Eeng 325 Lab 2

# Eeng 325 Lab 2 - Lab 2 DC Power Supply Alex Lemon October 8...

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Lab 2: DC Power Supply Alex Lemon October 8, 2007 1 Introduction Nearly all electronic circuits require direct-current (DC) power supplies. How- ever, because alternating-current (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 easily-transmitted 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 full-wave rectiﬁer Figure 1: A peak rectiﬁer utilizing a four-diode full-wave bridge rectiﬁer (i.e. remove the capacitor from Figure 1 ). Let the voltage through the load 1

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be 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 ideal-diode 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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## This note was uploaded on 07/19/2008 for the course EENG 325 taught by Professor Hurkoser during the Fall '07 term at Yale.

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Eeng 325 Lab 2 - Lab 2 DC Power Supply Alex Lemon October 8...

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