some question about AMC7135

AMC7135 is very famous in led driver and some people use it with 12v supply. the voltage drop is very small compare to LM317.
what is the other advantage from AMC7135?
the heat dissipation is also low?

if the voltage supply is 12v and running 3v 1A led,
1. zener drop 12v to 5v or 6v than constant current using LM317
2. zener drop 12v to 5v than constant current using AMC7135
which one is better? what is the advantage and disadvantage?

If the input voltage, and the voltage and current to the LED, are the same, then whatever you have in between the two will dissipate the same power. A lower minimum voltage drop would give you the ability to run the LED and it's controller at a lower voltage, but if you are using it at 12V in, then that advantage does not matter.

So, your choice is not one of efficiency, but of parts cost, capability and/or convenience.

renzen said:

AMC7135 is very famous in led driver and some people use it with 12v supply. the voltage drop is very small compare to LM317.
what is the other advantage from AMC7135?
the heat dissipation is also low?

Click to expand...

Never heard of this AMC7135 chip before.

No. The heat dissipation will be about the same.

The drop out voltage refers to the minimum voltage that can be across the OUT and GND pins and still have the device function properly.

The heat dissipation will be roughly the same bevause both the LM317 and the AMC7135 are linear. They act as a resistor, dissipating as much as they can/need to depending on how much current has to flow, and how much voltage has to be dropped.

renzen said:

if the voltage supply is 12v and running 3v 1A led,
1. zener drop 12v to 5v or 6v than constant current using LM317
2. zener drop 12v to 5v than constant current using AMC7135
which one is better? what is the advantage and disadvantage?

Click to expand...

There is no use for thiis other than sharing the heat dissipation over both devices rather than one. Keep in mind, you will need a GIANT zener diode and a GIANT resistor if you are going to do that!

Since the LED requires 3V @ 1A (3W), with a linear regulator, the input power will be 12V @ 1A. The regulator will have to drop 9V. With 1A running through it, the regulator will dissipate 9W of heat (1x9=9)

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Ultimately a switching regulator is best. MUCH more efficient, no giant heatsink needed. Cost is roughly the same or less, because there is no need for a big metal heatsink. Complexity is usually greater, however not too much.