S-87.1010 Electronics I / S-87.2113 Basic electronics, 1

S-87.1010 Electronics I / S-87.2113 Basic electronics, 1. Exercise / Marko Kosunen
1. a) Calculate the voltage over and current through load resistor RL in figures 1 and 2. Calculate
the current and voltage limits and draw a picture of them, when Rs << RL , Rs ≈ RL and
Rs >> RL .
b) Calculate the value of Rs , which gives the largest power to a load resistor RL . Draw a picture
of the load power as a function of RL .
Rs
+
J
RL
E
Rs
RL
−
Figure 2:
Figure 1:
Solution: a) Let’s define the equations for voltage over and the current trough load resistor in
figure 1:
IL =
E
Rs + RL
(1)
The voltage can be defined by voltage dividing
VL =
RL
·E
RL + Rs
(2)
Next, calculate the limits for load current and load voltage with the equations 1 and 2
Rs << RL :
E
→ IL ≈
RL
RL
·E =E
→ VL ≈
RL
(3)
(4)
(5)
Rs ≈ RL :
E
E
→ IL =
=
RL + RL
2RL
1
RL
·E = E
→ VL ≈
RL + RL
2
(6)
(7)
(8)
Rs >> RL :
E
→ IL ≈
Rs
RL
→ VL ≈
·E
Rs
(9)
(10)
(11)
1
Let’s define the equations for the load voltage and load current for figure 2. The load current
can be calculated by using current dividing:
IL =
Rs
·J
Rs + RL
VL = J · (Rs ||RL ) = J
(12)
Rs RL
RL + Rs
(13)
Next, calculate the limits for load current and load voltage with the equations 12 and 13
Rs << RL :
Rs
1
Rs
→ IL =
·J =
·J ≈
·J
Rs + RL
1 + RL /Rs
RL
Rs
Rs RL
≈J
≈ JRs
→ VL = J
RL + Rs
1 + Rs /RL
(14)
(15)
(16)
Rs ≈ RL :
RL
J
·J =
→ IL =
RL + RL
2
RL RL
RL
→ VL = J
=J·
RL + RL
2
(17)
Rs >> RL :
Rs
→ IL =
·J ≈J
Rs + RL
RL
Rs RL
= J RL
→ VL = J
≈ JRL
RL + Rs
+
1
R
(20)
(18)
(19)
(21)
(22)
s
We can see from figure 3a) that the non-ideal voltage source in figure 1 (voltage source E in
series with resistor Rs ) works as an almost ideal voltage source, when Rs << RL . In this case
the value of the source resistor Rs does not affect the load voltage. When Rs = RL , the load
voltage is half of the voltage from the non-ideal voltage source. When Rs >> RL , the load
voltage VL ≈ REs · RL and the load current IL ≈ REs ≈ current source J = REs .
We can see from figure 3c) that the non-ideal current source in figure 2 works as an ideal current
source when Rs >> RL . In this case the value of the source resistance Rs does not affect the
load current, almost all of the current will go trough the load resistor RL . When Rs = RL , the
current from the source is the same for both resistors and half of the current J. When Rs << RL ,
almost all of the current will flow trough the source resistance Rs and the voltage over the load
is ≈ J · Rs , which represents a good voltage source.
2
a)
b)
VL
VL
E
JRL
E
2
1
2 RLJ
1
RL
1
logRs
c)
RL
logRs
RL
logRs
d)
IL
E
RL
IL
E
2 RL
J
2
J
1
RL
1
logRs
Figure 3:
b) The load power (figure 1) will be
PL =
VL2
.
RL
(23)
The voltage VL over the load resistor can be calculated by voltage dividing (fig. 1)
VL =
RL
· Erms
Rs + RL
(24)
The load voltage VL could also be calculated from figure 2:
VL = Jrms · (Rs ||RL ) = Jrms ·
Rs RL
Rs + RL
3
(25)
By putting the equation 24 in equation 23 we get a value for the load power
L
Erms )2
( RLR+R
VL2
s
PL =
=
=
RL
RL
2
RL
E2
(RL +Rs )2 rms
RL
=
RL
E2 .
(RL + Rs )2 rms
(26)
Now we can draw a picture of the load power as a function of the load resistor. The figure 4
presents the load power we can get, when the load resistor changes between Rs /100 and 100Rs .
We can see, that the load power will approach zero when the load resistance is either very large
or very small compared to the source resistance. The load power seems to gain its maximum
when the load resistance is as large as the source resistance.
0.14
0.12
Output power [mW]
0.1
0.08
0.06
0.04
0.02
0
1
10
2
10
3
10
Load resistance [Ω]
4
10
5
10
Figure 4: The power consumed in the load.
