thanks for all the suggestions,
The gate resistor i replaced with a 1kΩ, measured Vgs: 20V, the mosfet is working comfortably (i suppose), & each string of LEDs, series with a resistor, no intension to modulate the current, but i still don't satisfy with the total current that passing thru the LEDs.
I recalled that at the beginning stage, when testing this circuit, before using heatsink to each LEDs, i fixed them all on an aluminium sheet (1mm thick, area 130 by 130mm) as dissipate heat, & the temperature rose till ~150°C! that might have effected the performance of the LEDs now.
i'm ordering another 4pcs of LEDs to test again, i'll keep you all inform by then.
thanks

 

Keep in mind that the forward voltage of the LEDs, in series with any current sharing resistors you added, may be limiting the current to less than the 1.8A that you are trying to achieve. Others have mentioned this, but I'm not sure you got the point.

 

Instead of cooking expensive LED's with a circuit still in the breadboard stage, you can calculate an equivalent resistor value and use one of those instead. They can take substantial abuse and are inexpensive too.

The 1k gate resistor will draw ~ 20ma at 20 volts of op amp output. The op amp won't be happy supplying that much current. The gate requires no current except to briefly charge the gate capacitors. Rather than using a voltage divider, limit the op amp supply voltage with a zener diode of 21 volts or so, just low enough to keep the op amp output just below 20 volts, the gate's limit. Feed the gate with one 100 ohm resistor from the op amp, with no divider at all to dampen any ringing. Now, all of the op amp's current will be be available to the gate to charge the gate capacitance instead of draining the current away through a voltage divider.

As eblc1388 observed in post #2, the reference voltage at point "A" is 291mv. This will cause the circuit to regulate the output current at 2.91 amps. This reference voltage must be 180mv to get 1.8 amps out. 24k and 180 ohms will give you 1.79 amps. To get exactly 280mv you could calculate in a small trimpot to make up for variances in the 24v supply, or, better yet, supply this voltage divider from the zener supply for the op amp.

 

Instead of cooking expensive LED's with a circuit still in the breadboard stage, you can calculate an equivalent resistor value and use one of those instead. They can take substantial abuse and are inexpensive too.

The 1k gate resistor will draw ~ 20ma at 20 volts of op amp output. The op amp won't be happy supplying that much current. The gate requires no current except to charge the gate capacitors. Rather than using a voltage divider, limit the op amp supply voltage with a zener diode to 21 volts or so, just low enough to keep the op amp output just below 20 volts, the gate's limit. Feed the gate with one 100 ohm resistor from the op amp, with no divider at all.

I thought the 1k was part of a 1k/10k voltage divider, in which case the DC load is 11k, not 1k, and will not load the op amp significantly.
I am concerned about the AC stability, as i mentioned in previous posts.

 

I dunno, I thought he kept the original 47 ohm R3 at the op amp output and changed the 10k gate resistor to 1k.

 

Keep in mind that the forward voltage of the LEDs, in series with any current sharing resistors you added, may be limiting the current to less than the 1.8A that you are trying to achieve. Others have mentioned this, but I'm not sure you got the point.

yah i workout by this way--> R series = E/I
where E = Vcc - V at LEDs - Vds - V at sensing resistor.
I = current flowing thru 1 string of LEDs.
so that workout the resistor is 1Ω.

 

yah i workout by this way--> R series = E/I
where E = Vcc - V at LEDs - Vds - V at sensing resistor.
I = current flowing thru 1 string of LEDs.
so that workout the resistor is 1Ω.

How do you know what the LED forward voltage is? The "datasheet" says 10-12V. If they are 11.4V or more, your circuit won't work as designed.

 

i measure across the series of LEDs to obtain voltage drop when on.

 

i measure across the series of LEDs to obtain voltage drop when on.

And what did you get for the forward voltage when they were biased at 900mA each?

 

voltage across LEDs string was 22.9V, this voltage was not at 0.9A, as observed from sensing resistor merely 0.2V, i worked back the current that flow thru LEDs should be ~0.6A

 

voltage across LEDs string was 22.9V, this voltage was not at 0.9A, as observed from sensing resistor merely 0.2V, i worked back the current that flow thru LEDs should be ~0.6A

I thought your current sense resistor was 0.1Ω. With 0.2V across it,that's 2 amps. With 2 parallel strings of LEDs, that would be 1A each. Where did I go wrong?

Did the 22.9V include the 1Ω ballast resistors?

 

my sense R was 0.15Ω, oops i noticed my schematic 0.1Ω, sorry about that.
from the scope 50mV/div, it covered 4 div so 0.2V across sense R.
the 22.9V measured anode to cathode of the series LEDs.
the workout value of ballast R, i base on expected max current flow, thru each string.

 

my sense R was 0.15Ω, oops i noticed my schematic 0.1Ω, sorry about that.
from the scope 50mV/div, it covered 4 div so 0.2V across sense R.
the 22.9V measured anode to cathode of the series LEDs.
the workout value of ballast R, i base on expected max current flow, thru each string.

When you add the voltage drops of the ballast resistors, you won't have enough left to regulate the current.

How much current do you really want through your LEDs?

 

i reworked out the ballast R is 0.8Ω, instead of 1Ω. Pls ref to attached pdf my workings.

 

Since your circuit has no regulation, here is a much cheaper way to do exactly the same thing.