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Why phase reversal ce transistor?

Updated: 11/4/2022
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12y ago

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... in CE config.. as Vo=Vcc-Ic Rc nw as Vcc is const. v cn say Vo is directly proportional 2 (const.-Ic Rc) so nw when d base voltage is increased Ib will inc. thus, Ic will inc. cuz Ic=beta Ib thus Ic Rc drop increases... as Vo is prop 2 (const.-Ic Rc) thus, Vo will decrease for inc. in Ic...thus d curve for d +ve half cycle in which Ib is increasing drawn in 180 deg phase showing Vo is dec. n vice versa 4 -ve half cycle...

nw in CC... as d o/p is taken frm emitter... Vo= Ie Re so der's no -ve sign..n for d +ve cycle Ie will incr(dat's leakage current) as der's no-ve sign it'll b in phase wid i/p..hence no phase shift as leakage current doesnt incr. so much thus,d o/p voltage Ie Re doesnt inc much thus, d o/p doesnt amplify much thus dis config knwn as emiter follower...

nw in CB,... d I/P is appiled at emitter n o/p is taken frm collector... so, 4 d +ve half cycle d EB junc will b less FB thus causing dec in Ib n thus, in Ic... so dec in Ic causes inc. in o/p voltage(cuz Vo=const.-Ic Rc)..n during -ve half cycle d junc will b more FB(inc in Ic) so o/p voltage will b decreasing..hence der's no phase shift...

for more info...

cont.

shrey.dhingra51@gmail.com

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