Given the circuit of Figure 6.30, assume the following component values: R1 = 20 kΩ, R2 = 100 kΩ, RC = 50 Ω, and RE = 100 Ω, the supply is VCC = 10 V. (1) Calculate IC, IB, VCE,VRE, and β. (2) Also calculate for the found value of β the BJT trans-conductance gm and rπ of the hybrid-π small signal

Question

Given the circuit of Figure 6.30, assume the following component values: [latex]R_1 = 20 kΩ, R_2 = 100 kΩ, R_C = 50 Ω, and R_E = 100 Ω[/latex], the supply is [latex]V_{CC} = 10 V[/latex]. (1) Calculate [latex]I_C, I_B, V_{CE},V_{RE}[/latex], and β. (2) Also calculate for the found value of β the BJT trans-conductance [latex]g_m and r_π[/latex] of the hybrid-π small signal model. At all times ignore the BJT Early effect; that is, [latex]r_0 → ∞[/latex]. Assume the reverse saturation current [latex]I_S = 7.11 × 10^{−15}[/latex] A at room temperature of 300 K and [latex]V_{BE} = 0.763 V.[/latex]

Accepted Answer

From Equation (6.16), repeated here for the reader’s convenience, we find for [latex]I_S = 7.11 × 10^{−15} A, v_{BE} = 0.763 V, and v_T = 0.026 V[/latex] that

[latex]I_C=I_Se^{\frac{v_{BE}}{v_T} }=39.5 mA[/latex] (6.78)

Using the above given values in the circuit of Figure 6.30, and applying Thèvenin to the resistor divider on the left of the BJT base node, we obtain

[latex]R_{Thev}=\frac{R_1R_2}{R_1+R_2} ;V_{Thev}=\frac{R_2}{R_1+R_2} V_{CC}[/latex]. (6.79)

Equations (6.79) were obtained just as Equations (6.32) and (6.33) were obtained, using the resistor divider method. For the Thevenized portion of the circuit we can write KVL equations, to find the base current:

[latex]R_{Thev}=\frac{R_1R_2}{R_1+R_2}[/latex] (6.32)

[latex]V_{Thev}=\frac{R_2}{R_1+R_2} V_{CC}[/latex] (6.33)

[latex]I_B=(V_{Thev}-V_{BE}-V_E)/R_{Thev}[/latex] (6.80)

where [latex]V_{Thev} = 8.33 V and R_{Thev} = 16,667 Ω[/latex] are calculated from Equations (6.79) using the given values of [latex]R_1, R_2, and V_{CC}[/latex]. Voltage [latex]V_E[/latex] is the voltage drop across resistor [latex]R_E[/latex]. This voltage is approximately equal to [latex]I_C R_E[/latex]; a more exact value is [latex]I_E R_E[/latex]. Since [latex]I_C ≈ I_E[/latex], the error in the approximation is small, because [latex]I_C + I_B = I_E[/latex], and the value of base current is quite small. Using the value of [latex]I_C[/latex] = 39.5 mA, we get that

[latex]V_E\approx I_CR_E=0.0395 imes 100=3.95 V[/latex]

Plugging the value of [latex]V_E = 3.95 V[/latex] into Equation (6.80) we obtain: [latex]I_B = 217 µA[/latex]. Since we calculated the collector current [latex]I_C[/latex] and base current [latex]I_B, β = I_C/I_B = 182[/latex]. Referring one more time to the circuit of Figure 6.30 we can see that the collector-emitter voltage:

[latex]V_{CE}=V_{CC}-I_{C}R_C-V_E[/latex]

Plugging the corresponding values, we obtain that

[latex]V_{CE}=10-(0.0395 imes 50)-3.95=4.08 V[/latex]

For part b of this example, since [latex]g_m = I_C/V_T and r_π = β/g_m[/latex], we obtain that

[latex]g_m=1.52 S and r_π=119 Ω[/latex]

The results for part (a) are regrouped and presented here:

[latex]I_C=39.5 mA[/latex]

[latex]I_B=217 µA[/latex]

[latex]V_{CE}=4.08 V[/latex]

[latex]V_{RE}=3.95 V[/latex] and

β=182

and for part b):

[latex]g_m=1.52 S[/latex]

and

[latex]r_π=119 Ω.[/latex]

Question 6.13: Given the circuit of Figure 6.30, assume the following compo...

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