BJT transistor analysis: common emitter confusion

[thegreengineer]

Recently I started studying semiconductors and analogue electronics. First I studied the diode and its I-V characteristics and analysis in DC circuits, as well as in AC circuits.
Now I started with BJT transistors. I was explained that transistors act as electronic switches. My problem is not on how transistors work. For explaining this further consider this diagram of a transistor in common emitter configuration.
https://scontent-dft4-2.xx.fbcdn.net/v/t1.0-9/14067588_1769551276662742_1192159594276821135_n.jpg?oh=5b8e91a0eddc348493fad870a70a27d3&oe=58597241
In this case I got two DC voltage sources of 9 V each, the two 330 Ω resistors, and a NPN BJT transistor. The transistor has a β=210.
According to what I was taught, β represents the current gain in common emitter and it ranges from 50 to 400 (it is a dimensionless quantity). The current gain represents the ratio of output current divided by input current. In this configuration, the emitter is common to the input and output sides (connected to ground). The input is the one p-n junction that is direct biased; whereas the output is the other p-n junction that is reverse biased. So, according to this, the input current is the base current and the output current is the collector current. So using the gain (β) definition, we have:
$\beta =\frac{I_{C}}{I_{B}}$.
I previously said that β=210. So if I replace β in the equation and I isolate for collector current I get that the collector current is 210 times the base current.

So let's suppose that the I'm being asked to calculate the collector current considering the circuit above. What I did was the following, since I don't know base current, I don't know collector current either. So what I did is performing KVL (Kirchoff's Voltage Law) on the mesh formed by the 9 V battery, the 330 Ω resistor (both connected in series to the base) and the p-n junction created by the base-emitter junction. This last one, as I was told, resembles a silicon diode, therefore there's going to be a voltage drop of 0.7 V approximately. Doing this I get:

$\sum V=9-330(I_{B})-0.7=0$

When I isolate for base current, I get:
$I_{B}=0.02515$ measured in Amps.

Remember I said that β=210, so using the previous definition for β I get:
$I_{C}=\beta I_{B}=(210)0.02515=5.28$ measured in Amps.

When I checked back the circuit and ran the simulator I got the base current correctly:
https://scontent-dft4-2.xx.fbcdn.net/v/t1.0-9/14079477_1769566206661249_3855909599223329391_n.jpg?oh=7a3d57d997c320e4f5a144cdd2fadada&oe=58549EF2
Since 0.02515 A= 25.15 mA.
However, when I checked the collector current I got a very different result, way less the current I calculated:
https://scontent-dft4-2.xx.fbcdn.net/v/t1.0-9/14051724_1769566949994508_7493816127739306855_n.jpg?oh=ae617ad533c50ad61ab5510d374d22a0&oe=5847C9CE
My question is, if the result is not consistent, then that means I did something wrong, what is it?

 

[rbelli1]

Look at the collector resistor and battery. Why would you expect 5+ amps through that resistor? What is the voltage across the resistor in the simulation given 27.24mA? Where does this voltage come from?

BoB

 

[LvW]

Marcus - you simply have assumed that the BJT would operate in the so-called "active" amplifying region (where the beta-value may be used).. But that is not the case.
Do you know why not?.

 

[thegreengineer]

No. Why not?

 

[Jony130]

Notice that Rc resistor and DC collector voltage (Vcc) will limit the maximum current that can flow in collector. The maximum collector that can flow in this circuit is Ic_max = 9V/330Ω = 0.027A = 27mA. So in this case the BJT is in saturation region and Ic = β * Ib do not hold anymore.
The BJT is just like a water tap. What is happening is that the base current is "controlling" the amount of current that Vcc supplies (can supply). Try increase in simulation Vcc voltage (2000V) or reduce Rc resistor value or increase the RB value .

 

[LvW]

For Ic=5.28A, the voltage drop at the collector resistor can be calculated (theoretically) to Vc=5.28*330=1742 Volt!