Base resistor R2 for voltage bias

Question

Can someone please explain to me how they arrived at 38.8k ohms for the R2 from this site, under the voltage divider bias technique... I just can't seem to figure it out. Pls help out voltage divider calculation

Answer 1

From the top

Follow along:

^{simulate this circuit – Schematic created using CircuitLab}

On left, is the starting schematic. I left the wire at the top of \$R_1\$ and \$R_{_\text{C}}\$ intact, to start, because you must convince yourself that I'm still correct when I remove that top wire. I've had people argue with me that it's not the same schematic after I remove that wire. I want to be sure that you don't feel like arguing. It's an important step towards understanding the process. Make sure you feel comfortable with this choice.

Next, I convert the base pair to its Thevenin equivalent. The page talks about this, so I won't spend a lot of time on it. But, here again, you must be sure that you are comfortable with this idea. It's a true idea. But make sure you agree with it and feel comfortable with this choice, as well.

Finally, I just throw away the collector resistor. It's not important to figuring this out. The reasoning is that the collector acts like a current source. But a current source, ideally, has \$\infty\$ impedance. (*) Essentially, this just means an open wire or open circuit. Put another way, it doesn't matter right now.

(It does matter in the sense that you will later need to verify that the BJT is not saturated. But for now you just have to assume that's true. And if that's true, then we can just remove the collector resistor and not worry about it. Thinking about it will just get in your way of understanding the process.)

Their \$R_{_\text{B}}\$ (same as their \$R_{_\text{TH}}\$)

The KVL of the right-most schematic is:

$$V_{_\text{TH}}-I_{_\text{B}}\cdot R_{_\text{TH}}-V_{_\text{BE}}-I_{_\text{E}}\cdot R_{_\text{E}}=0\:\text{V}$$

For these purposes do also recall that \$I_{_\text{Q}}=I_{_\text{E}}=I_{_\text{C}}=\beta\cdot I_{_\text{B}}\$. So the above can be restated and then solved for \$I_{_\text{B}}\$:

$$\require{cancel}\begin{align*} V_{_\text{TH}}-I_{_\text{B}}\cdot R_{_\text{TH}}-V_{_\text{BE}}-\beta\cdot I_{_\text{B}}\cdot R_{_\text{E}}&=0\:\text{V} \\\\ V_{_\text{TH}}-V_{_\text{BE}}&=I_{_\text{B}}\cdot R_{_\text{TH}}+\beta\cdot I_{_\text{B}}\cdot R_{_\text{E}} \\\\ V_{_\text{TH}}-V_{_\text{BE}}&=I_{_\text{B}}\cdot \left(\cancel{I_{_\text{B}}}\cdot R_{_\text{TH}}+\beta\cdot \cancel{I_{_\text{B}}}\cdot R_{_\text{E}}\right) \\\\&\therefore \\\\ I_{_\text{B}}&=\left[\frac{V_{_\text{TH}}-V_{_\text{BE}}}{R_{_\text{TH}}+\beta\cdot R_{_\text{E}}}=\frac{I_{_\text{Q}}}{\beta}\right] \end{align*}$$

This leaves us with two unknowns, \$V_{_\text{TH}}\$ and \$R_{_\text{TH}}\$, but only one equation with which to work it out. That's a problem.

They resolved it by simply deciding on a value for \$V_{_\text{TH}}\$ and then letting \$R_{_\text{TH}}\$ (aka \$R_{_\text{B}}\$) fall out from there.

$$\begin{align*} \frac{V_{_\text{TH}}-V_{_\text{BE}}}{R_{_\text{TH}}+\beta\cdot R_{_\text{E}}}&=\frac{I_{_\text{Q}}}{\beta}=\frac{I_{_\text{E}}}{\beta} \\\\ \beta\cdot\left(V_{_\text{TH}}-V_{_\text{BE}}\right)&=I_{_\text{E}}\cdot\left(R_{_\text{TH}}+\beta\cdot R_{_\text{E}}\right) \\\\ \beta\cdot\frac{V_{_\text{TH}}-V_{_\text{BE}}}{I_{_\text{E}}}&=R_{_\text{TH}}+\beta\cdot R_{_\text{E}} \\\\ \beta\cdot\frac{V_{_\text{TH}}-V_{_\text{BE}}}{I_{_\text{E}}}-\beta\cdot R_{_\text{E}}&=R_{_\text{TH}} \\\\ R_{_\text{B}}=R_{_\text{TH}}&=\beta\cdot\left(\frac{V_{_\text{TH}}-V_{_\text{BE}}}{I_{_\text{E}}}- R_{_\text{E}}\right) \end{align*}$$

And that's where they got their equation for \$R_{_\text{B}}\$.

But it assumes that \$V_{_\text{TH}}\$ can be guessed at and provided as input to the process.

And they didn't guess well, in my opinion.

Accepting their \$V_{_\text{BB}}\$ value (same as their \$V_{_\text{TH}}\$)

But let's assume they were right to pick \$V_{_\text{BB}}=1.5\:\text{V}\$ and not argue.

