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I have simulated the following circuit using partsim.com: enter image description here

I don't understand why R1 and R2 function as a simple voltage divider. I would expect R1 and R4 to to provide parallel resistance (keeping in mind the ~0.7 voltage drop of Vbe). Can someone explain what I'm missing?

Further, seeing R4 as (1 + beta) R4 from the R1, R2 node does not seem to explain the behavior when the R4 resistance is lowered since R1 and R2 don't act like a voltage divider then:

enter image description here

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  • \$\begingroup\$ "...R1 and R4 to provide parallel resistance..." What exactly do you mean by this? \$\endgroup\$ – Hearth May 20 '17 at 20:13
  • \$\begingroup\$ I mean I would think the current going through R1 would split between R2 and R4 to some extent. \$\endgroup\$ – Campbell May 20 '17 at 20:15
  • \$\begingroup\$ Even though you're in a simulator use realistic component values. 5 and 10 ohms is just too low. If you build this in the real world, the 2n2222 would be damaged instantly. Make all resistors 1000 times (yes one-thousand times) larger: so 5 kohm and 10 kohm, then the 2n2222 transistor model will also behave more like what is expected of it. \$\endgroup\$ – Bimpelrekkie May 20 '17 at 20:16
  • \$\begingroup\$ I mean I would think the current going through R1 would split between R2 and R4 to some extent Yes that is true but how much do you expect to flow through R2 and how much through R4 ? The current through R4, is that all coming from R1 or not ? What is the main factor which makes BJTs so useful ? They amplify... ?? \$\endgroup\$ – Bimpelrekkie May 20 '17 at 20:18
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    \$\begingroup\$ @FakeMoustache I think I get his question, though. He isn't understanding that collector current also occurs in the emitter resistor. He has over-simplified things in his head, is all. If you took away the collector resistor and took away \$R_2\$ then you'd get his question. He just needs to gain the nuance that there will be collector current and that this will impact the emitter voltage and that this will impact his viewpoint. That, plus Thevenin to include \$R_2\$ as well, in the base side. \$\endgroup\$ – jonk May 20 '17 at 21:07
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I'm going to avoid arguing over specific resistor values, for now, and just focus on what I perceive as the arc of your question. I think you are wondering about this, to begin:

schematic

simulate this circuit – Schematic created using CircuitLab

If it were only this case, then you could compute the current as:

$$I=\frac{V_{CC}-V_{BE}}{R_1+R_2}$$

where \$V_{BE}\$ is the base-emitter voltage that occurs at the computed current. (There may be an iteration or two before you nail it down well, unless you use the product-log function.)

But now what happens if the circuit is changed:

schematic

simulate this circuit

At this point, there is another source of current (through the collector) that will also flow via \$R_4\$. Assuming that the NPN BJT is still in its active region (where \$\beta\$ is reasonably large), then the NPN BJT will arrange things differently, so that \$I_C=\beta\cdot I_B\$ and \$I_E=\left(\beta+1\right)\cdot I_B\$. And you cannot ignore the collector current, anymore -- especially because it is so much larger than the base current.

In this case, we'd find from the Kirchhoff voltage law that:

$$\begin{align*} V_{CC} - I_B\cdot R_1 - V_{BE} - I_E\cdot R_4 &= 0\:\textrm{V} \\\\ V_{CC} - V_{BE} &= I_B\cdot R_1 + I_E\cdot R_4 \\\\ V_{CC} - V_{BE} &= I_B\cdot R_1 + \left(\beta+1\right)\cdot I_B\cdot R_4 \\\\ V_{CC} - V_{BE} &= I_B\cdot\left[ R_1 + \left(\beta+1\right)\cdot R_4\right] \\\\ I_B=\frac{V_{CC} - V_{BE}}{R_1 + \left(\beta+1\right)\cdot R_4} \end{align*}$$

And from that you could compute the collector and emitter currents, work out the voltages, etc.

