# Circuit Analysis with an LED

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.

• Can I say well done for an excellently-written and detailed question :-) A good example for OPs. Commented Mar 24, 2020 at 21:54
• A 9V battery that puts out only 6.8V is a nearly-dead battery. It is surprising that the 7805 regulator is giving you 5.0V at its output. Commented Mar 24, 2020 at 21:58
• Thank you @TonyM :) Commented Mar 24, 2020 at 22:04
• @glen_geek, it is still powering the circuit, I am assuming it is at the end of it's life. I used to get a Vout of 5.00V, then 4.80, 4.40, and now 4.20V, but I don't think the problem with the regulator, I will try to re-design the circuit again without the regulator and post the findings in an edit to my post Commented Mar 24, 2020 at 22:27