Q2
a)
Figure 1
Figure 1 is a schematic diagram of a multistage amplifier. Answer questions below by
referring to the schematic. Use root means square (rms) value in calculation.
i)
Calculate the output voltage, Vout if the input voltage given is
300 2 sin 2π 100tmV .
ii)
 R   100k Ω 
A1 = 1 + 2  =  1 +
 = 11
R1  
10k Ω 

 R   100k Ω 
A2 = 1 + 4  = 1 +
 = 1.5
 R3   200k Ω 
 R   70k Ω 
A3 = 1 + 6  = 1 +
 = 2.4
 R5   50k Ω 
ATOTAL = A1 A2 A3 = 11×1.5 × 2.4 = 39.6
(
Vout ( rms ) = ATOTAL × Vin ( rms ) = 39.6 300 2 sin100tmV
)
Vout ( rms ) = 11.88sin 100tV
iii)
What will happen when R6 opens? Draw the input and output waveforms.
A3= open loop gain which is very high. Therefore any small input will be
amplified by the open loop gain at A3 and will cause the output to be clipped at
saturation voltage of the amplifier.
iv)
Determine new output voltage, Vout if the value of R6 changes to 100kΩ.
New _ value _ of _ R3 = 100k Ω
Therefore
 R   100k Ω 
A3 = 1 + 6  = 1 +
=3
50k Ω 
 R5  
ATOTAL = A1 A2 A3 = 11× 1.5 × 3 = 49.5
Vout ( rms ) = Vin ( rms ) × ATOTAL
Vout ( rms ) = 300sin100tmV × 49.5 = 14.85sin100tV
v)
Name the type of multistage amplifier shown in Figure 1.
- Non-Inverting Multistage Amplifier
b) Controlled sources can be categorized based on its circuit configurations. It can be
configured as voltage/current to control current/voltage or vice versa. Answer the
questions below.
i)
Name 4 type of controlled sources configuration circuit.
-
Voltage Controlled Voltage Source (VCVS)
Voltage Controlled Current Source (VCIS)
Current Controlled Voltage Source (ICVS)
ii)
Current Controlled Current Source (VCVS)
Design a VCVS amplifier to meet the specifications given.
Inverting Configuration
Closed Loop Gain, Acl=32
Input Resistance, R1/i=1kΩ
iii)
Design a VCIS amplifier to meet the specifications given.
Non-Inverting Configuration
Closed Loop Gain, Acl=63
Input Resistance, R1/i=2kΩ
iv)
Design an ICVS amplifier to meet the specifications given.
Inverting Configuration
Input Current =1 mAp-p
Output voltage =6Vp-p
v)
Design an ICIS amplifier to meet the specifications given.
Inverting Configuration
Input Current =2 mAp-p
Output Current =66 mAp-p
R2 =1kΩ, RL =1kΩ
c)
Figure 3
Based on the circuit shown in Figure 3;
i)
Derive the equation to calculate overall closed loop gain,
instrumentation amplifier (IA).
Vout
of the
V1 − V2
R 

 R 
Vout1 =  1 +
 V1 + 
 V2
 aR 
 aR 
R 

 R 
Vout 2 = 1 +
 V2 + 
 V1
 aR 
 aR 
Vout = Vout1 − Vout 2

R 
R 
 R   
 R  
Vout = 1 +
 V1 + 
 V2  − 1 +
 V2 + 
 V1 
 aR    aR 
 aR  
 aR 
 2R 
 2R 
Vout = 1 +
 V1 −  1 +
 V2
 aR 
 aR 
 2R 
Vout = [V1 − V2 ] 1 +

 aR 
Vout
 2R 
= 1 +
[V1 − V2 ]  aR 
ii)
Closed loop gain of the IA is set to 40dB and all internal resistances have
equal value of 2kΩ, determine the value of external resistance, aR.
Acl = 40dB = 100(real _ number )
Acl = 100 = 1 +
2R
aR
2
99
External resistance ( aR ) is the multiplication of internal resistance ( R ) with the factor of 2.
a=
 2 
Rexternal = aR =   × 2k Ω = 40.40Ω
 99 
iii)
Draw the IA output waveform based on the specifications below.
V1 = 100sin 2π 1000tmV
V2 = 80 sin 2π 1000tmV
DC Offset = + 5V
V1 = 100sin 2π 1000tmV
V2 = 80 sin 2π 1000tmV
Vout = (V1 − V2 ) ×100 = (100sin 2π 1000tmV − 80 sin 2π 1000tmV ) ×100 = 2 sin 2π 1000tV
With DC Offset = + 5V
Vout = Vout +VDC = 2sin 2π 1000tV + 5V