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Lecture 11 OUTLINE pn Junction Diodes (cont’d) – Narrow-base diode – Junction breakdown Reading: Pierret 6.3.2, 6.2.2; Hu 4.5.

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Presentation on theme: "Lecture 11 OUTLINE pn Junction Diodes (cont’d) – Narrow-base diode – Junction breakdown Reading: Pierret 6.3.2, 6.2.2; Hu 4.5."— Presentation transcript:

1 Lecture 11 OUTLINE pn Junction Diodes (cont’d) – Narrow-base diode – Junction breakdown Reading: Pierret 6.3.2, 6.2.2; Hu 4.5

2 Introduction Lecture 11, Slide 2 The ideal diode equation was derived assuming that the lengths of the quasi-neutral p-type & n-type regions (W P ’, W N ’) are much greater than the minority-carrier diffusion lengths (L n, L p ) in these regions.  Excess carrier concentrations decay exponentially to 0.  Minority carrier diffusion currents decay exponentially to 0. In modern IC devices, however, it is common for one side of a pn junction to be shorter than the minority-carrier diffusion length, so that a significant fraction of the “injected” minority carriers reach the end of the quasi-neutral region, at the metal contact. Recall that  p =  n = 0 at an ohmic contact  In this lesson we re-derive the diode I-V equation with the boundary condition that  p = 0 at a distance x c ’ (rather than  ) from the edge of the depletion region. EE130/230A Fall 2013

3 Excess Carrier Distribution (n side) From the minority carrier diffusion equation: For convenience, let’s use the coordinate system: So the solution is of the form: We have the following boundary conditions: Lecture 11, Slide 3 xc'xc' 0 x’x’’ 0 EE130/230A Fall 2013

4 Applying the boundary conditions, we have: Therefore Since this can be rewritten as We need to take the derivative of  p n ’ to obtain the hole diffusion current within the quasi-neutral n region: Lecture 11, Slide 4EE130/230A Fall 2013

5 Thus, for a one-sided p + n junction (in which the current is dominated by injection of holes into the n-side) with a short n-side: Lecture 11, Slide 5 Evaluate J p at x=x n (x’=0) to find the injected hole current: EE130/230A Fall 2013

6 Therefore if x c ’ << L P : For a one-sided p + n junction, then: and Lecture 11, Slide 6EE130/230A Fall 2013

7  p n is a linear function:  J p is constant (No holes are lost due to recombination as they diffuse to the metal contact.) Lecture 11, Slide 7 Excess Hole Concentration Profile x'x' 0 x'cx'c 0 slope is constant pn(x)pn(x) If x c ’ << L P : EE130/230A Fall 2013

8 General Narrow-Base Diode I-V Define W P ‘ and W N ’ to be the widths of the quasi-neutral regions. If both sides of a pn junction are narrow (i.e. much shorter than the minority carrier diffusion lengths in the respective regions): Lecture 11, Slide 8 x JNJN xnxn -xp-xp J JPJP e.g. if hole injection into the n side is greater than electron injection into the p side: EE130/230A Fall 2013

9 Summary: Narrow-Base Diode If the length of the quasi-neutral region is much shorter than the minority-carrier diffusion length, then there will be negligible recombination within the quasi-neutral region and hence all of the injected minority carriers will “survive” to reach the metal contact. – The excess carrier concentration is a linear function of distance. For example, within a narrow n-type quasi-neutral region:  The minority-carrier diffusion current is constant within the narrow quasi-neutral region. Shorter quasi-neutral region  steeper concentration gradient  higher diffusion current x pn(x)pn(x) xnxn 0 location of metal contact (  p n =0) WN’WN’ EE130/230A Fall 2013Lecture 11, Slide 9

10 pn Junction Breakdown A Zener diode is designed to operate in the breakdown mode: Breakdown voltage, V BR VAVA Lecture 11, Slide 10EE130/230A Fall 2013 C. C. Hu, Modern Semiconductor Devices for Integrated Circuits, Figure 4-10

11 Review: Peak E -Field in a pn Junction x xnxn -xp-xp E(x)E(x) E (0) where N is the dopant concentration on the lightly doped side For a one-sided junction, Lecture 11, Slide 11EE130/230A Fall 2013

12 Breakdown Voltage, V BR If the reverse bias voltage (-V A ) is so large that the peak electric field exceeds a critical value E CR, then the junction will “break down” (i.e. large reverse current will flow) Thus, the reverse bias at which breakdown occurs is Lecture 11, Slide 12EE130/230A Fall 2013

13 Avalanche Breakdown Mechanism if V BR >> V bi High E -field: Low E -field: E CR increases slightly with N: For 10 14 cm -3 < N < 10 18 cm -3, 10 5 V/cm < E CR < 10 6 V/cm Lecture 11, Slide 13EE130/230A Fall 2013 R. F. Pierret, Semiconductor Device Fundamentals, Figure 6.12

14 Tunneling (Zener) Breakdown Mechanism Dominant breakdown mechanism when both sides of a junction are very heavily doped. V A = 0V A < 0 Typically, V BR < 5 V for Zener breakdown EcEc EvEv Lecture 11, Slide 14EE130/230A Fall 2013 C. C. Hu, Modern Semiconductor Devices for Integrated Circuits, Figure 4-12

15 Empirical Observations of V BR V BR decreases with increasing N V BR decreases with decreasing E G Lecture 11, Slide 15EE130/230A Fall 2013 R. F. Pierret, Semiconductor Device Fundamentals, Figure 6.11

16 V BR Temperature Dependence For the avalanche mechanism: – V BR increases with increasing T, because the mean free path decreases For the tunneling mechanism: – V BR decreases with increasing T, because the flux of valence-band electrons available for tunneling increases Lecture 11, Slide 16EE130/230A Fall 2013

17 Summary: Junction Breakdown If the peak electric field in the depletion region exceeds a critical value E CR, then large reverse current will flow. This occurs at a negative bias voltage called the breakdown voltage, V BR : where N is the dopant concentration on the more lightly doped side The dominant breakdown mechanism is avalanche, if N < ~10 18 /cm 3 tunneling, if N > ~10 18 /cm 3 Lecture 11, Slide 17EE130/230A Fall 2013


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