YEAR 12 PHYSICS ELECTRONICS REVISION
Question 1.
_
Output
+
110 V rms 60 Hz
Fuse
50uF
0V
2.2k
0V
0V
10:1
Ripple Factor: r =
Transformer
VPeak =
Vr  pp 
V DC
VP N P

VS N S
2 VRMS
Approx. peak to peak ripple voltage =
1
Vpeak(rect)
fRC

1 
V p ( rect )
Approx DC Voltage = 1 
 2 fRC 
Calculate the ripple factor for this filtered bridge rectifier.
Vp(pri) = 1.414Vrms = 1.414(110V) = 155.56V
Vp(sec) = 155.56 ÷ 10 = 15.56V
Vp(rect) = Vp(sec) – 1.4V = 15.56V-1.4V = 14.16V
 1 


1
V p ( rect )  
14.16V  0.215V
Vr ( pp)  
6 
fR
C
(
60
Hz
)(
22000

)(
50

10
)


 L 



1 
1
V p ( rect )  1 
14.16V  14.05V
VDC  1 
3
6
2
fR
C
(
2

(
60
Hz
)(
22

10

)(
50

10
F
))


L


r
Vr ( pp)
VDC

0.215
 0.0153
14.05
Explain why the fuse is necessary and what else could be added to the circuit to
ensure safety. Justify your reasons.
The fuse is necessary because the initial current will flow through the capacitor
(as it is now DC) with no resistance causing an extremely high current called a
surge current which could destroy the diodes.
A small resistor could be placed between the rectifier and the capacitor. It must
be smaller than the load resistor.
Determine the peak to peak ripple Voltage if the capacitor is increased to 100µF and
the load resistor to 1.2kΩ.


1
14.16V  1.97V
Vr ( pp)  
6
 (60 Hz )(100  10 )(1200) 
Question 2.
In your second experiment, you calculated time constants for various RC circuits.
Considering your results think about the following circuit:
S1
S2
R = 2.0kΩ
R
C = 10μF
V
C
Calculate the time constant if S1 is closed and then calculate what would happen to
the time constant if a second capacitor was placed in series with the other.
τ = RC = 0.02s
Assuming the second capacitor is also 10μF:
CT = 5 × 10-6F
τ = 0.01s
If S1 is opened and S2 is closed, describe the current flow in the circuit.
The capacitor would discharge conventional current upwards through the
resistor in a clockwise direction.
Explain why S1 and S2 should not be closed at the same time.
Because there is no resistance in the section that contains S2 there would initially
be a huge current running through S2, effectively a short circuit, which would
probably destroy the wire.
Question 3.
In your band-pass filter experiment, you had great difficulty getting a clear sound,
unless you used multiple low and high pass RC filters.
You were instructed to use the frequency you discovered during the Fourier analysis
to calculate the size of the resistance required.
As you discovered in your research, the response at the cut off frequency is 70.7% of
the Vin; and using multiple RC combinations (each one called a pole) meant a drop off
in signal intensity of 70.7%. This required an amplification, in our case supplied by an
old speaker.
If we had designed the filter using frequencies (fc1, fc2) either side of your specific
frequency, it could be possible to have a full response for the specific frequency as
shown in the graph below.
Vout/Vin Vs Frequency
The Quality Factor (Q) of a band pass filter is the ratio of the centre frequency to the
bandwidth (BW). If Q < 10 the filter is said to be a wide band and Q > 10 it is said to
be narrow band
BW = fc2 – fc1
and the frequency about which the pass-band is centred is
called the centre frequency, f0.
f 
f c1 f c 2
A crude way of improving Q is to use multiple filters as we tried.
You are to redesign your circuit with a Q value of 15. Calculate the required
resistance and draw the new circuit showing a low and high pass filter.
Q = 15 = f0 ÷ BW e.g. f0 = 700Hz
∴ BW = 46.67Hz
∴ fc1 ≈ 676.67Hz, fc2 ≈ 723.34Hz
∴ fc1 is the frequency for the high pass filter and fc2 for the low pass.
If the capacitor used was 0.1μF
 1 
∴ f c1  

 2RC 
 1  

1
  
  2352
Rc1  
6
 2f c1C   2 (676.67 Hz )(0.1  10 F ) 


1
  2200
∴ Rc 2  
6
2

(
723
.
34
Hz
)(
0
.
1

10
F
)


0.1uF
2200
Ohms
2352Ω
0.1uF
2200Ω
2352
Ohms
0V
Output
0V
0V
What would be the bandwidth if the quality was tripled? BW = 700 ÷ 45 ≈ 15.56Hz