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Datasheet: http://diodes.com/datasheets/AL6562.pdf

I am trying to understand how this IC works block by block and I am confused about over-voltage protection in the error amplifier - specifically how it will work when I'm regulating current and not voltage. Unfortunately, the datasheet does not specifically talk about current regulation and instead talks about voltage regulation when it comes to the error amplifier, naturally fed by a voltage divider.

If you look at the linked schematic, R4 is the input to the error amplifier. From the schematic, I think it is easy to see that this there to regulate the current. My output current is 50 mA and the error amplifier's reference is 2.5 V, therefore R4 should be 50 Ohms.

The datasheet talks about over-voltage protection (OVP) in this section (Page 10). It states that for OVP, the IC monitors the current into the COMP pin and if it exceeds 40 uA, it will trigger the OVP. As 40 uA is the current threshold, this is used to set the voltage threshold via the voltage divider like so: dV/40uA = R My question is, if I am using the IC to regulate current and not voltage how would OVP behave?

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This is based only on my reading of the data sheet:

OVP is largely unimportant in the LED driving example in figure 1 as there is no load change situation in formal use that corresponds to a sudden load reduction in the voltage output example in figures 2.

The one likely exception is if one of the string of series diodes fails OC (open circuit) and Vout rises rapidly. In that case cimp and inv are isolated from Vo and the OVP mechanism seems not to apply.

A "work around" that would seem liable to work well would be to bridge the diode string with a zener diode plus series resistor with the zener voltage higher than the string operating voltage and the series resistor + zener combination dimensioned to pass more current than the LED string in normal operation. If an LED fails OC then the OVP mechanism will now operate as in data sheet fig 1.

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