V V I I0 dpn h+ p+ n e- n p e- x h+ Lp if W large → holes recombine W Reduce W Thus: A forward biased p+n diode is a good hole injector A reverse biased np diode is a good minority carrier collector Since gradient of dpn x=W is zero, hole diffusion current is also zero Magnitude of hole diffusion current cannot be derived from this layer.Magnitude of hole diffusion current at x=W is almost 0.Magnitude of hole diffusion current at x=W is same as at x=0.

V V I I0 dpn h+ p+ n e- n p e- x h+ Lp W What is the magnitude of the hole diffusion current at the edge x=W of the “green” region? V V I I0 dpn h+ p+ n e- n p e- x h+ Lp If W large → holes recombine W Thus: A forward biased p+n diode is a good hole injector A reverse biased np diode is a good minority carrier collector Excess hole concentration reduces exponentially in W to some small value. Recombination of excess electrons will occur and excess will be np0 at end of layer.No recombination of excess holes will occur.Recombination of excess holes will occur and excess will be large at end of layer.Recombination of excess holes will occur and excess will be 0 at end of layer.V V I0 I h+ p+ n e- n p e- h+ If W large, then? W Thus: A forward biased p+n diode is a good hole injector A reverse biased np diode is a good minority carrier collector Remember forward bias characteristics p+.By increasing the doping in the p region only.How can we make a hole injector from a pn diode?.V I h+ p n e- How can we increase the minority carrier concentration near the depletion region edge? V I e- p n h+ I V I0 Test: True-False pn If we only increase then |I0| will still increase. If minority carrier concentration can be increased near the depletion region edge, then I0 will increase.If minority carrier concentration can be increased what will happen to I0?.V I e- p n h+ I np and/or pn V I0 Test: Multiple choice Caused by minority carriers swept across the junction.I0 small I V I0 How can we make a BJT from a pn diode?.Because the current consist of minority carriers injected across the depletion region.Because the bias across the depletion region is small.Test: Multiple choice Why is the reverse bias current of a pn diode small? V I p n I V I0 How can we make a BJT from a pn diode? Holes and electrons determine device characteristics Three terminal device Control of two terminal currents Bipolar Junction transistor Amplification and switching through 3rd contact

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