I want to design a simple circuit with a 9V Alkaline Battery (it currently have only 6.80V), a Voltage Regulator (uA7805C), a Blue LED (151051BS04000 from Wurth Electronik) and a current limiting resistor.
The circuit is shown below using the Circuit Simulator
simulate this circuit – Schematic created using CircuitLab
According to the Blue LED's datasheet Vf = 3.2V @ If = 20mA, and this is the I-V graph for it
The uA7805C will output a steady voltage of 4.8V - 5.2V
Assuming the VR Vout = 4.80V, Vf = 3.2V, hence i = 20mA.
- i = (Vout - Vled)/R,
- Vout - Vled = i*R,
- R = (Vout - Vled)/i,
- R = (4.80 - 3.20)/(20m) = 0.08 KOhm = 80 Ohm
I have an 82 Ohm resistor (5% tolerance, 0.25W), recalculating i, i = 19.5 mA ~ 20mA, there won't be a change in Vf.
When I built the circuit using a breadboard and measured used the oscilloscope (multimeter wasn't available) to measure voltage drops across the components, I found Vout = 5.0V, Vled = +2.96V, and Vr = +2.04V.
calculating i, i = Vr/R = 2.04/82 = 26.2mA
Here is what I don't understand, Vled = +2.96V, and according to the graph If = 2.5mA at Vf = +2.96V, but I calculated i = 26.2mA, and according to the graph Vf should be 3.2V
Kindly help me with this problem.