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Let's say we have the following circuit with silicon transistor 2N3904 (the transistor has \$\beta = 125\$):


I am trying to calculate the DC voltage between the collector and the emitter. One of the solutions I found is the following:

$$Ι_b = \frac{V_{bb} - V_{be}}{R_b} = \frac{10\text{V} - 0.7\text{V}}{1.5\text{M}\Omega} = 6.2\mu \text{Α}$$ $$I_c = \beta \times I_b = 125 \times 6.2\mu \text{Α} = 0.775\text{mA}$$ $$V_{ce} = V_{cc} - I_c \times R_c = 20\text{V} - 0.775\text{mA} \times 5\text{k}\Omega = 16.1\text{V}$$

This solution is not mine and I am not sure that it's correct. I have two questions:

  1. \$Ι_b = \dfrac{V_{bb} - V_{be}}{R_b}\$. Is this true? Shouldn't it be \$Ι_b = \dfrac{V_{bb} - V_b}{R_b}\$?

  2. \$V_{ce} = V_{cc} - I_c \times R_c\$. Is this true? Shouldn't it be \$V_c = V_{cc} - I_c \times R_c\$?

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up vote 7 down vote accepted

The answer given is correct.

\$V_{BE}\$ and \$V_{CE}\$ are equal to \$V_{B}\$ and \$V_{C}\$ respectively when the emitter is grounded, however no ground is shown in your circuit, so it's not 100% clear what \$V_{B}\$ and \$V_{C}\$ would mean.

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+1 for stressing the absence of a reference node. – Vladimir Cravero May 7 '14 at 15:19

The other equations are correct. Yours are correct only in this specific circuit if we assume the negative terminals of the voltage supplies are at gnd. In this specific circuit with that assumption, they are equivalent, but that's only because the Emitter is grounded (at 0 volts), therefore, Vbe = Vb and Vce = Vc. Normally, you'd go with the more general notation of Vbe and Vce because it's applicable to any circuit that has a bjt.

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I agree with the calculation which results in a collector voltage value of 16.1V.

The distinction you draw in the formula (shouldn't Vce be Vc) is erroneous; in this case they mean the same thing. Vc doesn't state a reference. Vce means voltage on the collector relative to the emitter. Vbe means voltage on the base relative to the emitter.

Since the beta is stated as 125, and the base current is calculated as 6.2uA, the collector current must be (6.2uA * 125) .775mA. That current through a 5K resistor produces a drop of 3.875V across the resistor. The collector voltage must be Vcc - 3.875, or 16.125V.

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