# BJT biasing in amplifier circuit

While reading the book I got stuck in a line, which says:

For a given value of supply voltage $V_{CC}$, the higher the value we use for $V_{BB}$, the lower will be the sum of voltage across $R_C$ and the collector base junction $V_{CB}$.

Now, my question is, if I increase $V_{BB}$, $I_B$ will increase, $I_C=\beta\cdot I_B$ will also increase. Then, voltage across $R_C$ will increase, which contradicts the book. What am I getting wrong here? And what is the correct solution? • What you're getting wrong is that the book says that the sum of two variables will decrease, whereas your contradiction consists of the claim that one of the two variables increases. – Kaz Apr 24 '14 at 23:22

The resistors in this case are preventing the maximum allowed current from flowing from the collector to the emitter. If you think about $R_c$ being an open, then there would be 0 current from collector to emitter no matter how much current you get from $R_b$ to $R_e$. As you can see from this example, beta is only a maximum value if it had plenty of voltage and current.
The driving factor in the case you have is that $V_{be}$ is always ~0.7 V because that junction acts like a diode. So if you increase the voltage on $V_b$, then $V_e$ has to follow it. $V_e$ will always be ~0.7V lower than $V_b$. With $V_e$ increasing, there's less room/voltage between $V_{cc}$ and $V_e$ to drop across $R_c$ and $V_{ce}$. The current through $R_c$ then has to follow Ohm's Law. With less voltage across it, less current will fall through it. In this case, it will not source enough current to match up with the maximum amount of current that could go through $V_{ce}$ given the BJT current gain equation.