# Voltage between collector and emitter (in transistor circuit) [2N3904]

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

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

$$Ι_b = \frac{V_{bb} - V_{be}}{R_b} = \frac{10V - 0.7V}{1.5MΩ} = 6.2μΑ$$ $$I_c = β \times I_b = 125 \times 6.2μΑ = 0.775mA$$ $$V_{ce} = V_{cc} - I_c \times R_c = 20V - 0.775mA \times 5kΩ = 16.1V$$

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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$V_{BE}$ 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 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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