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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

schematic

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 enter image description here

  • 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.

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    \$\begingroup\$ Can I say well done for an excellently-written and detailed question :-) A good example for OPs. \$\endgroup\$
    – TonyM
    Commented Mar 24, 2020 at 21:54
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    \$\begingroup\$ 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. \$\endgroup\$
    – glen_geek
    Commented Mar 24, 2020 at 21:58
  • \$\begingroup\$ Thank you @TonyM :) \$\endgroup\$
    – Embedded_Dude
    Commented Mar 24, 2020 at 22:04
  • \$\begingroup\$ @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 \$\endgroup\$
    – Embedded_Dude
    Commented Mar 24, 2020 at 22:27

1 Answer 1

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All your calculations are correct. The graph is a typical I-V curve for this LED, the drop will be higher for some parts, and less for others. The table lists the maximum voltage drop at 20 mA to be 3.4 V. They did not list a minimum voltage. This is a common practice, they don't want to reject parts because they are "too good" (low voltage drop is normally considered a good thing).

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  • \$\begingroup\$ But I thought the table lists the typical Forward Voltage (3.2V) and the maximum Forward Voltage (3.2V), at Forward Current = 20 mA. That makes sense the graph gives the maximum/typical Vf values at certain If currents, but not the minimum. \$\endgroup\$
    – Embedded_Dude
    Commented Mar 24, 2020 at 22:50
  • \$\begingroup\$ @Forat - the graphs in datasheets are usually the typical values for a typical device at a specific temperature. Not only can a specific device be different but the specific device will probably vary depending upon the conditions, especially temperature. \$\endgroup\$
    – Kevin White
    Commented Mar 25, 2020 at 0:58
  • \$\begingroup\$ @KevinWhite Yeah, I got that, but it might make circuit analysis regarding diodes abit confusing \$\endgroup\$
    – Embedded_Dude
    Commented Mar 25, 2020 at 1:06
  • \$\begingroup\$ @Mattman994 Just an afterthought, we is having a low voltage drop at the LED a good thing? if we have a 5V source connected in series with an 82 Ohm resistor and the Blue LED with Vf = 3.2V, I = Vr/R = 1.8/82 = 22mA, the power dissipated into the circuit, P =( 5*22)mW = 110mW, if we change Vf to 2.96V, and the Vr to 2.04V, I = 2.04/82 = 25mA, P = (5*25)mW = 125mW \$\endgroup\$
    – Embedded_Dude
    Commented Mar 25, 2020 at 1:18
  • \$\begingroup\$ @Forat - Multiple LEDs are often put in series, so in general, a low voltage drop is better. For your design, the wide possible range of the voltage drop makes it difficult to predict the current. But, unless you have critical brightness requirements, don't worry too much about it. A 50% change in current does not appear to the human eye to be 50% different in brightness because the eye response is logarithmic. \$\endgroup\$
    – Mattman944
    Commented Mar 25, 2020 at 1:30

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