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Unformatted text preview: P1034 ' '
(a) L; IS on, L; IS on, and Q, is off. V = 7.5 volTs and I — 0 (b) QDeDsD4I/I on on on an 0 0
on on on on 2 V 2 mA
10 Off on on Off 5 V 5 mA Off on On off 5 v 5 mA owes: The ploT of Vversus Vin is: ok. When The source VOITage Is negu. Ivc, diode 03 is on and The oquuT VAT) is zero. For source volToges beTween 0 and 10 V, none of The diodes conducTs and V,(T) = v;( 3‘). Finally when The source volTage exceeds 10 V, D; is on and D; is in The breakdown region so “0,55 Refer To Figure P1055 in The bo P1057 50* Refer To The circui’r shown in Figure P1060 in The book. If The ou’rpuf
voltage afternst To become less than -5 V, The Zener diode breaks down
and current flows, charging The capacitance. Thus The negative peak is
clamped To —5 V. The input and oquuT waveforms are: l‘ y;
IO 1‘ (MS)
.40 1'5 ...
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This note was uploaded on 05/15/2010 for the course ELECTRONIC 101 taught by Professor . during the Spring '10 term at Kadir Has Üniversitesi.
- Spring '10