V
avg
=
V
P
/π = 5/π = 1.6 V, PIV = 5 V
V
avg
=
V
P
/π =

50/π = 15.9 V,
PIV = 50 V
1.6
Determine the peak and average power delivered to
R
L
in figure below.
V
rms(sec)
= 0.5 ×
V
rms(pri)
= 0.5 × 120 V rms = 60 V rms
V
P(out)
=
V
P(sec)
–
0.7 =
√2 × 60 V rms
–
0.7 = 84.14 V
V
avg(out)
=
V
P(sec)
/
π
= 26.78 V
P
avg
=
V
avg(out)
2
/
R
L
= 26.78
2
/220 = 3.26W
1.7
Find the average value of each voltage in figure below.
a)
V
avg
=
V
p
/ π = 5 V / π = 1.59 V
b)
V
avg
= 2
V
p
/ π = 200 V / π = 63.7 V
c)
V
avg
= (2 (10 V) / π) + 10 = 16.4 V
d)
V
avg
= (2 (40 V) / π)
 15 = 10.5 V
1.8
Analyze the circuit in figure below using practical model.
(a) What type of circuit is this?
(b) What is the total peak secondary voltage?
(c) Find the peak voltage across each half of the secondary.
(d) Sketch the voltage waveform across
R
L
.
(e) What is the peak current through each diode?
(f) What is the PIV for each diode?
(a)
Centertapped fullwave rectifier
(b)
V
(sec)
= 120 V rms / 4 = 30 V rms,
V
P(sec)
= 1.414 (30 V rms) = 42.4 V
(c)
V
P(sec)
/ 2 = 42.4/2 = 21.2 V
(d)
(e)
I
P
=
V
P
/
R
L
= 20.5 / 1k = 20.5 mA
(f)
PIV = 2
V
P(out)
+ 0.7 V = 2(20.5) + 0.7 = 41.7 V
1.9
Draw the output waveform for the bridge rectifier in figure below.
1.9
Refer to the figure below, answer all the questions using practical model.
(a) Determine the transformer turn ratio, peaktopeak ripple and dc output voltages if
the transformer has a 40 V rms secondary voltage rating, and the voltage has a
frequency of 60 Hz.
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 Summer '16
 Dr. Yip
 Electric charge, Rectifier, voltage waveform