggestion is merely a guess.

That said, I still don't really like it...

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This circuit is almost for sure not going to work for a reason nobody has mentioned.

When current flows through the LED's, they are going to turn on. The IC probably will not be damaged because you are keeping the clamp current under a reasonable level, but that fault current will turn on the LED's for sure. So what you have is a circuit where the state of P0 does not really matter. In fact, it doesn't even matter whether the IC driver chip is powered on. Whenever 12V is energized, the diodes will be on. If that is what you want, then you don't need to connect the IC at all, so I assume that is not what you want.

You can fix it by adding a BJT or discreet MOSFET to drive the LED's. Suggested BJT is 2N3904 or 2N2222 type. Suggested MOSFET is BSS138 type.

You seem pretty smart, so I have not gone on at length, but if you have any questions or want me to elaborate, just ask, and I will be happy to do so.

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    \$\begingroup\$ Actually I take it back. Adam also pointed out the fundamental problem. \$\endgroup\$ – mkeith Nov 14 '14 at 17:17

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