The simplified circuit shown is used to modulate the output of the laser diode with a small time-varying input voltage signal (∆vin). The laser diode should be operated at its optimal DC operating conditions (i.e. Id=10 mA when Vd=2.2 V).
Determine the values of R1, R2and Rc, which bias the transistor in the middle of its operating range and allow the laser diode to operate at its optimal conditions.
LD
Figure 5.30 Circuit symbol for a laser diode.
Rc Ic
Ib R1
Vcc= 20 V
Ve = Vd Vc
Vb
b c
e C
1 µF
(a) (b) (c)
(d) (e)
A
Assume the transistor has a current gain of 100.
Hint: It may be useful to revise the transistor amplification and biasing sections of Chapter 4.
Solution
Since Id=Ie≈Ic, we require Ic=10 mA.
To operate the transistor in the middle of its range, we require Vc≈ Vcc
= ×20
=10 V
and Ve≈ Vcc. Fortunately Ve=Vd=2.2 V, which is approximately equal to Vcc.
Hence Rc=
=
=
=1 kΩ
and Ib= Ic=0.1 mA (since current gain is 100).
For correct operation of the transistor and R1/R2voltage divider, we want the current through R2(I) to be at least a factor of 10 greater than Ib; hence:
I=10 Ib
=10 ×0.1 ×10–3
=1 mA But:
Vb=Vbe+Vd
=0.7 +2.2
=2.9 V Hence:
R2=
=
=2.9 kΩ
We can now calculate the value of R1from the R1/R2voltage divider:
Vb=Vcc
( )
2.9 =20
( )
2.9R1+(2.9 ×2.9 ×103) =20 ×2.9 ×103 2.9R1=(20 −2.9) ×2.9 ×103 2.9R1=17.1 ×2.9 ×103
R1=17.1 ×103 R1=17.1 kΩ 2.9 ×103
R1+2.9 ×103 R2
R1+R2
2.9 1 ×10−3 Vb
I
1 100
10 10 ×10−3
20 −10 10 ×10−3
Vcc−Vc
Ic 1
10 1 10 1 2 1 2
1 The light emission from an LED can be varied by:
A varying the current passing through the diode B varying the illumination of the diode
C varying the reverse-biased voltage across the diode D all of the above.
2 Which one of the following statements is true about the difference between LDs and LEDs?
A LDs always emit more light than LEDs at the same drive current.
B LDs can be directly modulated whereas LEDs cannot.
C LDs emit light over a narrower range of wavelengths than LEDs.
D The output of LDs spreads out over a broader range of angles than LEDs.
3 Write down one advantage and one disadvantage of using a phototransistor instead of a photodiode in an optical detection circuit.
4 When the LDR shown in the diagram is in darkness, it has a resistance (RLDR) of 2 MΩ.
a Calculate the total resistance of the series circuit when the LDR is in darkness.
b Calculate the voltage drop across R2(Vout) when the LDR is in darkness.
c The LDR is now illuminated with a light source and its resistance decreases. Determine the resistance of the LDR if Voutnow is 6 V.
5 An LDR is used to determine whether the light intensity is adequate for a person working at a desk in an office. The LDR has the characteristics shown in Figure 5.13, and the voltage divider circuit shown below is used.
a Determine whether the light level at the desk is above the minimum acceptable level (approximately 100 mW m–2) if the output voltage of the circuit is 4 V.
b What is the intensity of the light radiation at the desk?
6 If the supply voltage (Vcc) is 6 V and the photocurrent (Iph) is 50 nA, determine:
a the voltage across the photodiode
• Light-dependent resistors (LDRs) are semiconductors whose resistance depends upon illumination. Their response is non-linear and slow.
• Photodiodes are pn-junction semiconductor devices which, in photoconductive mode (i.e. reverse biased), generate a photocurrent that depends upon illumina-tion. The response is linear and fast.
• Phototransistors are BJT devices that, when biased correctly, can generate a collector current that depends upon illumination. They have a high optical gain. The response is linear, and the response time moderate.
• Ep=hf =
• The light-emitting diode (LED) is a pn-junction device and the intensity of its emitted light is directly propor-tional to its forward-biased current. Its light output can be modulated rapidly, and it has a narrow range of emission wavelengths.
• The laser diode (LD) is a pn-junction device and the intensity of its emitted light is directly proportional to its forward-biased current. The device has special current-confining geometry and very high dopant levels and emits laser light (when the current is above a given threshold). Its output can be modulated rapidly, and it has a very narrow range of emission wavelengths.
hc λ
Vout Vcc
R2 6.8 kΩ 9 V
LDR
Vcc
20 kΩ
Vout 12 V
R
LDR 5.2 QUESTIONS
5.2 SUMMARY OPTICAL TRANSDUCERS
7 The photodiode shown in the circuit above has the same I–V characteristics as that shown in Figure 5.16.
Assume that the photodiode is to be operated in photoconductive.
a Determine the maximum current that can flow through RL.
b Determine the maximum light intensity that can be
reliably measured by the circuit.
8 The LED in the circuit has a switch-on voltage (Vs) of 2.0 V and an operating current of 20 mA. Determine the
value of RLfor the LED to be operating correctly.
9 All LEDs in circuits (i) and (ii) are identical. Each LED has a switch-on voltage (Vs) of 2 V and draws a current of 20 mA for optimal light production. (If the current is much smaller, the LED light output is too dim. If the current is much larger, the LED overheats and burns out.)
a For each circuit determine the RLthat gives optimum operation for all LEDs.
b Which combination LEDs ((i) or (ii)) emits the most light with these particular values of RL?
c Assuming circuits ((i) and (ii)) have exactly the same ideal battery power supply, determine which circuit will emit light (i.e. LEDs operating at optimum level) for the longest time. Explain your answer.
10 The following graph shows the resistance–temperature characteristics of a thermistor. A circuit uses this thermistor as part of a temperature sensor which can activate an LED whenever the temperature rises above a certain level.
a What is the resistance of the thermistor at 20°C?
b The potential difference across the LED at 20°C is 2.5 V and the current through it is 11 mA. What is the value of R?
c The LED is activated by a minimum potential difference of 2.0 V across it which gives a current through it of 4.8 mA. What is the minimum temperature that will activate the LED?
Vd
Vcc
RL Iph = 50 nA
(50 MΩ) 6 V
Vout Vcc
RL 500 kΩ (+10 V)
Vc
RL
9 V
Vc
RL Vc 9 V
RL 9 V
(i) (ii)
0 0.5 1.0 1.5 2.0
10 20 30 Temperature (C)
40 50
10 V
R T
LED
PHYSICS IN ACTION