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I'm mainly having trouble with the wording in datasheets.

So I know diode protection can be useful in a situation where you're using a large capacitor at the output of a regulator and the fault case of the input shorting to ground occurs. The diode keeps the current from going through the regulator(correct?)

So when looking at regulator datasheet, if it's not explicitly stated that it has reverse current protection, does that usually mean I should use diode protection there?

The datasheet mentions it has protection against reverse output voltage and reverse input voltage. I'm not sure if those mean the same thing as having reverse current protection.

In addition to the normal protection features associated with monolithic regulators, such as current limiting and thermal limiting, the device protects itself against reverse input voltages and reverse output voltages.

But then I also see this:

A parasitic substrate diode exists between OUT and IN of the LT3015. Therefore, do not drive OUT more than 0.3V below IN during normal operation or during a fault condition.

I'm guessing that statement is enough to say that diode protection will be needed from output to input?

The datasheet in question would be the LT3015. http://cds.linear.com/docs/en/datasheet/3015fb.pdf

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  • \$\begingroup\$ If you reverse the voltage, that reverse diode going from one rail to another would go kablooey. \$\endgroup\$ – Bradman175 Aug 16 '16 at 2:42
  • \$\begingroup\$ Don't use the diode unless the datasheet says it is needed. If linear thinks it is needed, they will show it in the datasheet or discuss it in depth. 0.3V is a pretty tight limit. A silicon diode is out of the running. A Schottky might do it, depending on the current. Can you come up with a fault condition that will cause Vin to be less than vout? \$\endgroup\$ – mkeith Sep 15 '16 at 4:40
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Under normal operation, the input voltage of an LT3015 should always be more negative than the output voltage, e.g. -12V in, -5V out. What the datasheet is saying is that if the output is somehow driven 0.3V lower than the input, e.g. -6V at the output, -5.7V at the input, then a parasitic diode that's part of chip substrate may/will start conducting. You would then have current flowing in an unexpected way, and this may damage your regulator or other connected circuitry. For example, if you hooked the regulator up like you would a positive voltage regulator, then this:

schematic

simulate this circuit – Schematic created using CircuitLab

effectively becomes this:

schematic

simulate this circuit

Under these conditions, your regulator has a good chance of functioning as an short-lived electric heater, releasing smoke or otherwise failing. That said, if you keep your polarities correct, this is unlikely to happen otherwise.

When the datasheet is talking about reverse input/output voltage protection, it means that the output can be ABOVE ground without damaging the regulator, i.e. "If IN is left open circuit or grounded, OUT can be pulled above GND by 30V"; "In dual supply applications where the regulator’s load is returned to a positive supply, OUT can be pulled above GND by 30V and still allow the LT3015 to start up and operate." That sort of configuration would look like this.

schematic

simulate this circuit

While both supplies are connected, the regulator provides regulation, current protection and thermal protection. If V2 is disconnected, the regulator will safely stop conducting, and if the regulator is shut down, the OUT pin can rise above ground without damage, and the regulator can later be turned on (without disconnecting power supplies). The important thing is that the IN pin should always be more negative than the OUT, ADJ, and SHDN pins to keep current from flowing through the parasitic diode. If you're concerned about protecting the regulator from reverse polarity, a diode in series with your power supply should suffice, though remember that the diode will increase the dropout voltage by ~1V.

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  • \$\begingroup\$ Thanks for the indepth answer. I have an addtional question. So, a diode from the Vin to the input of the regulator will protect against input shorts. But what about an output short? Would a diode from output to input or output to ground be needed in that situation? @Alex \$\endgroup\$ – Dmoles Aug 16 '16 at 16:48
  • \$\begingroup\$ @Dmoles An output short circuit is automatically handled by the regulator's overcurrent protection, no diode required. The one caveat is that under certain conditions you may need to cycle the input power after removing the fault. See the "Overload Recovery" section on pg.17-18 of the datasheet for details. \$\endgroup\$ – Alex Aug 16 '16 at 21:02

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