Lab 2: DC Power Supply
Alex Lemon
October 8, 2007
1 Introduction
Nearly all electronic circuits require directcurrent (DC) power supplies. How
ever, because alternatingcurrent (AC) power can be transmitted much more
eﬃciently, the most readily available power supplies (i.e. wall sockets) use al
ternating current. Therefore, it is very important to be able to convert the
easilytransmitted AC power into the DC power necessary for most electronics.
2 Peak Rectiﬁer
The ﬁrst part of the DC power supply uses four diodes to create what is referred
to as an unregulated DC power source. An unregulated DC signal is centered
at some DC oﬀset that is much larger than an AC ripple with which the signal
oscillates about the DC oﬀset.
Figure 1
shows the peak rectiﬁer that I used
for this stage of the power supply. First, consider only the fullwave rectiﬁer
Figure 1: A peak rectiﬁer utilizing a fourdiode fullwave bridge rectiﬁer
(i.e. remove the capacitor from
Figure 1
). Let the voltage through the load
1
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View Full Documentbe chosen with its positive terminal between diodes
D
1
and
D
2
and its negative
terminal between diodes
D
3
and
D
4
. Let
v
A
be the voltage at the node between
D
1
and
D
2
and let
V
B
be the voltage at the node between
D
3
and
D
4
. Finally,
let’s choose to use the idealdiode model for simplicity. First, consider the case
where the ﬁrst and third diodes are conducting and the second and fourth diodes
are oﬀ. The voltage drops across the two conducting diodes must be equal to
zero.
v
D
1
=
v

in

v
A
= 0
v
D
2
=
v
+
in

v
B
= 0
Combining these equations and using the deﬁnitions of
v
in
and
v
out
.
(
v
A

v
B
) + (
v
+
in

v

in
) = 0
v
out
+
v
in
= 0
v
out
=

v
in
Using the condition for these diodes to be in the on state also gives a condition
on
v
in
such that the rectiﬁer will operate in this regime. In particular, we
know that the current through the ﬁrst and third diodes must be positive (note
that the current passing through the diodes is the same as the current passing
through the load). This condition is equivalent to
i
D
=
v
A

v
B
R
L
=
v
out
R
L
>
0
.
This implies that
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 Fall '07
 HurKoser
 Direct Current, voltage regulator, Rectifier

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