Dear Team,

Please see the attached image which shows the circuit of driving LED using an opamp. We are fixing current through LED using opamp. I am clear about how they obtained the current equation.Isense=Vin/Rsense because opamp is in negative feedback

My question is why they are not considering the forward voltage(Vf) of the LED anywhere. For LED to turn on Certain Vf is required.
If we fix the forward current will it set the Vf automatically?, Or do I need to select a "V+" which is always greater than Vf.

Regards
Hari

 

Hard to tell. what value did you have in mind for Rsense. If it is large enough to limit the current to a reasonable value for the LED, The there will be three voltage drops:

1. Vf for the LED
2. Vce for the transistor
3. Vin

That also implies that the range of Vin is not arbitrary or unlimited.

 

Hi Papabravo

My Rsense will be 10 ohms.
I can write KVL for "V+" =Vf+Vce+Vin.
From this equation, I can determine the minimum voltage required for V+(Voltage at the collector of Transistor).
Please correct me if I am wrong.

Regards
Hari

 

If we fix the forward current will it set the Vf automatically?, Or do I need to select a "V+" which is always greater than Vf.

V+ must be greater than \(Vf_{LED} + V_{CE} + V_{RSENSE} \)

If the opamp can saturate the transistor, Vce can be neglected. Since we don't know what opamp you're using, it's supply voltage, or the range of Vin, we can't be more specific.

 

10 Ω is not enough resistance to limit the maximum current which is what you need to do.
Example: Let us say that V+ = 12 VDC and the maximum current we want is 20 mA. The required resistor is 600 Ω. You can split that between the 10 Ω sense resistor and the nearest 1% value so the sum of the two resistors will be as close to 600 Ω as possible.

Edit: turns out that won't work. Back to your original circuit, you need to choose Rsense so it limits the maximum current in the LED, given the supply voltage. For +12V, Rsense ≈ 400 Ω should do the trick, with Vsense ≈ 8 VDC.