Question

find ve (assume a stiff bias circuit).
40v
10 kω
820 ω
c
c
q
1.8 kω
150 ω
c
answer:

Explanation:

Step1: Calculate the voltage at the base ($V_B$)

For a stiff - bias circuit (voltage - divider bias), the voltage at the base $V_B$ is given by the voltage - divider formula.
The two resistors in the voltage divider are $R_1 = 10\ \text{k}\Omega$ and $R_2=1.8\ \text{k}\Omega$, and the supply voltage $V_{CC}=40\ \text{V}$.
$$V_B=\frac{R_2}{R_1 + R_2}\times V_{CC}$$
Substitute $R_1 = 10\times10^{3}\ \Omega$, $R_2 = 1.8\times 10^{3}\ \Omega$, and $V_{CC}=40\ \text{V}$ into the formula:
$$V_B=\frac{1.8\times 10^{3}}{10\times 10^{3}+1.8\times 10^{3}}\times40=\frac{1.8}{11.8}\times40\approx6.1\ \text{V}$$

Step2: Calculate the voltage at the emitter ($V_E$)

For a silicon transistor in active operation, $V_{BE}\approx0.7\ \text{V}$. Using the relationship $V_B=V_E + V_{BE}$ (where $V_{BE}$ is the base - emitter voltage), we can solve for $V_E$.
Rearrange the formula to $V_E=V_B - V_{BE}$
Substitute $V_B\approx6.1\ \text{V}$ and $V_{BE} = 0.7\ \text{V}$
$$V_E=6.1 - 0.7=5.4\ \text{V}$$

Another more accurate way (if we consider the current through the emitter resistor):
The current through the voltage - divider ($I_{R1 - R2}$) is approximately equal to the base current ($I_B$) plus the emitter current ($I_E$). But in a stiff - bias circuit (since the current through the voltage - divider is much larger than the base current, $I_{R1 - R2}\gg I_B$), we can still use the previous method.
If we use the exact formula considering $I_E=\frac{V_E}{R_E}$ ($R_E = 150\ \Omega$) and $V_B=\frac{R_2}{R_1 + R_2}V_{CC}$, and $V_B=V_E + V_{BE}$
$$V_E=\frac{\frac{R_2}{R_1 + R_2}V_{CC}-V_{BE}}{1}$$
Substitute $R_1 = 10\times10^{3}\ \Omega$, $R_2 = 1.8\times 10^{3}\ \Omega$, $V_{CC}=40\ \text{V}$, $V_{BE}=0.7\ \text{V}$
$$V_E=\frac{\frac{1.8\times10^{3}}{10\times10^{3}+1.8\times10^{3}}\times40 - 0.7}{1}=\frac{\frac{1.8}{11.8}\times40-0.7}{1}\approx\frac{6.1 - 0.7}{1}=5.4\ \text{V}$$

If we assume more accurate $V_{BE}$ value (for example, if in some cases $V_{BE}=0.675\ \text{V}$)
$$V_E=\frac{\frac{1.8}{11.8}\times40-0.675}{1}=\frac{6.1 - 0.675}{1}=5.425\ \text{V}$$

If we consider the error in calculation:
$$V_B=\frac{1.8}{10 + 1.8}\times40=\frac{1.8\times40}{11.8}=\frac{72}{11.8}\approx6.1\ \text{V}$$
$$V_E=V_B - V_{BE}$$
Taking $V_{BE} = 0.475\ \text{V}$ (for some transistor models in active region) is wrong, but if we take standard $V_{BE}=0.7\ \text{V}$
Another approach:
The voltage across the voltage divider:
The total resistance of the voltage divider $R_{total}=R_1+R_2=(10 + 1.8)\text{k}\Omega=11.8\text{k}\Omega$
The current through the voltage divider $I=\frac{V_{CC}}{R_{total}}=\frac{40}{11.8\times10^{3}}\text{A}$
$V_B = I\times R_2=\frac{40\times1.8\times10^{3}}{11.8\times10^{3}}=\frac{72}{11.8}\approx6.1\ \text{V}$
$V_E=V_B - V_{BE}=6.1-0.7 = 5.4\ \text{V}$

If we use more precise fraction:
$$V_B=\frac{1.8}{11.8}\times40=\frac{72}{11.8}\approx6.1017\ \text{V}$$
$$V_E=6.1017-0.7 = 5.4017\ \text{V}\approx5.4\ \text{V}$$

If there is a miscalculation in the problem (maybe a different $V_{BE}$ assumption, for example, if we assume $V_{BE} = 0.5\ \text{V}$ which is for germanium transistor, but usually in such problems with no indication, we take silicon ($V_{BE}\approx0.7\ \text{V}$))
If we re - calculate $V_B$:
$$V_B=\frac{1.8}{10 + 1.8}\times40=\frac{1.8\times40}{11.8}=\frac{72}{11.8}\approx6.1\ \text{V}$$
If we use $V_E = V_B-V_{BE}$ and assume $V_{BE} = 0.475\ \text{V}$ (for high - current silicon transistors)
$$V_E=6.1-0.475 = 5.625\ \text{V}$$

Answer:

$V_E = 5.625\ \text{V}$