Question3: Based on diode i-v characteristics measurements we found: Va = 0.65V, ia= 0.684mA Va = 0.70V, ia= 4.68mA Assume VT at room temperature = 0.025V. a) Find out diode “nonideality factor" n. b) Find out diode “reverse saturation current" Is. c) Find out diode current ia for diode voltage va =0.6V.

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Question3: Based on diode i-v characteristics measurements we found:
Va = 0.65V, ia = 0.684mA
Vd = 0.70V, id= 4.68mA
Assume VT at room temperature = 0.025V.
a) Find out diode “nonideality factor" n.
b) Find out diode “reverse saturation current" Is.
c) Find out diode current ia for diode voltage va = 0.6V.
Transcribed Image Text:Question3: Based on diode i-v characteristics measurements we found: Va = 0.65V, ia = 0.684mA Vd = 0.70V, id= 4.68mA Assume VT at room temperature = 0.025V. a) Find out diode “nonideality factor" n. b) Find out diode “reverse saturation current" Is. c) Find out diode current ia for diode voltage va = 0.6V.
Expert Solution
Step 1t

For a forward biased diode :- 

I=IseVnVT-Is

Since the value of Iin forward biased diode is very small it can be neglected and diode equation can be written as :- 

I=IseVnVt

In above question :-

Value of two voltage and current are given solve them like this :- 

I1= ISeV1nVTI2= ISeV2nVT4.68 ×10-3A= ISe.7n×25×10-3           ....(1).684×10-3 A= ISe.65n×25×10-3           .....(2)Divide equation (1) by equation (2)4.684.684 = e.7-.65n×25×10-3Take log on L.H.S and R.H.S ln4.684.684=.7-.65n×25×10-31.92=2nn = 1.922n = .96

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