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Diode Solutions

# Diode Solutions - assumption No D4 must be ON& I 300 ±...

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Tutorial 2 – Sept. 20, 21, and 25, 2006 The following pertain to the 8 questions on diodes. 1. 4 equations, 4 unknowns (I, V 1 , V 2 , m) ° V V V 657 . 1 2 1 = + ± V I k k I V OUT 657 . 1 ) 67 . 1 ( 5 ) 67 . 1 ( = - = - ² T nV V S e I I 1 = ³ T nV V S e I m I 2 = ° Eqn. 2 yields I = 2mA directly ° multiply ² x ³ : T S nV V V e I m I 2 1 2 2 + = , where V 1 + V 2 is known from ° ° m = ) 25 ( 25 . 1 657 . 1 2 2 m e I I S …= 37.56 (pick a real integer : 37 or 38) ° now solve for V 1 , V 2 with eqns 3 and 4 V 1 = 0.885V, V 2 = 0.772V 2. we must make assumptions and test them ° is D3 off? Try it! ° get k V k V out out 1 8 ) 7 . 0 ( 8 8 + - = - V V out 6 . 5 - = , so node B = -6.3V

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° D2 OFF would be obviously inconsistent ° D2 ON puts node A at ground – not good for D3 OFF! Therefore, D3 is ON ° node A is at V out ° Try D2 OFF: V V k V k V k V k V out out out out out 127 . 3 8 8 3 8 8 4 8 1 8 ) 7 . 0 ( - = ° - × = - + - = + - ° is D2 OFF? YES – solution is consistent 3. a)
b) By symmetry, I D1,2,3,4 = I ref / 2 and r d = (2nV T )/ I ref c) 2r d ||2r d = r d ° v OUT /v S = R L /( R S + r d + R L ) 4. a) D2, D3 OFF by inspection D1 and D4 cannot both be ON D5 is most likely ON ° V O = 1.3V ° is D1 ON? Yes: I 300 ± = (4 – (1.3-0.7))/300 = 11.3mA ° not good, D4 is off by

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Unformatted text preview: assumption! No: D4 must be ON & I 300 ± = (4 – 2)/(300 + 700) = 2mA Check D1: 1.3V & 3.4V = OFF as assumed b) I = 1.3V/400 ± = 3.25mA c) I 300 ± = 2mA 5. a) D1, D5 OFF by inspection I D2 = I 1 , I D3 = I 2 , I D4 = I 1 + I 2 & these 3 diodes are ON b) r d2 = nV T / I 1 = 100 & , r d4 = nV T /(I 1 + I 2 )= 25 & c) v out /v in = (R 1 ||r d4 )/ (R 1 ||r d4 + r d2 ) = 0.196V/V 6. a) I D3 = I If D7 is ON then D8 is off (has 0.2V reverse bias) D8 ON is impossible… turns off D7 ± no current flow in D8! D7 ON requires D4, D5 and D6 ON too ± I D4,5,6,7 = I/2 D1 and D2 are OFF b) c) v out /v in = 0.294V/V by analysis 7. slope is ½ by voltage divider limited to -3V when D2 is ON, to 3V when D1 is ON, and gain of ½ when both are OFF 8. DC analysis: I R1 = 5V/R 1 I R2 = 4.3V/R 2 r d = nV T R 2 /4.3 R in = R 1 || R 2 (1+ nV T /4.3)...
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