Protect the LED with a series diode or not?

[fryingpan]

Let's say you plug in the power supply the wrong way round. Center positive, barrel negative (which should be the right way  ).

LEDs can be damaged by reverse voltages in excess of 5V. The current, though, will by and large flow through the resistors, into the Schottky diode. Therefore my question is: do I need a series diode before the LED? (It's just that I want to minimise the size of the stripboard I'm using, a 1590B enclosure is only so large).

(R2 could be larger. At 9V, this allows for around 3mA. At 18V, it's 8mA. Ideally there should be some Zener regulation to essentially clamp the voltage at the LED's anode and ensure the same brightness, but this particular design doesn't really benefit much from higher power voltages anyway, 9V it's the intended supply).

[ElectricDruid]

The other part of the question is "How big a deal is it?". Assuming you did fry the LED, it's not actually that hard or expensive to replace. There's usually *some* special component somewhere in a circuit that you'd really rather wasn't the one that got fried. But that's not the LED.

One solution would be to use the 1N5817 diode in series to protect the *whole* circuit, and then put the LED+resistor on the safe side of that diode. No?

[fryingpan]

The other part of the question is "How big a deal is it?". Assuming you did fry the LED, it's not actually that hard or expensive to replace. There's usually *some* special component somewhere in a circuit that you'd really rather wasn't the one that got fried. But that's not the LED.

One solution would be to use the 1N5817 diode in series to protect the *whole* circuit, and then put the LED+resistor on the safe side of that diode. No?

I designed it this way so as to have R1 in between the diode and the rest of the circuit. It probably doesn't matter much, but the current inrush into the LED might cause popping. (To really cure this you actually need two more components). With that resistor in between you should (should) "isolate" the LED from the rest of the circuit, as far as popping is concerned.

Replacing an LED is no biggie. But if I can avoid this, why not?

[R.G.]

You could protect the LED with a parallel diode, reversed polarity from the LED, too.
A bigger question is why not protect the whole circuit from reversed voltages.

[Rob Strand]

Replacing an LED is no biggie. But if I can avoid this, why not?

It will probably survive. 

There was a thread about LED reverse breakdown a few years back.  Some decades ago I remember LED's which did show reverse breakdown at about 5V.  However when the thread came up AMZ mentioned he saw higher voltages. I rechecked a number of modern LEDs and the breakdown voltage was much higher, IIRC 20V.  As a sanity check I then dug through my ancient and dim LEDs and I could only find one or two that showed breakdown around 5V.  You might be able to find a thread.    (As a caveat I didn't check white and blue LEDs.)
This thread,
https://www.diystompboxes.com/smfforum/index.php?topic=116192.0

[merlinb]

GaAs red LEDs can handle huge reverse voltages. For others it's a dice roll.

[ElectricDruid]

An alternative solution - replace the LED in your original design with a bipolar Red/Green LED. Since the LED will now conduct in both directions, it will never be subjected to over-voltage with reverse power, and as a bonus, if the LED lights up red, you know the power supply is wired wrong!

[fryingpan]

You could protect the LED with a parallel diode, reversed polarity from the LED, too.
A bigger question is why not protect the whole circuit from reversed voltages.

The idea is: keep the led "isolated" (not really, but 100 ohm might help) from the rest of the circuitry so that when you turn on the circuit there are no popping sounds (unlikely, it's a low gain circuit, but still). The Schottky diode after the resistor (a Schottky would limit reverse voltage to about 350mV) can't come before because otherwise the current draw would be massive. But anyway, the bipolar LED seems like the best solution. (I was going to go with an orange LED but it's not important).

[jorg777]

It looks like you're thinking of an LED like an incandescent bulb, which has a big inrush current as it heats up.  LEDs aren't like that.  The whole circuit will have an inrush, mainly due to any power supply decoupling and filtering capacitors.  The LED won't add any noticeable pop.