L16 Part2 Darlington_1

# L16 Part2 Darlington_1 - EE 311 Lecture 16 &amp; EE0 311...

This preview shows pages 1–4. Sign up to view the full content.

1 EE 311 Lecture 16 & EE0 311 Lecture 15 Part 2 6.11.2 The Darlington Ref: Basic SS Amp Eqs. Slide 6 Ref: EE 311 L12 & EEO 311 L11 Part 2 Slide 9 R10 & R20

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 V B1 = 0 VDC V E2 = V B1 - V BE1 - V BE2 = 0 - (V BE1 + V BE2 ) I E2 = [V E2 - (- V EE )] / R E = [V EE - (V BE1 + V BE2 )] / R E I B2 = I E2 / ( β 2 + 129 = I E1 Design for I E2 = 100 mA (h FE2 = 100) 6.11.2 Darlington Configuration
3 I B2 = I E2 / ( β 2 + 129 = 100 mA / (100 + 1) = 1 mA I E1 = I B2 = 1 mA = I C1 (h FE1 = 70) I C = I s e (VBE/VT) or V BE = V T ln(I C / I S ) V BE2 - V BE1 = V T [ln(I C2 /I S ) - ln(I C1 /I S )] = V T ln(I C2 /I C1 ) = V T ln(I C2 /I B2 ) = V T ln( β 2 ) V BE2 - V BE1 = V T ln( β 2 ) = 25 mV * ln (100) = 115 mV Assume V BE2 = 0.700 V Then V BE1 = 0.700 - 0.115 = 0.585 V Let V EE = 15 V R E = (V EE - V BE2 - V BE1 ) / I E2 = (15 - .7 - .585) / 100 mA =13.7V/100mA= 0.137 k = 0.14 k R INQ1 = r x1 + r π1 + ( β 1 + 129 R INQ2 R INQ2 = r x2 + r π2 + ( β 2 + 129 * (R E || R L ) R L = 100 = .1 k

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This document was uploaded on 09/28/2011.

### Page1 / 5

L16 Part2 Darlington_1 - EE 311 Lecture 16 &amp; EE0 311...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online