To confirm the result, we can calculate the derivative of the load power and set it to be zero.
The load power can also be written as
2
PL = RL · (RL + Rs )−2 · Erms
.
(27)
In calculating the derivative, the following rules may be applied:
D(f · g) = f · Dg + g · Df,
(28)
Df n = n · f n−1 · Df
(29)
In equation 28 f and g are function depending on the same variable. When applying the derivative rule to the load power equation we can see that
f = RL
(30)
4
and
g = (RL + Rs )−2 .
(31)
The derivative of f is
Df =
∂f
∂RL
=
= 1.
∂RL
∂RL
(32)
The derivative of g can be gotten by using the equation 29.
∂g
∂RL
∂(RL + Rs )−2
=
∂RL
Dg =
= −2(RL + Rs )−3 ·
(33)
(34)
∂(RL + Rs )
∂RL
(35)
= −2 · (RL + Rs )−3
−2
.
=
(RL + Rs )3
(36)
(37)
With these we can calculate the derivative of the load power in regards of RL and then find the
zero of the function.
∂PL
2
= D(f · g) · Erms
∂RL
(38)
2
= (f · Dg + g · Df ) · Erms
−2
2
+ (RL + Rs )−2 · 1) · Erms
= (RL ·
(RL + Rs )3
1
−2RL
2
+
) · Erms
= 0.
=(
3
2
(RL + Rs )
(RL + Rs )
(39)
(40)
(41)
The derivative is zero, when Erms is zero or when the expression between the parentheses is
zero. Because vs,rms is a constant, we can only look at the expression between the parentheses.
−2RL
1
+
=0
3
(RL + Rs )
(RL + Rs )2
2RL
1
=
2
(RL + Rs )
(RL + Rs )3
Multiply both sides of the equations with (RL + Rs )2 .
5
(42)
(43)
1=
2RL
(RL + Rs )
(44)
RL + Rs = 2RL
(45)
Rs = RL
(46)
Based on the picture drawn and the calculation above, the maximum power is gained when
Rs = RL .
2. a) Modify the current amplifier circuit of Fig. 5 to equivalent voltage amplifier, transconductance amplifier and transresistance amplifier circuits.
b) What is the open-circuit voltage gain Avo , short-circuit transconductance Gm and open-circuit
transresistance Rm of the amplifier, if the current gain β = 100, Rin = 1kΩ, Rout = 100Ω,
Rs = 50Ω and RL = 100Ω.
Rs
i in
+
vin
vs
vout
Rout
βi in
Rin
RL
−
Figure 5:
Solution:
a)
Rout
Rs
+
Rs
+
vin
E
+
Avovin
Rin
-
RL
E
-
vin
Rin
Gmvin
Rout
RL
-
a)
b)
Rout
Rs
+
+
E
vin
Rmiin
Rin
-
RL
-
c)
Figure 6:
In the figure 6a) is the voltage amplifier, where the voltage is controlled by voltage. In the figure
6
6b) is the transconductance amplifier, where the current source is controlled by voltage. In the
figure 6c) is the transresistance amplifier, where the voltage source is controlled by current.
b) Let’s calculate the open circuit voltage gain with the help of the figures 5 and 6a).
The open-circuit voltage gain has the be the same for both of the circuits (open-circuit, when
RL = ∞):
Avo vin = βiin Rout
vin
iin =
Rin
Rout
Avo vin = βvin
Rin
Rout
Avo = β
= 10
Rin
(47)
(48)
(49)
(50)
The transconductance can be calculated with the help of figures 5 and 6b). In both circuits the
short-circuit currents have to be the same (short-circuit, when RL = 0):
βiin = Gm vin
vin
β
= Gm vin
Rin
β
Gm =
= 0, 1S
Rin
(51)
(52)
(53)
The transresistance can be calculated with the help of figures 5 and 6c). Both of the circuits
have to have the same open-circuit output voltage:
βiin Rout = Rm iin
(54)
Rm = βRout = 10000Ω
(55)
7
3. a) Calculate the power gain, transducer power gain, and available power gain of the circuit in
Fig. 5. Give the results also in decibels. Is the load RL matched?
c) If two identical amplifiers of the a-part are connected in cascade, what is the available power
gain of the amplifier chain in decibels?
