Download A LED Exercise

A LED Exercise
NAME ________________
<Watch for correct symbol translations! μ is mu (micro), Ω is ohm, Δ is delta, → is a right arrow, • is a bullet.>
1. Initially the power supply is set at Vs = 6V. What value
is required for the resistor so that the LED operates at 60
mA?
LED characteristic
100
90
80
70
Current (mA)
An LED which has the characteristics shown in this graph
is to be used in the circuit below in which both VS and R
can be varied. For this LED the switch on voltage (VD) is
1.7 volt which produces a current of 10 mA at which point
the LED will just glow dimly. Let us say that the diode
operates best at 60 mA, but will fail if the current exceeds
80 mA for too long.
60
50
40
30
20
10
0
0
0.5
1
1.5
2
2.5
Voltage
2. In a similar way, find the minimum value of the resistor that could be used without
damaging the LED.
3. If a current of 20 mA is flowing and the resistor is 400 Ω, what is the supply voltage?
4. If we find a voltage across the 200 Ω resistor of 2.0 V what is the supply voltage and will the LED be
glowing?
5. If there is a voltage drop of 6.5 V across the resistor (R is not 200 Ω this time) and 2.1 V across the LED,
what is the supply voltage and the resistance of R?
6. If the supply voltage is 8 V and R is 300 Ω, what will be the current in the LED?
PTO...
The transistor amplifier
NAME ______________________
The circuit shown is a typical transistor amplifier circuit.
Signal voltages at Vin are amplified by ‘transistor action’
and appear at Vout.
Assume that Vcc = 10 V and that it is desired that Vc with
no signal should be 6 V and Ve should be 1 V. The current
gain of the transistor is 100. When Vbe is 0.7 V the base
current is 15 μA.
Show your working for the problems below.
When the transistor is correctly biased as above what is:
1. The base voltage (Vb)?
2. The collector current (Ic)?
3. The emitter resistor (Re)?
4. The collector resistor (Rc)?
5. The collector-emitter voltage Vce ?
6. A rule of thumb suggests that R2 should be about 10 times the value of Re. If this rule is used, what should
be the values of R1 and R2 ?
7. What would be the current through the R1 and R2 combination? Would these values for the resistors be
suitable in terms of keeping a stable base current?
8. It is found that a momentary increase in Vb of 0.1 V produces an increase in the base current of 2 μA. What
change will this produce in Vc ? Describe the changes that will occur in Vout. And so what is the voltage
amplification of the circuit?
PTO...
ANSWERS – LED test and Transistor Amplifier:
Note on powers of ten and units:
Remember that in V = IR if I is in mA and R in kΩ then V will be in volts as the milli and kilo cancel each
other.
It is easy to make mistakes in converting between units. Make sure you know how to do this!
Note that 1 μA = 0.001 mA, 10 μA = 0.010 mA, 100 μA = 0.100 mA and 1000 μA = 1 mA etc. (If you use three
decimal places the 0.123 mA represent the 123 μA, etc. – but be careful about the number of zeros!)
LED Test
1. VR = 6 – 2.1 = 3.9 V and so R = 3.9/0.060 = 65 Ω
2. VR = 3.85 V, So R = 3.85/0.080 = 48 Ω
3. VR = 8 V & VD = 1.85 V so VS = 9.85 V
4. I = 2/0.2 = 10 mA and so VD = 1.7 V (or 1.75 V) so VS = 2.0 + 1.7 =3.7 V
5. VS = 6.5 + 2.1 = 8.6 V so from graph I = 65 mA (or 60 → 70 mA) so R = 6.5/65 = 100 Ω [6.5/65 = 0.100kΩ
= 100 Ω] or 93 → 108 Ω depending on current taken from graph.
6. At VD = 2.0 V, I = (8 – 2)/0.3 = 20 mA → VD = 1.85
At VD = 1.85 V, I = 6.15/0.3 = 20.5 mA → VD = 1.85 so 20.5 mA is the answer.
TRANSISTOR AMPLIFIER test
Some general notes:
• Remember that in these amplifiers the base-emitter voltage is always close to Vbe = 0.7 V when the transistor
is biased for linear amplification. But the voltage across the emitter resistor must be added to the zero line to
obtain the actual ‘base voltage’ Vb.
• The relation between the collector current Ic, collector voltage Vc, and the collector (or load) resistor, Rc , is
given by ΔV = Ic  Rc where ΔV is the difference between Vcc (or VS) and Vc. Also remember that the collector
current is related to the base current by the current gain (AI).
• Although the current through the emitter is actually equal to Ic + Ib it is very closely equal to Ic as Ib is very
small by comparison – as the current gain (AI = Ic/Ib) is around 100 or 200 for a normal transistor.
• The voltage gain, by comparison, depends on the actual configuration of the circuit. In particular, it depends
on the ratio of Rc/Re (the smaller Re the greater the voltage gain). We don’t need to worry about how or why this
is. Any questions about voltage gain would most likely be of the form of question 8 in this test.
• There are various ways of looking at the voltage divider and relating R1 and R2 to the voltages involved.
Fundamentally, as the current through these resistors is assumed to be the same, the voltage across them is
proportional to the resistance, so for example: R1/R2 = ΔV1/ΔV2 but also ΔV1 + ΔV2 = Vcc
• When we deal with actual ‘signal’ voltages, we are dealing with changes in Vin and Vc (or Vout), that is ΔVin
and ΔVc (ΔVout). These are often represented as lower case letters eg. vin and vout and the corresponding currents
as iin and iout. Thus Av = vout/vin and AI = iin/iout.
Answers
1. Vb = Ve + Vbe = 1.0 + 0.7 = 1.7 V
2. As Ic = AI  Ib = 100  15 = 1500 μA = 1.5 mA
3. Re = Ve/I = 1.0/1.5 = 0.667 kΩ = 667 Ω
4. As there is a 4 V drop (10 – 6) across the resistor, with a current of 1.5 mA, then R = 4/1.5 = 2.67 kΩ
5. This is the difference between Vc and Ve = 6 – 1 = 5.0 V
6. In this case: R2 = 6.67 kΩ (10 times Q3) and so R1 = R2 (ΔV2/ΔV1) where ΔV1 = 10 – 1.7 = 8.3 V and so R1 =
6.67  8.3/1.7 = 32.6 kΩ
7. R1+2 = 39.3 kΩ and so I = 10/39.3 = 0.254 mA = 254 μA which is about 17 times the base current and so the
base current will make little difference to the Vb.
8. As the current gain is 100, the difference of 2 μA in the base current will produce a change of ΔI = 200 μA =
0.2 mA in the collector current. In turn this will produce a change in voltage across the RC of ΔVc = ΔI  R =
0.2  2.67 = 0.53 V. This will lower the Vc and so it will drop to 5.47 V. Vout will also briefly drop by 0.53 V
from its normal DC level–whatever that was–but will return to that level after the brief drop. This resulted from
a change in the base voltage of 0.1 V and so the voltage gain is AV = ΔVout/ΔVin = 0.53/0.1 = 5.3