In the figures attached below, figure (b) is the Thevenin equivalent of (a).
It is said that for a given value of the supply voltage \(V_{CC}\), the higher the value we use for \(V_{BB}\), the lower will be the sum of voltages across \(R_C\) and the collector-base junction (\(V_{CB}\)). Can you please clarify why the sum of the two voltages will be lower?
My thought is as follows:
For a fixed \(V_{CC}\), I guess we can use a higher \(V_{BB}\) by fixing the value of \(R_2\) while reducing the value of \(R_1\), then \(R_B\) will be smaller,
\(R_B = \left( \frac{R_1R_2}{R_1+R_2} \right)\)
For the loop containing \(R_C\) abd \(R_B\),
\(V_{CB} + I_CR_C - I_BR_B = V_{CC} - V_{BB}\)
Why can we tell that \(V_{CB}+I_CR_C\) will be smaller when \((V_{CC}-V_{BB})\) is smaller?
It is said that for a given value of the supply voltage \(V_{CC}\), the higher the value we use for \(V_{BB}\), the lower will be the sum of voltages across \(R_C\) and the collector-base junction (\(V_{CB}\)). Can you please clarify why the sum of the two voltages will be lower?
My thought is as follows:
For a fixed \(V_{CC}\), I guess we can use a higher \(V_{BB}\) by fixing the value of \(R_2\) while reducing the value of \(R_1\), then \(R_B\) will be smaller,
\(R_B = \left( \frac{R_1R_2}{R_1+R_2} \right)\)
For the loop containing \(R_C\) abd \(R_B\),
\(V_{CB} + I_CR_C - I_BR_B = V_{CC} - V_{BB}\)
Why can we tell that \(V_{CB}+I_CR_C\) will be smaller when \((V_{CC}-V_{BB})\) is smaller?
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