Q.Define $\alpha_{dc}$ and $\beta_{dc}$. Derive the relation between them.

[!TLDR] $\alpha_{dc}=I_C/I_E$ (common-base gain), $\beta_{dc}=I_C/I_B$ (common-emitter gain); from $I_E=I_B+I_C$, they satisfy $\beta_{dc}=\dfrac{\alpha_{dc}}{1-\alpha_{dc}}$ and $\alpha_{dc}=\dfrac{\beta_{dc}}{1+\beta_{dc}}$. [!ANSWER] $\alpha_{dc}=I_C/I_E$, $\beta_{dc}=I_C/I_B$; …

Definitions: the DC current amplification factor in the COMMON EMITTER configuration is $\beta_{dc}=\dfrac{I_C}{I_B}$ -- the ratio of collector current to base current -- typically lying between about 20 and 200 for general-purpose transistors. The DC current gain in the COMMON BASE configuration is $\alpha_{dc}=\dfrac{I_C}{I_E}$ -- the ratio of collector current to emitter current -- always a fraction just below 1, since $I_C$ is always slightly less than $I_E$. Derivation of the relation between them: start from current conservation at the base node, $I_E=I_B+I_C$ --- (1). Divide throughout by $I_E$: $1=\dfrac{I_B}{I_E}+\dfrac{I_C}{I_E}=\dfrac{I_B}{I_E}+\alpha_{dc}$, so $\dfrac{I_B}{I_E}=1-\alpha_{dc}$ --- (2). Now form $\beta_{dc}=\dfrac{I_C}{I_B}$: dividing numerator and denominator of this by $I_E$ gives $\beta_{dc}=\dfrac{I_C/I_E}{I_B/I_E}=\dfrac{\alpha_{dc}}{1-\alpha_{dc}}$ --- (3), using the results from (2) above. Equivalently, solving equation (3) for $\alpha_{dc}$ in terms of $\beta_{dc}$: from $\beta_{dc}(1-\alpha_{dc})=\alpha_{dc}$, i.e. $\beta_{dc}=\alpha_{dc}(1+\beta_{dc})$, so $\alpha_{dc}=\dfrac{\beta_{dc}}{1+\beta_{dc}}$ --- (4). Equations (3) and (4) are the two equivalent forms of the same relation, let …

Define $\alpha_{dc}=I_C/I_E$ and $\beta_{dc}=I_C/I_B$, then divide the current-conservation equation $I_E=I_B+I_C$ through by $I_E$ and by $I_B$ respectively …

Mixing up which ratio is which (writing $\alpha_{dc}=I_C/I_B$ and $\beta_{dc}=I_C/I_E$, the reverse of the correct definitions) -- $\alpha$ is conventionally the common-BASE gain (referenced to $I_E$) and $\beta$ the common-EMITTER gain (refere …