Ignoring odd arrangements, there are two cases that have very different results with BJTs: (1) operated as an amplifier (active mode); and, (2) operated as a switch (saturated mode).

Technically, these can be analyzed using a single, rather complex model. (See the DC version of the Ebers-Moll model.) But for ease of analysis, these are instead usually broken into two different modes with two different behaviors.

If you already know the circuit's purpose, then you can just take the assumptions for that purpose and start there. If the numbers work out then you are done. If you don't know the circuit's purpose (or it is a test to see if you can work things out, either way), then you usually start with the active mode assumptions and see if the results are consistent. If so, done. If not, you switch over to the saturated mode assumptions and re-compute. (Of course, you could do it in the other order.)

One or the other will result in a consistent result and you will know the mode of operation and quantities that will be close to what will be measured.

(The following does not apply to Darlington arrangements.)

When writing below, I'll be providing magnitudes (unless you see a specific minus sign.) You need to apply them appropriately, depending upon NPN or PNP situations....

active mode assumptions

• The collector acts like a current sink (NPN) or source (PNP) with infinite impedance.

  (This is wrong due to the Early Effect. But since the Early Effect generates only about a 1% error, it can usually be ignored.)

• The bipolar will have a relatively unchanging ratio during operation, \$\beta=\frac{I_{_\text{C}}}{I_{_\text{B}}}\$.

  (While any two different bipolars from the same family of devices will have different values, they will be within 30-50% of each other. For small signal devices without a datasheet you can expect \$100 \le \beta\le 350\$. But check the datasheet for details, if available. For large signal devices without a datasheet you can expect \$10 \le \beta\le 60\$. But check the datasheet for details, if available. \$\beta\$ will depend upon the device temperature and quiescent operating current. But given some specific temperature and quiescent operating current, you can expect this value to remain relatively flat (fixed.))

• The bipolar will have a relatively unchanging \$V_{_\text{BE}}\$ at any given quiescent operating current.

  (While any two different bipolars from the same family of devices will have different values, yet again, they will be within about \$50\:\text{mV}\$ of each other when operating at the same quiescent point. Without a datasheet and without knowing the operating point you can expect \$450 \le V_{_\text{BE}}\le 950\$ at room temperature. But check the datasheet for details, if available. \$V_{_\text{BE}}\$ will depend upon the device temperature and quiescent operating current. But given some specific temperature and quiescent operating current, you can expect this value to remain relatively flat (fixed.) Over wide temperature changes, you can expect to see something in the area of about \$-2\frac{\text{mV}}{^\circ\text{C}}\$ change. Over wide operating current changes and with small signal devices, you can expect to see about \$60\:\text{mV}\$ change per decade change in collector current. For large signal devices this may be more like \$100\:\text{mV}\$ change per decade change, but they usually start at a lower \$V_{_\text{BE}}\$, too. Again, look at the datasheet to get a better idea.)

One enters into analysis of an existing circuit with some idea of \$V_{_\text{BE}}\$ (for some estimated collector current and operating temperature) and \$\beta\$, and makes adjustments later as calculations reveal their results.

saturated mode assumptions

• The collector acts like a voltage source with zero impedance.

  (This, again, isn't strictly true. Usually, there are some small bulk impedances at the collector, emitter, and base.)

• \$V_{_\text{BE}}\$ will be marginally higher than in active mode -- perhaps \$50\:\text{mV}\$ higher, maybe a little more.

• \$V_{_\text{CE}}\$ is relatively constant and small. Typically the saturated \$V_{_\text{CE}}\le 200\:\text{mV}\$.

example circuit

None of the values are specified below. Just the topology, for now:

^{simulate this circuit – Schematic created using CircuitLab}

Three resistor values need to be specified, along with three ideal voltage source values. Once those are known, the situation can be worked out.

In the following, I'll start by assuming active mode and room temperature.

• \$V_{_\text{CC}}=10\:\text{V}\$
• \$V_{_\text{EE}}=-10\:\text{V}\$
• \$V_{_\text{BB}}=-8\:\text{V}\$
• \$R_{_\text{B}}=10\:\text{k}\Omega\$
• \$R_{_\text{C}}=1\:\text{k}\Omega\$
• \$R_{_\text{E}}=100\:\Omega\$
• \$\beta\approx 150\$
• \$V_{_\text{BE}}\approx 700\:\text{mV}\$ at \$I_{_\text{C}}=2\:\text{mA}\$

KVL provides \$I_{_\text{B}}=\frac{V_{_\text{BB}}-V_{_\text{BE}}-V_{_\text{EE}}}{R_{_\text{B}}+R_{_\text{E}}\cdot\left(\beta+1\right)}\approx 51.8\:\mu\text{A}\$. So \$I_{_\text{C}}=\beta\cdot I_{_\text{B}}\approx 7.77\:\text{mA}\$ and \$I_{_\text{E}}\approx 7.82\:\text{mA}\$. Also, \$V_{_\text{B}}\approx -8.52\:\text{V}\$, \$V_{_\text{E}}\approx -9.22\:\text{V}\$, and \$V_{_\text{C}}\approx 2.23\:\text{V}\$.

Suppose, instead, we happened to pick out a bipolar with \$\beta=200\$, instead. Then, \$I_{_\text{B}}\approx 43.2\:\mu\text{A}\$, \$I_{_\text{C}}\approx 8.64\:\text{mA}\$ and \$I_{_\text{E}}\approx 8.68\:\text{mA}\$, \$V_{_\text{B}}\approx -8.43\:\text{V}\$, \$V_{_\text{E}}\approx -9.13\:\text{V}\$, and \$V_{_\text{C}}\approx 1.36\:\text{V}\$.

As you can see, some but not much change. And regardless, the bipolar actually has the collector voltage above the base voltage. So it is in active mode, as assumed.

Note: The current is almost 5 times higher than what I assumed for the base-emitter voltage. This would mean that \$V_{_\text{BE}}\approx 741\:\text{mV}\$, instead. A few more details would change. (You can work it out.) But this circuit would still be in active mode and the analysis would be close to what you'd experience if you built this on a solderless breadboard.

Now suppose everything else stays the same except that \$V_{_\text{BB}}=-5\:\text{V}\$, instead? A quick check says that \$I_{_\text{B}}\approx 171\:\mu\text{A}\$ and \$I_{_\text{C}}\approx 25.7\:\text{mA}\$. This would set \$V_{_\text{B}}\approx -6.71\:\text{V}\$ and \$V_{_\text{C}}\approx -15.7\:\text{V}\$! This last value, \$V_{_\text{C}}\approx -15.7\:\text{V}\$, immediately tells you that it just cannot be right and therefore this must be a saturated situation, instead.

This means you now ignore \$\beta\$. It's very wrong and cannot be used. Instead, you now apply the saturated \$V_{_\text{CE}}\le 200\:\text{mV}\$ and \$V_{_\text{BE}}\approx 750\:\text{mV}\$, to start out. If you solve this out you will get \$I_{_\text{C}}\approx 18\:\text{mA}\$ and \$I_{_\text{E}}\approx 18.2\:\text{mA}\$. The collector and emitter voltages will be very close to each other, as expected.

My question is how does the resistor at C affect the current at B.

If the collector resistor gets large enough, it will force the collector voltage down very close to the emitter voltage, forcing the bipolar into saturation. Make it a lot smaller and the bipolar will move back into active mode.