VEB 25 mr V 0.3= VEB + VÓ VEB 0.3 V l014 UEB/ 26x10-3 e 3x 1013= UEB t VEB/26 x/0-3 In(3x10"): In (UEB +(e e) VEB/26x03
VEB 25 mr V 0.3= VEB + VÓ VEB 0.3 V l014 UEB/ 26x10-3 e 3x 1013= UEB t VEB/26 x/0-3 In(3x10"): In (UEB +(e e) VEB/26x03
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What steps did they do to get VEB to be 3 volts? Are you supposed to subtract VEB before taking the ln of both sides? I put my process so you can see where I got stuck.
![The image contains a mathematical derivation focused on determining the value of \( V_{EB} \), which is approximately 0.3 V. The steps are shown below:
1. The initial equation is given as:
\[
\frac{0.3 \, V}{10^{-14}} = V_{EB} + 10^{-4} e^{\frac{V_{EB}}{25 \, \text{mV}}}
\]
2. Simplifying the equation, the term \( \frac{10^{-14}}{10^{-14}} \) cancels out, resulting in:
\[
3 \times 10^{13} = V_{EB} + e^{\frac{V_{EB}}{26 \times 10^{-3}}}
\]
3. Taking the natural logarithm of both sides:
\[
\ln(3 \times 10^{13}) = \ln\left(V_{EB} + e^{\frac{V_{EB}}{26 \times 10^{-3}}}\right)
\]
4. Highlighted analysis questions around the expression:
\[
\ln\left(e^{\frac{V_{EB}}{26 \times 10^{-3}}}\right) \, ?
\]
5. The boxed conclusion from the derivation:
\[
V_{EB} \approx 0.3 \, V
\]
The derivation process involves logarithmic transformations and simplifications of exponential equations, ultimately leading to determining the voltage \( V_{EB} \) with an approximation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5362cbea-9427-4b2b-99cb-57a208edd853%2F0a402508-922c-414c-917c-c81761537dd7%2Fkkylgo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains a mathematical derivation focused on determining the value of \( V_{EB} \), which is approximately 0.3 V. The steps are shown below:
1. The initial equation is given as:
\[
\frac{0.3 \, V}{10^{-14}} = V_{EB} + 10^{-4} e^{\frac{V_{EB}}{25 \, \text{mV}}}
\]
2. Simplifying the equation, the term \( \frac{10^{-14}}{10^{-14}} \) cancels out, resulting in:
\[
3 \times 10^{13} = V_{EB} + e^{\frac{V_{EB}}{26 \times 10^{-3}}}
\]
3. Taking the natural logarithm of both sides:
\[
\ln(3 \times 10^{13}) = \ln\left(V_{EB} + e^{\frac{V_{EB}}{26 \times 10^{-3}}}\right)
\]
4. Highlighted analysis questions around the expression:
\[
\ln\left(e^{\frac{V_{EB}}{26 \times 10^{-3}}}\right) \, ?
\]
5. The boxed conclusion from the derivation:
\[
V_{EB} \approx 0.3 \, V
\]
The derivation process involves logarithmic transformations and simplifications of exponential equations, ultimately leading to determining the voltage \( V_{EB} \) with an approximation.
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