Note that hooking up the collector has a huge impact!

Now the last step:

schematic

simulate this circuit

Adding a second resistor to the base portion of the circuit provides a voltage divider and allows us to provide a different effective voltage (other than the supply rail) while at the same time keeping some equivalent base resistor. It adds one degree of design freedom and is very nice to have.

The resulting calculations are quite similar, so long as you substitute in at the right places. You know that \$R_{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$ and that \$V_{TH}=V_{CC}\cdot \frac{R_2}{R_1+R_2}\$. So:

$$\begin{align*} V_{TH} - I_B\cdot R_{TH} - V_{BE} - I_E\cdot R_4 &= 0\:\textrm{V} \\\\ V_{TH} - V_{BE} &= I_B\cdot R_{TH} + I_E\cdot R_4 \\\\ V_{TH} - V_{BE} &= I_B\cdot R_{TH} + \left(\beta+1\right)\cdot I_B\cdot R_4 \\\\ V_{TH} - V_{BE} &= I_B\cdot\left[ R_{TH} + \left(\beta+1\right)\cdot R_4\right] \\\\ I_B=\frac{V_{TH} - V_{BE}}{R_{TH} + \left(\beta+1\right)\cdot R_4} \end{align*}$$

You can see that the calculation is quite similar, except that you've substituted in the Thevenin equivalents, where appropriate.

All of the above also shows you that \$R_4\$ has to be multiplied up to a much larger value when computing the base current. This happens because of all the collector current (a function of base current in active mode) adds to the base current before flowing in \$R_4\$.

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"I mean I would think the current going through R1 would split between R2 and R4 to some extent."

In that case, good question! The fact that you're asking this shows that you're actually thinking about it.

The answer is that yes, the current does get split, but it's more complicated than that, because the current into the base is not in general the same as the current out of the emitter. In forward active mode, the resistor is effectively magnified by 1 + β from the point of view of the base node. This is because the current through the resistor is the emitter current, Ie, not the base current Ib. Since Ie = (1 + β)Ib, what is 'seen' at the base node is this:

schematic

simulate this circuit – Schematic created using CircuitLab

Now consider what happens if you take a large resistor and put it in parallel with a small one. The resistance of the pair is going to be much closer to that of the smaller resistor--so this "magnified" resistor will have much less of an effect on the output than you might initially think.

Note that this is a fairly simplified explanation and may contain some errors besides--I'm a bit rusty on the details.

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  • \$\begingroup\$ Or wait... if I change R4 to 1K instead of 10 K (I changed all resistances by a factor of 1000 earlier seeking a solution) then Vc saturates at 1.2 V and R1 and R2 don't act as a voltage divider (Vr2 = 1.84 V). I don't see how the resistance seen at the R1/R2 node can be R4* (1 + beta) given this. \$\endgroup\$ – Campbell May 20 '17 at 21:04
  • \$\begingroup\$ When Vc becomes too low and the NPN saturates, the ratio Ic/Ib will no longer be equal to beta. Ic/Ib might be 10 instead of 100 (assuming beta is 100). To see that beta of 100, Vc needs to be high enough so that the NPN is not in saturation meaning that the NPN determines what Ic will be. In saturation the NPN is fully on and Rc (R3 in your example) determines the current Ic. \$\endgroup\$ – Bimpelrekkie May 20 '17 at 21:26
  • \$\begingroup\$ and R1 and R2 don't act as a voltage divider More accurate would be to say: you cannot ignore the influence of Ib on the voltage divider anymore. Ib loads it so much that the voltage drops below the expected value. \$\endgroup\$ – Bimpelrekkie May 20 '17 at 21:29
  • \$\begingroup\$ That's one of those simplifications I mentioned; I left that bit out for simplicity's sake. @FakeMoustache has it right; saturation throws things off. \$\endgroup\$ – Hearth May 20 '17 at 21:31

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