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I was trying to solve a problem from SEDRA-SMITH book. I got stuck so I refer to the solution. When Studying how the author solved the problem I got even more confused.

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Let's refer first to the given vbe and ic in the question. As per the explanation in the earlier chapter of the book these givens are instantaneous values.

sedra

Here's the first part of the solution. This is only the part I need to understand.

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the part of the solution that I don't quite understand is the equation for vbe1. I suppose this is a DC value. An uppercase letter is used for the symbol and the subscript. But if that is so, the equation does not agree with the equation for the instantaneous vbe that is

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Following the above setup the equation for vbe1 should be something like this

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One other thing that confuses me is the use of 4.8 mA in the second term of the VBE1 equation. Isn't that supposed to be Is? How is Ie equal to Is? From what I understand Is is a very small quantity. Why is Ie used in this case?

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Null
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Paumdrdo
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  • Where does Q3 connect to? What has D12.1 got to do with this - is it important? If not remove that paragraph because I along with a few others don't need to waste time reading it if it is irrelevant. – Andy aka May 27 '20 at 12:22
  • Q3 is connected to -Vcc. – Paumdrdo May 27 '20 at 12:26

1 Answers1

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Simply, they use a Large-Signal analysis.

The small-signal analysis should only be used to find voltage gain only.

As for the \$V_{BE}\$ equation. It's strange that you never see this equation before in the book? Especially that is it in chapter 12. But ok.

First, think you need to do is to rearrange the Shockley equation and solve it for \$V_{BE}\$.

$$I_C = I_S \times e^{\frac{V_{BE}}{V_T}}$$

$$\frac{I_C}{I_S} = e^{\frac{V_{BE}}{V_T}}$$ $$\ln\left(\frac{I_C}{I_S}\right) =\frac{V_{BE}}{V_T}$$ $$V_T\ln\left(\frac{I_C}{I_S}\right) =V_{BE}$$ $$V_{BE} = V_T\ln\left(\frac{I_C}{I_S}\right)$$

Now let's define \$ΔV_{BE}\$ as the difference between the actual \$V_{BE1}\$ voltage and some reference \$V_{BE2}\$ voltage for a different collector current.

$$ΔV_{BE} = V_{BE1} - V_{BE2} = V_T\ln\left(\frac{I_{C1}}{I_S}\right) - V_T\ln\left(\frac{I_{C2}}{I_S}\right)=V_T \left(\ln\frac{I_{C1}}{I_S}- \ln\frac{I_{C2}}{I_S} \right) $$ $$ΔV_{BE} = V_T \ln\left(\frac{I_{C1}}{I_S} \times \frac{I_S}{I_{C2}}\right)=V_T \ln\left(\frac{I_{C1}}{I_{C2}}\right)$$ $$ΔV_{BE} = V_T \ln\left(\frac{I_{C1}}{I_{C2}}\right)$$

Therefore, if we know the \$V_{BE1}\$ at \$I_{C1} = 1mA\$ we can calculate the \$V_{BE2}\$ value at different current \$I_{C2}=4.8mA\$.

\$V_{BE2} = V_{BE1} + ΔV_{BE} \$

The end of the story.

G36
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