Bipolar junction transistors have no input resistance.

An input resistance can be defined for ports of n-port networks, such as the common-emitter configuration you've shown, with the two terminals connected to \$V_{in}\$ being one such port. If you connect an ideal voltage source \$V_{in}\$ to such a port, resulting in a current flow \$I_{in}\$, the input resistance is defined to be

$$ R_{in} := \frac{V_{in}}{I_{in}}.$$

Applied to your common-emitter circuit, we get \$ R_{in} = V_{in} / I_B \$.

For DC inputs, \$R_{in}\$ varies strongly with \$V_{in}\$ (in this case) - the input resistance will be large for voltages below the BE-diode's forward voltage and very low for voltages above the forward voltage. You call this the large signal input resistance.

When used as an amplifier, one would bias the BJT suitably with a DC voltage to put it into the linear region and superimpose a small AC signal to be amplified. In this case, the small signal input resistance \$r_{in}\$ is of interest. You can calculate \$r_{in}\$ from a small-signal model of a BJT and the various transistor parameters that can depend on the biasing conditions. Any textbook on BJTs or amplifier circuits should cover that.

As for your circuit diagram: I think it is misleading. \$R_{in}\$ should probably not be in there. It is voltage-dependent and thus not an actual resistor, and if you considered it a voltage-dependent resistor, it would also incorporate the BE diode. Just leave it out.

Regarding the (common-emitter) current gain \$\beta\$: it is roughly constant only in the linear region of the BJT. (If it were not, there would be no linear region, so this is just a tautology.)

Bipolar junction transistors have no input resistance.

An input resistance can be defined for ports of n-port networks, such as the common-emitter configuration you've shown, with the two terminals connected to \$V_{in}\$ being one such port. If you connect an ideal voltage source \$V_{in}\$ to such a port, resulting in a current flow \$I_{in}\$, the input resistance is defined to be

$$ R_{in} := \frac{V_{in}}{I_{in}}.$$

Applied to your common-emitter circuit, we get \$ R_{in} = V_{in} / I_B \$.

For DC inputs, \$R_{in}\$ varies strongly with \$V_{in}\$ (in this case) - the input resistance will be large for voltages below the BE-diode's forward voltage and very low for voltages above the forward voltage. You call this the large signal input resistance.

When used as an amplifier, one would bias the BJT suitably with a DC voltage to put it into the linear region and superimpose a small AC signal to be amplified. In this case, the small signal input resistance \$r_{in}\$ is of interest. You can calculate \$r_{in}\$ from a small-signal model of a BJT and the various transistor parameters that can depend on the biasing conditions. Any textbook on BJTs or amplifier circuits should cover that.

As for your circuit diagram: I think it is misleading. \$R_{in}\$ should probably not be in there. It is voltage-dependent and thus not an actual resistor, and if you considered it a voltage-dependent resistor, it would also incorporate the BE diode.

Regarding the (common-emitter) current gain \$\beta\$: it is roughly constant only in the linear region of the BJT. (If it were not, there would be no linear region, so this is just a tautology.)

Bipolar junction transistors have no input resistance.

An input resistance can be defined for ports of n-port networks, such as the common-emitter configuration you've shown, with the two terminals connected to \$V_{in}\$ being one such port. If you connect an ideal voltage source \$V_{in}\$ to such a port, resulting in a current flow \$I_{in}\$, the input resistance is defined to be

$$ R_{in} := \frac{V_{in}}{I_{in}}.$$

Applied to your common-emitter circuit, we get \$ R_{in} = V_{in} / I_B \$.

For DC inputs, \$R_{in}\$ varies strongly with \$V_{in}\$ (in this case) - the input resistance will be large for voltages below the BE-diode's forward voltage and very low for voltages above the forward voltage. You call this the large signal input resistance.

When used as an amplifier, one would bias the BJT suitably with a DC voltage to put it into the linear region and superimpose a small AC signal to be amplified. In this case, the small signal input resistance \$r_{in}\$ is of interest. You can calculate \$r_{in}\$ from a small-signal model of a BJT and the various transistor parameters that can depend on the biasing conditions. Any textbook on BJTs or amplifier circuits should cover that.

Regarding the (common-emitter) current gain \$\beta\$: it is roughly constant only in the linear region of the BJT. (If it were not, there would be no linear region, so this is just a tautology.)

Bipolar junction transistors have no input resistance.

An input resistance can be defined for ports of n-port networks, such as the common-emitter configuration you've shown, with the two terminals connected to \$V_{in}\$ being one such port. If you connect an ideal voltage source \$V_{in}\$ to such a port, resulting in a current flow \$I_{in}\$, the input resistance is defined to be

$$ R_{in} := \frac{V_{in}}{I_{in}}.$$

Applied to your common-emitter circuit, we get \$ R_{in} = V_{in} / I_B \$.

For DC inputs, \$R_{in}\$ varies strongly with \$V_{in}\$ (in this case) - the input resistance will be large for voltages below the BE-diode's forward voltage and very low for voltages above the forward voltage. You call this the large signal input resistance.

When used as an amplifier, one would bias the BJT suitably with a DC voltage to put it into the linear region and superimpose a small AC signal to be amplified. In this case, the small signal input resistance \$r_{in}\$ is of interest. You can calculate \$r_{in}\$ from a small-signal model of a BJT and the various transistor parameters that can depend on the biasing conditions. Any textbook on BJTs or amplifier circuits should cover that.

As for your circuit diagram: I think it is misleading. \$R_{in}\$ should probably not be in there. It is voltage-dependent and thus not an actual resistor, and if you considered it a voltage-dependent resistor, it would also incorporate the BE diode. Just leave it out.

Regarding the (common-emitter) current gain \$\beta\$: it is roughly constant only in the linear region of the BJT. (If it were not, there would be no linear region, so this is just a tautology.)