Current sources slide

# Current sources slide - β / 2. For β 0 = 100 and V A...

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A simple two-transistor current source.

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I c1 = I c2 Summing currents at the collector of Q 1 yields 0 2 1 1 = - - F C c ref I I I β and thus 2 1 2 1 C F ref C I I I = + = if β F is large, the collector current of Q 2 is nearly equal to the reference current: R V V I I on BE CC ref C ) ( 2 - = 2245
Simple current source with current gain.

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The emitter current of transistor Q 3 is equal to 2 2 1 3 2 C F F C F C E I I I I β = + = - The base current of transistor Q 3 is equal to ( 29 2 3 3 1 2 1 C F F F E B I I I + = + - = Finally, summing currents at the collector of Q 1 , we obtain ( 29 0 1 2 2 1 = + - - C F F C ref I I I Since I C1 and I C2 are equal, F F ref C o I I I + + = = 2 2 2 1
Widlar current source

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Summing voltages around the emitter-base loop, assuming that V A is infinite, and neglecting base currents results in 0 2 2 2 2 1 1 = - - R I I I In V I I In V C S C T S C T For identical transistors, I S1 and I S2 are equal, the equation becomes 2 2 2 1 R I I I In V C S C T = 0 2 2 2 1 = - - R I V V C BE BE and thus
Cascode current source with bipolar transistors.

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Thus, cascode current sources can boost the output resistance and equivalent open-circuit voltage approximately

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Unformatted text preview: β / 2. For β 0 = 100 and V A =130V, we have V V I r V A o o Thev 6500 2 2 = = = β At an output current I o = 1mA, we find that Ω = = M mA V R o 5 . 6 1 6500 N-Channel mirror In the most general case, the ration of i O to i I is + + -- = 1 2 1 2 2 1 2 2 1 2 1 1 1 K K v v V V V V L W W L i i DS DS T GS T GS I O λ + + = 1 2 2 1 2 1 1 1 DS DS I O v v L W W L i i If v DS2 = v GS1, then the ratio of i O /i I becomes = 2 1 2 1 L W W L i i I O + + = 1 2 1 1 DS DS I O v v i i...
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## This note was uploaded on 11/07/2009 for the course ELEC ece212 taught by Professor None during the Spring '09 term at York University.

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Current sources slide - β / 2. For β 0 = 100 and V A...

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