L8Part2 6.3.3 BJT Current Source_1

L8Part2 6.3.3 BJT Current Source_1 - 1 1 1 1 2 1 1 -+ + + =...

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1 Assume V A = infinity Assume Q1 & Q2 have identical areas, Then I S values are identical. V BE1 = V BE2 I C1 = I C2 I REF = I C1 + I B1 + I B2 = I C1 [1 + 2 / β ] I B1 = I C1 / β I B2 = I C2 / β I O = I C2 6.3.3 BJT Circuits A Simple Current Source β / 2 1 1 / / 2 + = = REF O REF C I I I I
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2 If Emitter-Base areas differ, then saturation currents differ. I C1 = I S1 e VBE1 / VT I C2 = I S2 e VBE2 / VT Since V BE2 = V BE1 I C2 / I C1 = I S2 / I S1 = m Early Voltage Finite V A β m m I I I mI I REF O O C C + + = = = 1 1 1 2  + + = + + = + + = m I I mI I I I I I I C REF C C C B B C REF 1 1
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Unformatted text preview: 1 1 1 1 2 1 1 -+ + + = A BE O REF O V V V m m I I 1 * 1 1 3 Current Steering I REF = (V CC + V EE- V EB1- V BE2 ) / R If All Q Have Same Area I 1 = I REF I 2 = I REF I 3 = 2 I REF I 4 = 3 I REF Figure 6.11 Generation of a number of constant currents of various magnitudes. Sedra/Smith...
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L8Part2 6.3.3 BJT Current Source_1 - 1 1 1 1 2 1 1 -+ + + =...

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