ECE 2100:
Professor Alyosha Molnar
Homework Assignment
#2 (
Solution
)
Due Feb 17, 2010
Areas covered:
Node, mesh analysis.
Thevenin, Norton equivalents.
Maximum power
transfer.
Superposition.
LEDs,
Photodiodes,
Photovoltaics.
1)
Consider two LED lighting systems, one using amber LEDs (
λ
= 600nm) and the
other using blue (
λ
= 450nm).
a.
What is the expected turn-on voltage of each?
V
TO
= 1235/
λ
, so
for
λ
= 600nm, V
TO
=
2.06V
for
λ
= 450nm, V
TO
= 2.74V
b.
In each case you put 3 diodes in series, and place a resistor in series with
them.
In both cases you want to limit current to 50mA when 12V is placed
across the whole string (Current sets the number of photons generated).
In
each case, choose an approximate resistor value to accomplish this.
For the amber LED, 3 in series will give a voltage of ~6.18V, so the resistor will
see 5.82V, which with a current of 50mA, means R=V/I = 5.82V/.05A =
120
Ω
For the blue LED, 3 in series will give a voltage of ~8.22V, so the resistor will
see 3.8 V, which with a current of 50mA, means R=V/I = 3.8V/.05A =
75
Ω
2)
Modeling an optical isolator:
Optical isolators are used to pass signals between
parts of a system that are operating at different voltages (for example in hospital
rooms where everything is isolated on a floating ground to avoid discharges).
The
basic circuit is shown in Fig. 3a: it uses a resistor in series with an LED, whose light
couples to a photodiode.
Specifically, light intensity is proportional to LED current
(L = k
1
I
LED
), and photodiode current is proportional to absorbed light (I
PD
= k
2
L).
0 < k
1
, k
2
<1.
An I-V curve of the input port looks like that shown below (Fig 2b).
I
PD
, when plotted vs I
LED
gives the curve below it (Fig 2d).
a.
For Vin > V
TO,
the turn-on voltage of the LED, this circuit can be modeled
as shown.
Based on the curves shown, find values for V
TO
and A
I
and Rs.
The x-intercept of plot 1 is V
TO
=
2V
,
The slope of plot 2 is A
I
=
0.001
One over the slope of plot 1 (above V
TO
) is 1V/10mA =
100
Ω
b.
For, Vin = V
TO
+
Δ
Vin, what must R
L
be to ensure Vo =
Δ
Vin?
I
LED
= (Vin – V
TO
)/R
S
for Vin > V
TO
So
I
LED
= (
Δ
Vin)/R
S
= (
Δ
Vin)/ 100
Ω
I
PD
= A
I
·I
LED
= 0.001(
Δ
Vin)/ 100
Ω
=(
Δ
Vin)/ 100k
Ω
Vo = R
L
·I
PD
= R
L
· (
Δ
Vin)/ 100k
Ω
We want Vo =
Δ
Vin, so
R
L
=100k
Ω