1
\$\begingroup\$

Why Built-in Potential of White,Red,Green and Blue LEDs are in the order:
$$V_W > V_B > V_G > V_R$$
where $$V_W \implies Built-in \quad Potential \quad of \quad White \quad LED $$

My Approach:
We know: $$\lambda_R>\lambda_G>\lambda_B$$
$$\therefore E_{gR}< E_{gG} < E_{gB} $$ as Band gap Energy (Eg) is inversely proportional to wavelength (lambda) of light

Now how to correlate Buit-in potential of White LED ?? any hints or suggestions please....?

\$\endgroup\$

3 Answers 3

3
\$\begingroup\$

The dies used in "White" LEDs are actually blue, with rare earth phosphor or phosphors to deliver longer wavelength yellow and red light (depending on the desired color temperature).

Electrically, they behave as blue LEDs.

\$\endgroup\$
3
  • \$\begingroup\$ so how its built-in potential is greater than Blue LED?? \$\endgroup\$
    – Suresh
    Feb 17, 2018 at 11:33
  • \$\begingroup\$ "so how its built-in potential is greater than Blue LED??" - it may not be. I tested a bunch of equivalent blue and white LEDs (same manufacturer, same series, all purchased at the same time) and their voltage drops were identical (2.97~3.07V @10mA). \$\endgroup\$ Feb 17, 2018 at 14:32
  • \$\begingroup\$ There are different wavelengths of blue and different internal resistances. Also heterojunction LEDs appear to have a higher drop than simple semiconductors. \$\endgroup\$ Feb 17, 2018 at 20:17
2
\$\begingroup\$

The forward voltage drop of a LED is a function of the wavelength of light it produces. The more energetic shorter wavelengths require a higher voltage from the LED that produces them.

There are several different technologies for making a "white" LED. Some use a blue LED, with phosphors to convert some of the blue light to longer wavelengths. Others use a near UV LED, with all the visible light coming from re-emission from phosphors.

White LEDs therefore have the same or a bit higher voltage drop than blue LEDs.

There is yet another, less common, technology for white LEDs, which is multiple single-color LEDs in the same package. If the single LEDs are wired in series, these will have a even higher voltage drop due to that of the multiple LEDs adding up. However, I haven't seen one of these in a while. There are multi-color LEDs, but these give separate access to the individual red, green, and blue LEDs so that you can control the overall color. That's not the same thing.

\$\endgroup\$
1
\$\begingroup\$

Vf= Vt+If*Rs

I usually call Rs=ESR and I know all diode ESR = k/Pd for k= ~0.5 to 1 and depends on thermal design for Pd= Vmax*Imax rated @ 85'C

Ideal equation only covers Vt and not Pd or ESR.

\$Vf=Vt+If*k/Pd\$

use this with k= 1 for epoxy types and k=0.5 for ceramic SMD approximately

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.