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Looking at the equations of the common base amplifier circuit shown below: enter image description here

The input current 𝐼𝑒 should be given by the equation: $$I_e = (V_{in} - 0.7)/R_e$$

Then, the output current 𝐼𝑐 should be: $$I_c = \frac{\beta}{\beta + 1} I_e$$

Taking KVL on the loop at the right would give: $$V_{cc} - V_{cb} - I_cR_{load}= 0$$

So, this gives us that π‘‰π‘œπ‘’π‘‘ is directly propotional to (π‘‰π‘–π‘›βˆ’0.7), but it is also limited by 𝑉𝑐𝑐 such that it can't increase more than it to keep the third equation valid without getting a negative 𝑉𝑐𝑏. Now, my question is, I'm wondering what will happen if a big 𝑉𝑖𝑛(or a small 𝑅𝑒) were present such that 𝐼𝑐 is high and in turn it wants to make a voltage drop across π‘…π‘™π‘œπ‘Žπ‘‘ which is bigger than 𝑉𝑐𝑐(considering 𝑉𝑐𝑏 to be approximately 0), what will happen after that if 𝐼𝑒 was further increased?

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3 Answers 3

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You are correct : those equations are the small signal analysis.

When you run out of VCC, the transistor will saturate, and in that case Ib will increase (or an alternative view, beta will decrease, amounts to the same thing) until the equations work again.

Ultimately Ic will be controlled by Vcc and Rc and all the rest is base current.

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As \$R_E\$ is decreased, \$V_{out}\$ will increase, up to a point. When \$V_{out}\$ approaches \$V_{cc}-0.7V\$, the transistor will begin to enter saturation. \$V_{out}\$ will increase more slowly. \$V_{out}\$ will not increase much more than \$V_{cc}-0.2V\$. [Here I am using \$V_{cc}\$ to represent the voltage of the right hand side battery.]

If \$R_E\$ is made sufficiently low, and if \$V_{in}\$ is greater than 0.7V, then the base-emitter junction of the transistor will eventually have excessive current, and the transistor can be destroyed.

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Your question is basically just this:

schematic

simulate this circuit – Schematic created using CircuitLab

Obviously, at first, the drop across \$R_\text{LOAD}\$ is \$0\:\text{V}\$ and so \$V_\text{OUT}=V_\text{CC}\$. But as you "pull" more current out of \$Q_1\$'s emitter, there's an increasing voltage drop across \$R_\text{LOAD}\$ until the point where \$V_\text{OUT}\approx 0\:\text{V}\$ (because \$I_1\cdot R_\text{LOAD}\approx V_\text{CC}\$. This is about when the BJT just starts to go into saturation.

As you pull still more current out of the emitter, the collector will drop a little bit further. But only until it gets to about one diode drop below ground. Keep in mind to make the emitter go \$60\:\text{mV}\$ still more negative (and if \$I_1\$ pulls hard enough the emitter will become still more negative), 10 times as much current must be pulled out of the emitter. And the collector cannot go more negative than the emitter itself. The BJT will be in deep saturation when the collector is within a few hundred millivolts or less of the emitter voltage. At this point the collector is barely moving (just as the emitter is barely moving.)

The collector current, in very deep saturation, is quite limited now (by the load resistance) and effectively hits a barrier and cannot increase much further. The base current continues to increase until the base current actually exceeds the collector current and it continues on in that way.

At really high currents that no one actually uses (and an indestructable BJT) the Ohmic resistances pretty much take over, though. (Usually those are neglected in simplified cases like this.)

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