IB = "\\frac{V_{CC}-0.7}{R_B}"
Here, VCC is also base biasing voltage.
"\\frac{I_C}{\\beta}" = "\\frac{12-0.7}{22000}"
Given "\\beta" = 90
IC = "\\frac{101.7}{2200}"
IC = 46.23 mA ...(1)
12 = ICRC + VCE
12 = (0.04623)(100) + VCE
VCE = 7.38 V
IB = 0.514 mA
Comments
Leave a comment