7 Adding images edited Apr 6 '14 at 23:23 somers 6322 silver badges99 bronze badges I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. 6 Adding images edited Apr 6 '14 at 23:10 somers 6322 silver badges99 bronze badges I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. 5 added 83 characters in body edited Apr 6 '14 at 22:53 somers 6322 silver badges99 bronze badges I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. I'm having difficulty working out required values of resistors $$\R_C\$$ and $$\R_E\$$ in the circuit below. Can't seem to find any formulas on the web. I know how to find $$\R_1\$$ and $$\R_2\$$: $$R_1 = (\frac{R_2}{V_B} \times V_{CC})-R_2$$ $$R_2 = \frac{β_{DC} \times R_E}{10}$$ I'd appreciate any help you can give with this as I can't move forward with other calculations until I find values for $$\R_C\$$ and $$\R_E\$$. Technical specs of circuit: BJT BC108 Voltage divider bias used $$\I_{C(sat)} = 20mA\$$ $$\V_{CC} = 10V\$$ 10μF coupling capacitors for the input and output connections 100μF capacitor for the bypass capacitor $$\β_{DC} ≈ 200\$$ Voltage gain without bypass capacitor = 1.5 I plotted a DC Load Line and the Q-point is at $$\I_C = 10mA\$$ and $$\V_{CE} = 5V\$$. 4 Additional spec information edited Apr 5 '14 at 22:10 somers 6322 silver badges99 bronze badges Tweeted twitter.com/#!/StackElectronix/status/452129713545547776 occurred Apr 4 '14 at 17:04 3 MathJaXed the equations and improved overall formatting. edit approved Apr 4 '14 at 16:58 Ricardo 4,4471414 gold badges3838 silver badges7777 bronze badges 2 added 70 characters in body edited Apr 4 '14 at 16:51 somers 6322 silver badges99 bronze badges 1 asked Apr 4 '14 at 16:26 somers 6322 silver badges99 bronze badges