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[SOLVED] constant current source

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yoosefheidari

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hi every body
i have a NCP6354 and i want to use it as a constant current source to drive a 6w white led at 2000ma
help me how can i do it:thinker:
 

Hi,

Instead of voltage feedback you may try:
connect a 300mOhms shunt from FB to GND.
Connect the LED from OUT to the shunt.

The problem is, that the shunt will dissipate 1.2W at 2A.
Not that good efficiency...
Maybe you need some capacitive feedback to ensure stability.

Klaus
 
Hi
so can i add a current limit to prevent led burning?a 2000ma current limit
Click to expand...

I´m confused.
of course you can. But Why?
You´ve built a constant current source, it already limits the current. ... why do you need another current limiter?

Klaus
 
You could use a shunt feedback resistor as KlausST suggested, but add a single-supply type opamp (such as an LM324, LM358) with non-inverting gain to reduce the value of the shunt resistance and its dissipation.
For example a gain of +6 will reduce the value of the needed shunt resistor to 50mΩ and its dissipation to 0.2W for a 2A constant-current output.
You likely will have to add some compensation capacitance across the opamp feedback resistor for stability.
 
crutschow said:
You could use a shunt feedback resistor as KlausST suggested, but add a single-supply type opamp (such as an LM324, LM358) with non-inverting gain to reduce the value of the shunt resistance and its dissipation.
For example a gain of +6 will reduce the value of the needed shunt resistor to 50mΩ and its dissipation to 0.2W for a 2A constant-current output.
You likely will have to add some compensation capacitance across the opamp feedback resistor for stability.
Click to expand...
thank you and KlausST
i dont understand that op-amp
can you show me by schematic shunt resistor and op-amp?
 

yoosefheidari said:
.....i dont understand that op-amp
can you show me by schematic shunt resistor and op-amp?
Click to expand...
R1 is the shunt resistor.
The op amp has a gain of six, which will give an output current of 2A through the LED D1 using a 50mΩ shunt.

Capture.PNG
 
Last edited:
crutschow said:
R1 is the shunt resistor.
The op amp has a gain of six, which will give an output current of 2A through the LED D1 using a 50mΩ shunt.

View attachment 148121
Click to expand...

thank you

but i have a question
what is the difference between original schematic that we adjust output voltage and then specific current go across the led with this schematic that we adjust voltage between led and resistor and again same current go across led?
 

Hi,

but i have a question
what is the difference between original schematic that we adjust output voltage and then specific current go across the led with this schematic that we adjust voltage between led and resistor and again same current go across led?
Click to expand...

The diffference:
* Datasheet = constant voltage source
* this solution = constant current source (as requested)

Behaviour:
Datasheet: change the load resistance: voltage will remain constant, current will change
this solution: change the load resistance: current will remain constant, voltage will change

Klaus
 

yoosefheidari said:
...........
what is the difference between original schematic that we adjust output voltage and then specific current go across the led with this schematic that we adjust voltage between led and resistor and again same current go across led?
Click to expand...
Because if you use a voltage, you need a resistor in series with the LED to regulate the current, and that wastes power.
yoosefheidari said:
can i set op-amp gain higher around 12 and use a 25mΩ resistor instead 50 ?
Click to expand...
Yes.
 
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