LED current

[loga]

Hi,
I have a circuit shown below but I don't know how to calculate the LED current(I3). The Vs is 5V.
Any ideas?

Edit : I can solve this problem.

[Bloch]

[size=78%]I have a circuit shown below but I don't know how to calculate the LED current(I3). The Vs is 5V.[/size]
Any ideas?

I love to make you home work 
Show how you think it can be calculated.

[IanB]

Yes, that doesn't look like a practically useful circuit, it looks more like an exercise.

What have you tried so far, and where did you get stuck?

[loga]

It's not my home work. It's a circuit that I see on the internet years ago and looks like a simple circuit but I can't figure out why no need a resistor on the LED branch. I just want to know how LED affect to current I1.

If there is no LED I can solve my problem with KVL.

Vs = I1*R1 + Vf + I1*R2 + I1*R3
Vs = I1*(R1+R2+R3) + Vf
or
I1 = (Vs - Vf)/(R1+R2+R3)

But when the LED is present in the circuit and I1 <> I2 but I1 = I2+I3.

To IanB: this circuit come from $0.99 product from china.

[Psi]

I can't figure out why no need a resistor on the LED branch.

R1 and R3 limit the total current in the circuit.
The current through R2 further splits this current.
So the LED is limited by the remainder.

[ve7xen]

Assume that Vf of both diodes is fixed, but they are not equal. D1 as a 1N4001 will have a lower Vf than the LED, so the LED Vf sets the drop across R2+D1. These assumptions should be enough to make KVL solvable.

[AndyC_772]

Combine R1 and R3 into a single resistor, located in the existing position for R1.

Assume a constant forward voltage Vf[led] for the LED, and a different constant voltage Vf[d] for the diode.

Call the battery -ve terminal GND. Now you know the voltages (with respect to GND) on both sides of both diodes.

I2 = (Vf[led] - Vf[d]) / R2

I1 = (Vs - Vf[led])  / (R1 + R3)

I3 = I1 - I2

... and solve from there.

[loga]

Combine R1 and R3 into a single resistor, located in the existing position for R1.

Assume a constant forward voltage Vf[led] for the LED, and a different constant voltage Vf[d] for the diode.

Call the battery -ve terminal GND. Now you know the voltages (with respect to GND) on both sides of both diodes.

I2 = (Vf[led] - Vf[d]) / R2

I1 = (Vs - Vf[led])  / (R1 + R3)

I3 = I1 - I2

... and solve from there.

At first I think as your solution. I must assume a constant forward voltage Vf[led] for the LED and I can find I3.
But assume that we don't know Vf[led] but we know Vf[d] say 0.7V.

I simulated this circuit on LT Spice, I tested on breadboard but I can't  find I3. with myself.
That is why I asked how to calculate I3.

[AndyC_772]

You need to know Vf[led], it affects I3! You can't calculate one without knowing the other, or at least, the relationship between the two.