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ECE 663 P-N Junctions
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ECE 663 So far we learned the basics of semiconductor physics, culminating in the Minority Carrier Diffusion Equation We now encounter our simplest electronic device, a diode Understanding the principle requires the ability to draw band-diagrams Making this quantitative requires ability to solve MCDE (only exponentials!) Here we only do the equilibrium analysis
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ECE 663 P-N junction diode V I
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ECE 663 P-N junction diode V I I = I 0 (e qV/ kT -1) p n v I p 0 = q(n i 2 /N D ) (L p / p )
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ECE 663 P-N Junctions - Equilibrium What happens when these bandstructures collide? Fermi energy must be constant at equilibrium, so bands must bend near interface Far from the interface, bandstructures must revert
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ECE 663 Time < 0: Pieces separated
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ECE 663
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Gradients drive diffusion left
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ECE 663 Gradients drive diffusion
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ECE 663 But charges can’t venture too far from the interface because their Coulomb forces pull them back! + + - - + + + + - - - - - - - - - + + + + +
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ECE 663 Separation of a sea of charge, leaving behind a charge depleted region http://scott.club365.net/uploaded_images/Moses-Parts-the-Red-Sea-2-782619.jpg
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ECE 663 Resulting in a barrier across a depletion region
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ECE 663 E E Depletion Region
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ECE 663 How much is the Built-in Voltage? P side N side
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ECE 663 N a acceptor level on the p side N d donor level on the n side How much is the Built-in Voltage?
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ECE 663 Special case: One-sided Junctions One side very heavily doped so that Fermi level is at band edge. e.g. p + -n junction (heavy B implant into lightly doped substrate)
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ECE 663 How wide is the depletion region?
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ECE 663 Depletion Approximation-step junction x
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ECE 663 Depletion approximation-step junction Exponentials replaced with step-functions
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ECE 663 Doping Charge Density Electric Field Electrostatic Potential N A x p = N D x n = W D /(N A -1 + N D -1 ) K s 0 E m = -qN A x p = -qN D x n = -qW D /(N A -1 + N D -1 ) V bi = ½| E m |W D
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ECE 663 Depletion Width
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ECE 663 Maximum Field E m = 2qV bi /k s 0 (N A -1 +N D -1 )
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ECE 663 How far does W d extend into each junction? Depletion width on the n-side depends on the doping on the p-side Depletion width on the p-side depends on the doping on the n-side e.g. if N A >>N D then x n >>x p One-sided junction
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ECE 663
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P-N Junction with applied voltage
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ECE 663 Reverse Bias +Voltage to the n side and –Voltage to the p side: Depletion region will be larger
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ECE 663 Reverse Bias Band Diagram
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ECE 663 Reverse Bias depletion Applied voltage disturbs equilibrium E F no longer constant Reverse bias adds to the effect of built-in voltage
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ECE 663 Forward Bias + - Negative voltage to n side positive to p side More electrons supplied to n, more holes to p Depletion region gets smaller
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ECE 663 Forward Bias Depletion
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ECE 663 General Expression Convention = V appl = + for forward bias V appl = - for reverse bias
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ECE 663 Positive voltage pulls bands down- bands are plots of electron energy Fermi level is not constant Current Flow FnFn FpFp n = n i e (F n -E i )/kT p = n i e (E i -F p )/kT
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In summary A p-n junction at equilibrium sees a depletion width and a built-in potential barrier. Their values depend on the individual doping concentrations Forward biasing the junction shrinks the depletion width and the barrier, allowing thermionic emission and higher current. The current is driven by the splitting of the quasi-Fermi levels Reverse biasing the junction extends the depletion width and the barrier, cutting off current and creating a strong I-V asymmetry In the next lecture, we’ll make this analysis quantitative by solving the MCDE with suitable boundary conditions
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