(New page: <math> V_B = V_0 + I_{C(Q)} R_E </math> <math>V_{CC} - R_1( I_B + I_{R_2} ) = V_B </math> <math>I_B = \frac{\beta}{I_{C(Q)}}</math> & <math> I_{R_2} = \frac{V_B}{R_2} </math> Substitute...) |
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| − | <math> V_B = V_0 + I_{C(Q)} R_E </math> | + | Start with initial estimation of <math>V_B</math> |
| + | :<math> V_B = V_0 + I_{C(Q)} \times R_E </math> | ||
| − | <math>V_{CC} | + | Then examine the circuit to create this equation for <math>V_B</math>. This equation says that the voltage at the base will be <math>V_{CC}</math> minus the voltage dropped across <math>R_1</math> |
| − | + | '''(1)''' <math>V_B = V_{CC} - R_1( I_B + I_{R_2} ) </math> | |
| − | + | The base current for the desired Q is derived from the relationship between <math>\beta</math> and <math>I_C</math> | |
| − | <math> | + | '''(2)''' <math>I_B = \frac{I_{C(Q)}}{\beta}</math> |
| − | + | From a simple application of Ohm's law we find <math>I_{R_2}</math> | |
| − | <math> | + | '''(3)''' <math> I_{R_2} = \frac{V_B}{R_2} </math> |
| − | + | Substituting equations '''(2)''' and '''(3)''' into '''(1)''' yields: | |
| + | :<math>V_{CC} - R_1 \Bigg(\frac{I_{C(Q)}}{ \beta} + \frac{V_B}{R_2}\Bigg) = V_B </math> | ||
| + | Solve for <math>R_2</math>: | ||
| − | <math>R_2 = \frac{R_1 V_B}{V_{CC} - V_B - \frac{R_1 \beta}{I_{C(Q)}}} </math> | + | :<math>V_{CC} - V_B = \frac{R_1 I_{C(Q)}}{\beta} + \frac{R_1 V_B}{R_2} </math> |
| + | |||
| + | :<math>V_{CC} - V_B - \frac{R_1 I_{C(Q)}}{\beta} = \frac{R_1 V_B}{R_2} </math> | ||
| + | |||
| + | |||
| + | |||
| + | '''(4)''' <math>R_2 = \frac{R_1 V_B}{V_{CC} - V_B - \frac{R_1 I_{C(Q)}}{\beta}} </math> | ||
| + | |||
| + | Now we have the information we need to satisfy the limitation on <math> R_1 || R_2 </math> must be larger than some value X | ||
| + | |||
| + | :<math> X = \Bigg(\frac{1}{R_1} + \frac{1}{R_2}\Bigg)^{-1}</math> | ||
| + | |||
| + | Substitute '''(4)''' into the equation: | ||
| + | |||
| + | :<math> X = \Bigg(\frac{1}{R_1} + \frac{V_{CC} - V_B - \frac{R_1 I_{C(Q)}} {\beta} }{R_1 V_B} \Bigg)^{-1}</math> | ||
| + | |||
| + | Then do some intense algebra and solve for <math>R_1</math> | ||
| + | |||
| + | :<math> R_1 = \frac{\frac{V_{CC}}{V_B}}{\frac{1}{X}+\frac{I_{C(Q)}}{\beta V_B}}</math> | ||
Latest revision as of 23:18, 31 October 2008
Start with initial estimation of $ V_B $
- $ V_B = V_0 + I_{C(Q)} \times R_E $
Then examine the circuit to create this equation for $ V_B $. This equation says that the voltage at the base will be $ V_{CC} $ minus the voltage dropped across $ R_1 $
(1) $ V_B = V_{CC} - R_1( I_B + I_{R_2} ) $
The base current for the desired Q is derived from the relationship between $ \beta $ and $ I_C $
(2) $ I_B = \frac{I_{C(Q)}}{\beta} $
From a simple application of Ohm's law we find $ I_{R_2} $
(3) $ I_{R_2} = \frac{V_B}{R_2} $
Substituting equations (2) and (3) into (1) yields:
- $ V_{CC} - R_1 \Bigg(\frac{I_{C(Q)}}{ \beta} + \frac{V_B}{R_2}\Bigg) = V_B $
Solve for $ R_2 $:
- $ V_{CC} - V_B = \frac{R_1 I_{C(Q)}}{\beta} + \frac{R_1 V_B}{R_2} $
- $ V_{CC} - V_B - \frac{R_1 I_{C(Q)}}{\beta} = \frac{R_1 V_B}{R_2} $
(4) $ R_2 = \frac{R_1 V_B}{V_{CC} - V_B - \frac{R_1 I_{C(Q)}}{\beta}} $
Now we have the information we need to satisfy the limitation on $ R_1 || R_2 $ must be larger than some value X
- $ X = \Bigg(\frac{1}{R_1} + \frac{1}{R_2}\Bigg)^{-1} $
Substitute (4) into the equation:
- $ X = \Bigg(\frac{1}{R_1} + \frac{V_{CC} - V_B - \frac{R_1 I_{C(Q)}} {\beta} }{R_1 V_B} \Bigg)^{-1} $
Then do some intense algebra and solve for $ R_1 $
- $ R_1 = \frac{\frac{V_{CC}}{V_B}}{\frac{1}{X}+\frac{I_{C(Q)}}{\beta V_B}} $
