This is another tutorial on DC BJT analysis. This tutorial is important by design. How to set circuit parameters to operate in active mode, saturation mode or shutdown mode. For example, changing the base and collector resistances (RC and RB ) will change the base and collector currents . This change will directly affect the operation of the transistor. Design problems are a bit more complicated. Let's begin.
Example 1: Calculation of R B , RC and RE .
\ [I_ {CQ} = \frac {1} {2} I_ {C (sat)} \]
\ [I_ {C (sat)} = 8mA = 4mA \]
\ [V_C = 18 V \]
\[\beta = 110 \]
Apply KVL to the input circuit.
\ [- V_B + I_BR_B + V_ {BE} + I_ER_E = 0 \text {Equation 1} \]
[I_C = \beta I_B\]
\ [I_B = 36 \mu A \]
\ [I_E = I_C + I_B \]
\ [I_E = 4.036 mA \]
Calculate the RC .
\ [R_C = \frac {V_ {CC} -V_C} {I_ {CQ}} \]
\[R_C = 2.5k \omega\]
and RE determines the value of the Q point and the slope of the load line .
\ [I_ {C (sat)} = \frac {V_ {CC}} {R_C + R_E} \]
\ [R_C + R_E = \frac {28} {8m} = 3.5k \Omega\]
\ [R_C = 3.5k-2.5k = 1k \ Omega \]
Insert RE into equation 1 and calculate RB .
\ [- 28 + 36 \ mu R_B + 0.7 + 4.036 m * 1k = 0 \]
\[R_B = 646k\omega\]
example n. 2: Calculate V CC , RC , R B
From load lines,
\ [V_ {CC} = 20 V \]
\ [I_ {C (low)} = 8 mA \]
\ [I_ {BQ} = 40 \ mu LA \]
\ [I_ {C (sat)} = \frac {V_ {CC}} {R_C} \]
\[R_C = 2.5k \omega\]
Apply KVL to the input circuit for RB .
\ [- V_ {CC} + I_B R_B + V_ {BE} = 0 \]
\ [- 20 + 40 \ mu R_B + 0.7 = 0 \]
\[R_B = 482k\omega\]
example n. 3: Design of simple base NPN transistor with following parameters.
\ [I_E = 1.5 mA \]
\ [V_ {CB} = 7.5 V \]
\[V_{CC}=15 V\]
\ [V_ {CE} = - 5V \]
Apply KVL to the input circuit.
\ [- V_ {EE} + V_ {BE} + I_ER_E = 0 \]
\ [R_E = \frac {V_ {EE} -V_ {BE}} {I_E} \]
\ [R_E = \frac {5-0.7} {1.5 m} = 4.5 k \Omega\]
KVL at the output of the loop.
Note that the common base configuration has unique current gain and therefore I C = I E .
\ [- V_ {CC} + V_ {CB} + I_CR_C = 0 \]
\ [R_C = \frac {V_ {CC} -V_ {CB}} {I_C} \]
\[R_C = \frac {15-7.5} {1.5 m} = 5k \Omega\]
Example #4: Calculation of RC and RE .
\ [I_E = 2 mA \]
\[V_{CB}=9V\]
KVL in the input loop.
\ [- V_ {EE} + V_ {BE} + I_ER_E = 0 \]
\ [R_E = \frac {V_ {EE} -V_ {BE}} {I_E} \]
\ [R_E = \frac {10-0.7} {2m} = 4.7k \Omega \]
KVL at the output of the loop.
\ [- V_ {CC} + V_ {CB} + I_CR_C = 0 \]
\ [R_C = \frac {V_ {CC} -V_ {CB}} {I_C} \]
\ [R_C = \frac {20-9} {2m} = 5.5k \Omega\]
Example #5: Calculation of R B and RC .
Offset point value.
\[V_{CE}=6V\]
\ [I_C = 2 mA \]
\ [I_B = \frac {I_C} {\beta} = 20 \mu A \]
Find the possible values of the polarization point for β = 50 - 150.
Evaluate KVL and RB in the input loop.
\ [- V_ {CC} + I_BR_B + 0V_ {BE} = 0 \]
\ [R_B = \frac {V_ {CC} -V_ {BE}} {I_B} \]
\ [R_B = \frac {12-0.7} {2 \mu} = 565k \Omega \text {equation 1} \]
Estimate KVL and RC in the output circuit .
\ [- V_ {CC} + I_CR_C + V_ {CE} = 0 \ text { equation 2} \]
\ [R_C = \frac {V_ {CC} -V_ {CE}} {I_C} \]
\ [R_C = \frac {12-6} {2m} = 3k \Omega \]
For β = 50 and β = 150: calculate the minimum and maximum I C and V CE .
From equation 1:
\ [I_B = \frac {12-0.7} {565k} = 20 \mu LA \]
\[I_{C(min)} = \beta_{min}I_B\]
\ [I_ {C (min)} = 50 * 20 \ mu = 1 mA \]
\[I_{C(max)} = \beta_{max}I_B\]
\ [I_ {C (max)} = 150 * 20 \ mu = 3 mA \]
From equation 2:
\ [V_ {CE (min)} = V_ {CC} -I_ {C (max.)} R_C \]
\ [V_ {CE (min)} = 12-3 m * 3k = 3V \]
\ [V_ {CE (max)} = V_ {CC} -I_ {C (min)} R_C \]
\ [V_ {CE (max)} = 12-1 m * 3k = 9V \]
I C can vary between 1 mA and 3 mA.
V CE can vary between 3V and 9V.
Example 6: Find R B and R C .
\[V_{CE}=5V\]
\ [I_C = 1 mA \]
\[V_{CC}=15 V\]
\[\beta = 100\]
\ [I_B = 10 \ mu LA \]
Find the possible values of the polarization point for β = 30 - 150.
Evaluate KVL and RB in the input circuit.
\ [- V_ {CC} + I_BR_B + 0V_ {BE} = 0 \]
\ [R_B = \frac {V_ {BB} -V_ {BE}} {I_B} \]
\ [R_B = \frac {15-0.7} {10 \mu} = 1.4M \Omega \text {equation 1} \]
Estimate KVL and RC in the output circuit.
\ [- V_ {CC} + I_CR_C + V_ {CE} = 0 \ text { equation 2} \]
\ [R_C = \frac {V_ {CC} -V_ {CE}} {I_C} \]
\[R_C = \frac {15-5} {1m} = 10k \Omega\]
For β = 30 and β = 150: Calculate the minimum and maximum IC and VCE.
From equation 1:
\ [I_B = \frac {15-0.7} {1.4M} = 10 \mu LA \]
\[I_{C(min)} = \beta_{min}I_B\]
\ [I_ {C (min)} = 30 * 10 \ mu = 0.3 mA \]
\[I_{C(max)} = \beta_{max}I_B\]
\ [I_ {C (max)} = 150 * 10 \ mu = 1.5 mA \]
From equation 2:
\ [V_ {CE (min)} = V_ {CC} -I_ {C (max)} R_C \]
\ [V_ {CE (min)} = 15-1.5 m * 10k = 0V \]
\ [V_ {CE (max)} = V_ {CC} -I_ {C (min)} R_C \]
\ [V_ {CE (max)} = 15-0.3 m * 10k = 12V \]
I C can vary between 0.3 mA and 1.5 mA.
V after Christ