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The current source in the following picture is an ideal current source: enter image description here

It is asked to calculate the Collector current based on I1 and ß (Beta).

What that is weird to me is that I think Ic must be zero because of the short circuit connection between Base and Collector (Which means Voltage of BC diode is 0 and so the diode is off). So the question is that:

Why we have a current in the BC diode greater than 0? (And how can I calculate this current?)

Can you please add the small signal model of the circuit too?

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  • \$\begingroup\$ The diodes model of the transistor is not good enough for some types of analysis. This is one of them. Just solve the currents equation. \$\endgroup\$ – Eugene Sh. Jul 27 '17 at 22:12
  • \$\begingroup\$ Which current equations? \$\endgroup\$ – Abraham Jul 27 '17 at 22:15
  • \$\begingroup\$ That's normal, the BC junction is reverse biased (off) for active mode of a BJT. It is forward biased in saturation where the beta relation doesn't hold. \$\endgroup\$ – sstobbe Jul 27 '17 at 22:45
  • \$\begingroup\$ The voltage across the transistor is a "large signal" value (determined by the ebers-moll equation), voiding the use of a small-signal model. This does not really matter though. The T model can be used to solve for currents with Kirchoffs current laws (Janka's answer) \$\endgroup\$ – user55924 Jul 27 '17 at 23:07
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    \$\begingroup\$ Assume for a moment that all of the current went through the base and not the collector. Wouldn't that mean that the collector would be an easier path for the current than the base and therefore couldn't be zero? Or assume for a moment that all of the current went through the collector. Wouldn't that mean that without any base current to supply the required recombination current involved, that it could not have any collector current and therefore a logical conflict? Somewhere in between these two cases lies the truth. \$\endgroup\$ – jonk Jul 28 '17 at 2:43
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\$I_c\$ is always \$\beta \cdot I_b\$ in this simple model. Your idea of a "short circuit" between collector and base never exists for a BJT, because its function doesn't rely on voltages but currents.

schematic

simulate this circuit – Schematic created using CircuitLab

So, it looks like a "short circuit", but it isn't because the current distribution between the \$I_b\$ and \$I_c\$ paths is controlled by the current sources.

$$I_c = \beta \cdot I_b$$ $$\beta = \mathrm{const}$$ $$I_1 = I_c + I_b$$

$$\Rightarrow I_1 = \beta \cdot I_b + I_b = (1 + \beta) \cdot I_b$$ $$\Rightarrow I_b = \frac{1}{1+\beta} \cdot I_1$$ $$\Rightarrow I_c = \frac{\beta}{1+\beta} \cdot I_1$$

Set \$\beta = 1\$ in your thoughts. Base and collector path have identical properties now. You see how the fixed current \$I_1\$ splits into two equal parts.

To find out \$U_{be}\$, one has to look up this reduced Ebers-Moll model:

$$I_c = I_S \cdot e^\frac{U_{be}}{U_T}$$ $$I_S = \mathrm{const} ; U_T = \mathrm{const} $$

So

$$\frac{\beta}{1+\beta} \cdot I_1 = I_S \cdot e^\frac{U_{be}}{U_T}$$ $$\Rightarrow U_{be} = \ln(\frac{\beta}{1+\beta} \cdot \frac{I_1}{I_S}) \cdot U_T $$

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  • \$\begingroup\$ Would you please add the Small Signal model and I-V equation of this transistor too? \$\endgroup\$ – Abraham Jul 28 '17 at 7:19
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    \$\begingroup\$ I've added these. But you better look up the small signal model of a BJT in a book or your lecture notes. \$\endgroup\$ – Janka Jul 28 '17 at 14:12
  • \$\begingroup\$ Thank you dear Janka. I'm following some tutorial videos and some books. (Sedra.Smit book and Dr.Razavi videos), but some points are a little vague for me. \$\endgroup\$ – Abraham Jul 28 '17 at 15:26

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