Consider an NPN Silicon BJT which is to be designated with an Emitter Efficiency of ye = 0.995. To maintain a reasonable base resistance, the base is doped with Boron, with NA = 1 x 1016 cm-3. Given that L, = W, = 0.7µm, De = 12cm2s-1, D, = 30cm?s-1 and npo = 1.88 x 10 cm-3. Calculate the n-type doping concentration needed in the emitter.

Consider an NPN Silicon BJT which is to be designated with an Emitter Efficiency of ye = 0.995. To maintain a reasonable base resistance, the base is doped with Boron, with NA = 1 x 1016 cm-3. Given that L, = W, = 0.7µm, De = 12cm2s-1, D, = 30cm?s-1 and npo = 1.88 x 10 cm-3. Calculate the n-type doping concentration needed in the emitter.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Draw the n-p-n silicon bipolar junction transistor at zero bias and also in forward active mode with an aid of the energy band diagram. Please ensure to draw with careful detail, reflecting the doping density in each region.
a. Calculate the transistor’s Emitter current if it is driven by a small Base current of 7mA with a Collector current of 10mA b. Calculate the transistor’s Collector current if it is driven by a small Base current of 1.96mA with a Emitter current of 100mA c. Calculate the transistor’s Base current if it is driven by an Emitter current of 50mA with a Collector current of 49mA
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7. Calculate the a and ß when the emitter current of BJT is equal to 1 mA and the collector current is 0.995 mA.
A silicon (Si) npn bipolar transistor is doped with impurity concentrations of 10° cm, 10" cm, and 106 cm in the emitter, base and collector, respectively. The lengths of the emitter and collector length are 1 um and the base width is 500 nm. The device is operating at room temperature. (3) Determine the base current density in the transistor if the base emitter voltage VBs is forward biased with 0.5 V. Assume the base and emitter lengths are shorter than the diffusion length. (4) Explain why current gain is higher in npn transistors than in pnp transistors. (5) Sketch forward active lc vs. VE characteristics of the npn transistor as Vae increases with equal increments of voltage (e.g. 400 mv, 450 mv, 500 mv and 550 mV). You do not need to calculate values for collector current.
The BJT is said to be in O a. active region, if both the junctions are forward biased b. saturation region, if both the junctions are reverse biased. OC. saturation region, if collector junction is reverse bigsed. d. cut-off region, if both the junctions are reverse biased.
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