An npn silicon bipolar transistor at T=300 K has uniform...

An npn silicon bipolar transistor at $T=300 \mathrm{~K}$ has uniform dopings of $N_{E}=$ $10^{19} \mathrm{~cm}^{-3}, N_{B}=10^{17} \mathrm{~cm}^{-3}$, and $N_{C}=7 \times 10^{15} \mathrm{~cm}^{-3}$. The transistor is operating in the inverse-active mode with $V_{B E}=-2 \mathrm{~V}$ and $V_{\mathrm{AC}}=0.565 \mathrm{~V} .(a)$ Sketch the minority carrier distribution through the device. (b) Determine the minority carrier concentrations at $x=x_{\bar{B}}$ and $x^{\prime \prime}=0 .(c)$ If the metallurgical base width is $1.2 \mu \mathrm{m}$, determine the neutral base width.

Transcript

0:00 Hello students, welcome back.

00:01 So in this question, we have a graph something like this.

00:10 This graph is starting from here and goes up to a constant value.

00:17 So this is a graph of pnx and another graph starts from downwards and goes up to infinity.

00:26 This is a graph of npx.

00:29 Okay, so this is a minus xp value.

00:31 This is x equals to zero value and this is x -n value.

00:35 So here we have a, n p at constant and here we have p n at constant okay so this was the a part now in the b part now in the b part we know that n infinity n b not is equal to n i squared divided by capital n b if you put here the values it is equal to 1 .5 into 10 to whole square divided by 10 to the power 17 so the result would be 2 .25 into 10 to the power 3 centimeter per cube and p infinity will be equal to n i square divided by n c so this is equal to 1 .5 into 10 to the power 10 whole square divided by 7 into 10 to the power 15 so this is equal to 3 .21 into 10 to the power 4 per centimeter cube.

01:42 Now at x is equal to xb we have n b into xb is equal to n b.

01:52 B.

01:53 Exponent to the power minus v bc upon v i.

01:58 So if we put here the values we get 2 .25 into 10 to the power 3 exponent to the power 0 .565 divided by 0 .025 9.

02:15 So this is equal to 6 .7 into 10 to the power 12 per centimeter cube.

02:22 Now at x to the power n equals to 0 we have p c0 is equals to p infinity exponent of 0 .565 divided by 0 .0259.

02:40 That is 3 .21 into 10 to the power 4 and exponent of 0 .565 divided by 0 .0259...

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