Op amp based adjustable current source for Laser or LEDs with Multisim simulations

Constant current source using an op amp and a transistor (or a MOSFET):

We give a detailed analysis of the circuit: derivation of the formulas and Multisim simulations. The values in the circuit are for illustrations only. Below we give more practical design considerations:

1. The beauty of this circuit is that I_LED can be adjusted by adjusting Vi, and I_LED is amplified by R1 and RF. (see point 3 example below).
To adjust Vi, one way is to use a microcontroller. The microcontroller outputs a PWM signal whose duty cycle can be controlled by a push-button for dimming purpose: press the push-button once we get 90% of maximum power, press again we get 80% of the power, .......... press 10 times we return to maximum power again. This concept can be commercialized in a lighting system.

2. From the video, we have

I_LED=V_i*((RE+R1+RF) / (R1*RE) ) .......... Eq. 1

One special case of this circuit is to take RF=0 (just use a wire to replace RF) and R1=∞ (just take R1 off ) then we will end up with a more simple circuit, then Eq. 1 became

I_LED=V_i*((RE+R1+RF) / (R1*RE) )
=V_i*((RE+∞+0) / ( ∞ *RE) )
=V_i*((∞ / ( ∞ *RE) ) if R1 → ∞, then R1+RE≈R1, we just dropped RE
=V_i*(1/ ( RE) ) ...... Eq. 2

3. A calculation example:

Suppose Vi= 1V (from a headphone jack to control a spotlight in DJ nightclub). we take RE=2 Ω (small RE is better to make the power consumption of RE small, so that RE is not hot.) and we use a MOSFET to replace the transistor for large current.
Now we can take 4 choices for R1 and RF:

case 1: simple circuit from Eq. 2 , R1=∞, and RF=0 , then
I_LED=Vi/RE=1/2=0.5A=500mA (use SI units first, to get A, then change A to mA, please)

case 2: R1=500 Ω, and RF=100 Ω, from Eq. 1, we have
I_LED=V_i*((RE+R1+RF) / (R1*RE) ) (use SI units please)
=1*((2+R1+RF) / (R1*2) ) =1*((2+500 +100 ) / ( 500*2) ) =0.6A=600mA

case 3: R1=2Ω, and RF=0 Ω
I_LED=1*((2+R1+RF) / (R1*2) ) =1*((2+2+0 ) / ( 2*2) ) =1A=1000mA

case 4: R1=2Ω, and RF=2 Ω
I_LED= 1*((2+R1+RF) / (R1*2) ) =1*((2+2+2) / (2*2) ) =1.5A=1500mA