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If
$Vout$ is 10V, I can assume that$V_C$ = 1mA. -
$I_B ≈ I_C/\beta$ , so ($\beta = 100$ )$I_B = 10 \mu A$ ? So...would that equate to something like 10V through a 1M resistor to the base? That seems...not right. -
that if
$Vout$ is 10V, then$V_C$ = 1mA. Then, if$I_B ≈ I_C/\beta$ , so ($\beta = 100$ )$I_B = 10 \mu A$ ? -
$\frac{I_{C_2}}{I_{C_1}}=e^\left(\frac{\Delta V_{_\text{BE}}}{V_T}\right)$
$$=V_T\cdot\left(\ln\left(\frac{I_{{\text{C}2}}}{I{\text{SAT}}}\right)-\ln\left(\frac{I_{{\text{C}1}}}{I{\text{SAT}}}\right)\right)$$