Biophotonics_Concepts_to_Applications

modulated at a frequency f =ω/2π(where the ordinary frequency f is measured in
hertz and the angular frequencyωis given in radians per second), the optical output
power of the device will vary as

P(xÞ¼P 0 1 þðÞxsi^2

hi 1 = 2

ð 4 : 4 Þ

where P 0 is the power emitted at zero modulation frequency.

Example 4.3A particular LED has a 5-ns injected carrier lifetime. When no

modulation current is applied to the device, the optical output power is 0.250

mW for a specified dc bias. Assuming other factors that might affect the

modulation speed are negligible, what are the optical outputs at modulation

frequencies f of (a) 10 MHz and (b) 100 MHz?

Solution:

(a) From Eq. (4.4) the optical output at 10 MHz is

P(xÞ¼

0 :250 mW

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 þ½Š 2 pðÞ 10  106 ðÞ 5  10 ^92

q ¼ 0 :239 mW¼ 239 lW

(b) Similarly, the optical output at 100 MHz is

P(xÞ¼

0 :250 mW

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 þ½Š 2 pðÞ 100  106 ðÞ 5  10 ^92

q ¼ 0 :076 mW¼ 76 lW

Thus the output of this particular device decreases at higher modulation

rates.

The modulation bandwidth of an LED can be defined in either electrical or
optical terms. Normally, electrical terms are used because the bandwidth is deter-
mined via the associated electrical circuitry. Thus themodulation bandwidthis
defined as the point where the electrical signal power, designated by Pelec(ω), has
dropped to half its constant value as the modulation frequency increases. This is the
electrical 3-dB point; that is, the modulation bandwidth is the frequency at which
the output electrical power is reduced by 3 dB with respect to the input electrical
power.
In a semiconductor optical source the light output power varies linearly with the
drive current. Thus the values of currents rather than voltages (which are used in
electrical systems) are compared in optical systems. Tofind the relationship,first let
i(ω)=ioutbe the output current at frequencyωand let i(0) = iinbe the input current
at zero modulation. Then, because Pelec(ω)=i^2 (ω)R, with R being the electrical

4.3 Light-Emitting Diodes 103