Solution:
a)
Pi =
Vin2
Rin
(56)
Vs
Rs + Rin
Rout
iL =
βiin
RL + Rout
iin =
PL = RL i2L = RL
Pavs =
(57)
(58)
Rout
βiin
RL + Rout
2
=
2
β2
RL Rout
V2
(Rout + RL )2 (Rs + Rin )2 s
vs2
4Rs
(60)
Vout,open = Rout βiin
Pavo =
(59)
(61)
2
Vout,open
β 2 Rout
=
V2
4Rout
4(Rs + Rin )2 s
(62)
Power gain:
G=
2
PL
β2
RL Rout
=
= 250 ≈ 23.98dB
Pi
(Rout + RL )2 Rin
(63)
Transducer power gain:
GT =
2
PL
RL Rout
β 2 4Rs
=
= 45, 351 ≈ 16, 566dB
Pavs
(RL + Rout )2 (Rs + Rin )2
(64)
Available power gain:
GA =
Pavo
β 2 Rout Rs
= 45, 351 ≈ 16, 566dB
=
Pavs
(Rs + Rin )2
The circuit is matched, because RL = Rout and/or GA = GT .
8
(65)
b)
β 2 Rs Rout
Ga1 =
= 45.351 ≈ 16, 566dB
(Rs + Rin )2
2
β 2 Rout
= 82.645 ≈ 19, 172dB
Ga2 =
(Rout + Rin )2
(66)
(67)
Gatot = Ga1 + Ga2 = 16, 566dB + 19, 172dB ≈ 35.738dB
Rs
E
i in
+
−
(68)
i in2
Rin
bi in
Rout
vout
bi in2
Rin
Rout
RL
Other way to calculate, according to the figure above (in the figure b=β in the calculation):
2
Rout
Rout
Vs
= β2
Rout + Rin
(Rout + Rin )(Rin+Rs )
2 2
2 2
(β Rout ) Vs
=
(Rout + Rin )2 (Rin + Rs )2 4Rout
Vout,open = βiin2 Rout = βRout βiin
2
Vout,open
4Rout
V2
Pavs = s
4Rs
2
(β 2 Rout
)2 Rs
Pavo
=
≈ 3748, 06 ≈ 35, 74dB
GA =
Pavs
(Rout + Rin )2 (Rin + Rs )2 Rout
Pavo =
(69)
(70)
(71)
(72)
4. Signal source has a source resistance Rs of 0.5M Ω and its voltage amplitude without a load
is 30mVrms . Signal needs to be amplified so that we’ll get at least 0.5W to the load RL (100Ω).
There are three different voltage amplifier stages in use (fig 7)
1. Rin = 1M Ω, Avo = 10, Ro = 10kΩ
2. Rin = 10kΩ, Avo = 100, Ro = 1kΩ
3. Rin = 10kΩ, Avo = 1, Ro = 20Ω
a) Draw a circuit where signal source is connected to the amplifier which is driving a load
resistor. Replace the common voltage amplifier symbol in figure 7 with the voltage amplifier
circuit model.
b) Calculate the rms-amplitude appearing at the inputs of the amplifiers when different amplifiers 1-3 are connected to the signal source.
c) Use the given amplifiers to design an amplifier, which will drive at least a 0.5W power PL to
the load RL .
Solution:
9
Avo
Rin
Ro
Figure 7:
a)
voltage amplifier
Rs
vs
vin
Ro
+ A v
vo in
−
Rin
RL v o
Figure 8:
b) Vs = 30mVrms and Vin =
Rin
V
Rs +Rin s
In case 1: Vin = 20mVrms
In case 2: Vin = 0.59mVrms
In case 3: Vin = 0.59mVrms
Rs
vs
vin
Rin
c) Let’s first calculate the needed output voltage Vo , when RL = 100Ω and PL = 0.5W .
10
2
The load power PL = RVoL , where we can get the output voltage Vo =
So, the needed voltage gain is Av = VVos = 235.7
√
PL · RL ≈ 7.07Vrms .
Clearly we need at least the first and second amplifier stages or the voltage will not be enough.
Let’s draw the two stage voltage amplifier:
Rs
Ro1
vin
Rin
vs
Ro2
v2
vin2
Rin2
Av1Vin1
Av2Vin2
Let’s calculate the output voltage using the values of amplifier stages 1 and 2:
Rin2
Av1 Vin1
Rin2 + Ro1
Rin2
Rin
= Av2
Av1
Vs = 10Vrms
Rin2 + Ro1
Rin + Rs
Vo = V2 = Av2 Vin2 = Av2
(73)
(74)
Because this will drive a load, the voltage will be dropped severely, so it will not be able to
drive the needed power to the load. So we will need also the third stage.
Ro2
v2
vin3
Ro3
Rin3
RL
Av3Vin3
Vo =
RL
Rin3
V2 = 7.6Vrms
Ro2 + Rin3 Ro3 + RL
The power with this output voltage is PL =
Vo2
RL
(75)
= 0.5739W , which is enough.
11

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