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I am confused why in one of the KVL equations \$I_{C1}\$ was multiplied by \$R_{C1}\$. Doing a quick KCL, the current through resistor would be \$I_{C1}-I_{B2}\$.

However, if solved by basically neglecting \$I_{B2}\$, \$I_{B2}\$ is very small and to some extent, it makes sense we can somewhat neglect it. But I feel that's not the reasoning behind it.

Hence, why is it (from my perspective) that \$I_{B2}\$ is being neglected when solving for \$V_{C}\$ (when \$I_{C1}\$ and \$R_{C1}\$ are being multiplied)?

  • 1
    \$\begingroup\$ The voltage of the base 1st BJT (from left to right)aint 7.5V it is 5V \$\endgroup\$
    – Miss Mulan
    May 2, 2021 at 18:58
  • \$\begingroup\$ @MissMulan I am well aware of this. It is something that was also mentioned in the post I refrenced from \$\endgroup\$ May 2, 2021 at 19:40
  • \$\begingroup\$ @G36 Thank you, I think I finally understand :) \$\endgroup\$ May 2, 2021 at 19:41

2 Answers 2


There are a few quick techniques you can use that do not ignore the base current. But I'll discuss just one here.

My recommendation is that you put a little time into it, think about the results and whether or not it's worth it for the future or if you now want to decide that the technique the author used is "good enough."

Before I discuss it, let's assume you've made a decision about the desired quiescent collector current, \$I_\text{Q}\$, and the base's biasing voltage, \$V_\text{B}\$. These are related to each other as the emitter voltage is \$V_\text{E}=V_\text{B}-V_\text{BE}\$ and \$I_\text{Q}=\frac{\beta}{\beta+1}\frac{V_\text{E}}{R_\text{E}}\$.

It's common practice to figure that the biasing pair of resistors at the base should have a current "waterfall" through them that is about \$10\times\$ as much as the expected base current. And since the expected \$\beta\$ in small signal BJTs these days will be \$\beta\ge100\$, it follows that this "waterfall" current should be about \$I_\text{W}=\frac{I_\text{Q}}{10}\$. (Yeah, I'm using \$w\$ for "waterfall.")

Now you can compute:


This allows both the waterfall current and also the expected base recombination current to get through the high-side biasing resistor. Once it gets through it, the base will get its share, leaving only the waterfall current of \$\frac{I_\text{Q}}{\frac1{10}\left(\beta=100\right)}\$ to go through the low-side biasing resistor.

So now you can compute $$R_{\text{B}_2}=\frac{V_\text{B}}{\frac{I_\text{Q}}{\frac1{10}\left(\beta=100\right)}}$$

This isn't complicated. You want the waterfall current, plus a little for the base, going through the high-side biasing resistor. But only the remainder waterfall current going through the low-side biasing resistor.

This approach is slightly more accurate. But resistors have their own tolerance values and there are limited choices, anyway. So practice a few designs using the simpler technique and then using the above one. Compare them using Spice simulation, but using only available resistor values. You can play with different values of \$\beta\$ and you can even vary the actual resistor values using tolerances, too. See if you feel better going the extra mile, or not, after that. Once you have been through it a few times, I think you will have a solid idea in mind and know what to do from then on, in practice, and what to do when you are just doing "ideal" analysis.

By the way, KCL doesn't ignore the base current. But that adds another variable and then requires more equations. If you don't have a solver at-hand, this just makes things harder on paper (or drawing in sand with your fingers.) So that's why you take short-cuts.

Also the factor of 10 figures all over the place in engineering. That's an order of magnitude. For many uses, when you are within 10% or 90% of something, you are close enough. So that's why 10 is seen in all kinds of estimates. Most of the important things happen over just one order of magnitude. Get used to that, too.


You have already answered your own question.

When performing hand calculations we usually simplified a bit to speed up the calculation process. Thus, the author decided to neglect the \$I_{B2}\$. Also, the \$I_{B2}\$ current will be small compare to the \$I_{C1}\$ current. And such a simplification is justified because in the typical application when BJT's is working in the active region the base current will be beta times smaller than the collector current. And for small-signal BJT's the beta value will be large than \$\beta > 100 \$.


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