Identification: Is there a name for the following NPN BJT Current Source with three identical BJTs, which doubles the reference input current?
Analysis: If \$R_L = \frac{1}{2} R_1\$ as shown in the schematic, then the following is valid: $$I_{out} = 2 ( I_{in} - 3 I_B ) = 2 \frac{\beta}{\beta + 3} I_{in} \approx 2 I_{in} \ \text{for} \ \beta \gg 1$$
- Can one easily see that \$I_{out} = 2 ( I_{in} - 3 I_B )\$ is true?
- What are the values for \$I_{C,2}\$ and \$I_{C,3}\$ based on the above assumptions?
Are the following answers correct?
- Due to the identical BJTs and identical base currents, \$I_{C,2}\$ = \$I_{C,3}\$
- Hence, \$I_{out} = I_{C,2} + I_{C,3} = 2 I_{C,2} = 2 I_{C,3}\$
- Because of the mirroring characteristics, \$I_{C,1} = I_{C,2} = I_{C,3} = I_{in} - 3 I_B\$
- Result: \$I_{out} = 2 ( I_{in} - 3 I_B ) = 2 I_C\$
- Further transformations: $$ I_B = \frac{1}{\beta} I_C \rightarrow 2 I_C = 2 ( I_{in} - 3 I_B ) = 2 ( I_{in} - \frac{3}{\beta} I_C )$$ $$ I_C = I_{in} - \frac{3}{\beta} I_C \rightarrow I_C (1 + \frac{3}{\beta} ) = I_{in} \rightarrow I_C ( \frac{3 + \beta}{\beta} ) = I_{in} $$ $$ \hookrightarrow I_{out} = 2 I_C = \frac{2 \beta}{\beta + 3} I_{in} $$