It would a great pleasure for if any body explain me working of attached circuit .

 

Looking at the first circuit, the R5, R6 potential divider produces 0.197V at the non-inverting op-amp input. The negative feedback from R16 keeps the voltage across R16 at that same voltage. That means that the current through R16 is kept constant at 151mA and that same constant current is maintained through the LED.

 

It is a constant current source to drive an LED.

The + input has a known voltage of 820 / 20820 * 5V.

The opamp will drive the transistor until the - input matches the + input.

The - input is 1.3 time the current through the LED.

So, we have I * 1.3 = 820 * 5 / 20820

I = 256mA

Bob

 

I fear my calculator may be broke. I still make it 151mA.

 

Whoops, I did V * R to get I. Albert Hall is correct, 151 mA.

edit: At least we agreed on what the circuit does!

Bob

 

I am guessing C13 is decoupling capacitor.

 

And a very good guess it is.

 

Thank you all for your support .

 

Hi,

For the second circuit the LED current will be LED_VREF/R9 which is LED_VREF/15.
So if you have 1.5v for LED_VREF then the current is 1.5/15=0.1 amps.

 

Both circuits are what is called a transconductance amplifier, where the output current is directly proportional to the input voltage.

Iout = Vin / R9 or R16

Because the transistor base current also goes through the emitter resistor, there is an error term equal to 1 / the transistor current gain.

ak

 

Because the transistor base current also goes through the emitter resistor, there is an error term equal to 1 / the transistor current gain.

Of course using a MOSFET in place of the BJT will eliminate that source of error (if that error is important to the constant-current accuracy requirement).

 

Both circuits are what is called a transconductance amplifier, where the output current is directly proportional to the input voltage.

Iout = Vin / R9 or R16

Because the transistor base current also goes through the emitter resistor, there is an error term equal to 1 / the transistor current gain.

ak

Thank you so much.

Is LED in always ON state.
How opamp works.

 

Thank you so much.

Is LED in always ON state.
How opamp works.

How the op amp works?
Look at the design of most op amps. LM741 is good. We have a constant current circuit split between two legs and a current mirror at the bottom. Normally the current source is split equally between the two legs. When we apply a voltage between the two legs we upset this balance and generate an "error" signal that goes through a gain and output stage. Components on the output feed this signal back until both inputs are equal again. The components determine the output voltage and thus the gain.

(Op amps in 100 words or less.)