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1 Chapter 11-1 Detailed Quantitative Analysis pnp transistor, steady state, low-level injection. Only drift and diffusion, no external generations One.

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Presentation on theme: "1 Chapter 11-1 Detailed Quantitative Analysis pnp transistor, steady state, low-level injection. Only drift and diffusion, no external generations One."— Presentation transcript:

1 1 Chapter 11-1 Detailed Quantitative Analysis pnp transistor, steady state, low-level injection. Only drift and diffusion, no external generations One dimensional etc. Assumptions: General approach is to solve minority carrier diffusion equations for each of the three regions : The goal is to relate transistor performance parameters ( ,  T,  dc etc. ) to doping, lifetimes, base-widths etc. and

2 2 General Quantitative Analysis Under steady state and when G L = 0, and For the base in pnp, we are interested only in holes. we are going to take a simplified approach.

3 3 Review: Operational Parameters Base transport factor :  T = I C / I EP Collector to emitter current gain:  DC =  T  Collector to base current gain:  DC =  DC / (1 –  DC ) Injection Efficiency : I EP –I EN I BR –I BR –I BE

4 4 Review of P-N Junction Under Forward Bias + V EB  xBxB xExE  p B (0)  n E (0) n E0 p B0 P (emitter)N (base) 00 Area = Q p Area = Q n

5 5 Review of P-N Junction Under Forward Bias (cont.) I n = q A D E d  n/dx E = – (q A D E /L E )  n E (0) I p = – q A D B d  p/dx B = (q A D B /L B )  p B (0) Total current I = I P + (– I N ) (“ – ” because x E and x B point in opposite directions) = (q A D B /L B )  p B (0) + (q A D E /L E )  n E (0) = (q A D B /L B ) p B0 [exp (q V EB / kT) –1] + + (q A D E /L E ) n E0 [exp (q V EB /kT) –1] ≈ (q A D B /L B ) p B0 exp (q V EB /kT) + (q A D E /L E ) n E0 exp (q V EB /kT) Note ! I p and I n can also be calculated based on the fact that Q p has to be replaced every  B seconds  I p = Q p /  B and I n = Q n /  E and I E = I P + I N

6 6 Simplified Analysis Consider the carrier distribution in a forward active pnp transistor p B0 n E0 n C0 EmitterBaseCollector n C (0) p B (0) n E (0)

7 7 Simplified Analysis (cont.) n E0, p B0 and n C0 = equilibrium concentration of minority carriers in emitter, base and collector n E (0), p B (0) and n C (0) = minority carrier concentration under forward active conditions at the edge of the respective depletion layers  n E (0),  p B (0) and  n C (0) = Excess carrier concentration at the edge of the depletion layers

8 8 Simplified Analysis (cont.)  n E (0) = n E (0) – n E0 = n E0 [exp (q V EB / kT) – 1]  p B (0) = p B (0) – p B0 = p B0 [exp (q V EB / kT) – 1] By taking the slopes of these minority carrier distribution at the depletion layer edges and multiplying it by “qAD”, we can get hole and electron currents. Note that I n = q A D n (dn/dx) and I p = – q A D p (dp / dx)

9 9 Calculation of Currents Collector current, I C I c = q A D B (dp/dx B ) (slope must be taken at end of base ) = q A D B [p B (0) – 0] / W B = q A D B p B (0) / W B I c = q A (D B /W B ) p B0 exp (qV EB / kT) ---- (A) (only hole current if we neglect the small reverse saturation current of reverse biased C-B junction)

10 10 Calculation of Currents (cont.) Emitter Current, I E I E is made up of two components, namely I EP and I EN I EP = I c + current lost in base due to recombination = I c + excess charge stored in base/  B = I c + q A W B  p B (0) / (2  B )  q A (D B /W B ) p B0 [exp (qV EB / kT) ] + q A [W B /(2  B )] p B0 [exp (qV EB / kT)] --- (B) [ Assuming exp (qV EB / kT) – 1  exp (qV EB / kT) when V EB is positive, i.e forward biased. ]

11 11 Calculation of Currents (cont.) Emitter Current (cont.) I EN corresponds to electron current injection from base to emitter since E-B junction is forward biased. I EN = qA (D E / L E ) n E0 [exp (q V EB / kT) – 1 ]  qA (D E / L E ) n E0 [exp (q V EB / kT)] ----- (C)

12 12 Calculation of Currents (cont.) Base Current, I B -supplies electrons for recombination in base -supplies electrons for injection to emitter. I B = qA p B0 [W B / (2  B )] [exp (qV EB / kT) ] + qA(D E / L E ) n E0 exp (qV EB / kT) ( recombination) + (electron injection to emitter) Now we can find transistor parameter easily.

13 13 Calculation of Currents (cont.) Base transport factor,  T  T = I C / I EP (same as eq. 11.42 in text) Emitter injection efficiency,   = I EP / [ I EP + I EN ] = 1 / [ 1 + I EN / I EP ] = 1 / [ 1+ (C) / (B) ]

14 14 Calculation of Currents (cont.)  n E0 = n i 2 / N E … doping in emitter  p B0 = n i 2 / N B … doping in base  dc =   T  DC =  DC / (1–  DC )


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