We know that \$V_{_\text{TH}}=V_{_\text{BB}}=V_{_\text{CC}}\cdot\frac{R_2}{R_1+R_2}\$ and \$R_{_\text{TH}}=R_{_\text{B}}=R_1\cdot\frac{R_2}{R_1+R_2}\$.

One can then set, as they do, \$f=\frac{R_2}{R_1+R_2}=\frac{V_{_\text{TH}}}{V_{_\text{CC}}}\$, and find \$V_{_\text{TH}}=V_{_\text{CC}}\cdot f\$ (or, alternately, that \$f=\frac{V_{_\text{TH}}}{V_{_\text{CC}}}\$) and \$R_{_\text{TH}}=R_1\cdot f\$. Since they specify \$V_{_\text{TH}}\$ and also know \$V_{_\text{CC}}\$, \$f\$ is known and can be used.

So, of course \$R_1=\frac{R_{_\text{TH}}}{f}=R_{_\text{TH}}\cdot\frac{V_{_\text{CC}}}{V_{_\text{TH}}}\$. And you said you understood that much.

So this gets to the nub of your question, I guess. How to work out \$R_2\$ from the rest of the above.

Remember that \$R_{_\text{TH}}=R_1\cdot\frac{R_2}{R_1+R_2}\$?

$$\require{cancel}\begin{align*} R_{_\text{TH}}&=R_1\cdot\frac{R_2}{R_1+R_2} \\\\ R_{_\text{TH}}\cdot\left(R_1+R_2\right)&=R_1\cdot R_2 \\\\ R_{_\text{TH}}\cdot R_1+R_{_\text{TH}}\cdot R_2&=R_1\cdot R_2 \\\\ R_{_\text{TH}}\cdot R_1&=R_1\cdot R_2-R_{_\text{TH}}\cdot R_2 \\\\ R_{_\text{TH}}\cdot R_1&=R_2\cdot\left(R_1\cdot \cancel{R_2}-R_{_\text{TH}}\cdot \cancel{R_2}\right) \\\\&\therefore\\\\ R_2&=\frac{R_{_\text{TH}}\cdot R_1}{R_1-R_{_\text{TH}}} \end{align*}$$

Given that they computed \$R_{_\text{TH}}=33\:\text{k}\Omega\$ and then found \$R_1=220\:\text{k}\Omega\$, it follows that \$R_2=\frac{33\:\text{k}\Omega\,\cdot\, 220\:\text{k}\Omega}{220\:\text{k}\Omega-33\:\text{k}\Omega}\approx 38.82\:\text{k}\Omega\$.

I think that was the algebra you were looking for.

At this point, I'd like to leave the above behind and add something new.

Returning to their pick for \$V_{_\text{BB}}\$ (same as \$V_{_\text{TH}}\$)

I already said that I may disagree with their approach. So let's get into that.

Taking their (the web page's) assumptions for design purposes that \$\beta=100\$, that the quiescent current is \$I_{_\text{Q}}=I_{_\text{E}}=I_{_\text{C}}=1\:\text{mA}\$, that \$V_{_\text{CC}}=10\:\text{V}\$, and that \$R_{_\text{E}}=470\:\Omega\$.

From here, we can work out that we expect \$I_{_\text{B}}=\frac{I_{_\text{Q}}=1\:\text{mA}}{\beta=100}=10\:\mu\text{A}\$ and that \$V_{_\text{E}}=I_{_\text{Q}}\cdot R_{_\text{E}}=470\:\text{mV}\$. Add another \$V_{_\text{BE}}=700\:\text{mV}\$ to that to find \$V_{_\text{B}}=V_{_\text{E}}+V_{_\text{BE}}=1.17\text{V}\$.

Now, this is where they skipped a step and just told you that \$V_{_\text{BB}}=1.5\:\text{V}\$. (I disagree.) But why did they pick this particular value? And why do I think it's a problem?

First off, I think they just picked it out of a hat, so to speak. They just wanted to make \$V_{_\text{BB}}\$ a simple value.

I would have chosen differently: \$V_{_\text{BB}}=1.3\:\text{V}\$.

The reason is that there should not be more than about 10% of \$V_{_\text{B}}\$ dropped across \$R_{_\text{BB}}\$. In this case, 10% of \$1.17\:\text{V}\$ is \$117\:\text{mV}\$. So adding those together gives \$1.297\:\text{V}\$. Which I'd round to \$1.3\:\text{V}\$.

But not \$1.5\:\text{V}\$.

I think they were lazy in pulling a number out of the air. Or perhaps they just didn't want to develop the details and took a short-cut. Either way, that choice impaired the value of what they wrote.

Biasing Pair Stiffness

This gets into a discussion about stiffness of the base biasing pair of resistors.

Recall this from the earlier developed KVL of the right-most schematic:

$$\begin{align*} I_{_\text{B}}&=\left[\frac{V_{_\text{TH}}-V_{_\text{BE}}}{R_{_\text{TH}}+\beta\cdot R_{_\text{E}}}=\frac{I_{_\text{Q}}}{\beta}\right] \end{align*}$$

This left us with two unknowns, \$V_{_\text{TH}}\$ and \$R_{_\text{TH}}\$, as mentioned before. They resolved it by picking a convenient \$V_{_\text{TH}}\$.

But this can be better resolved by establishing a new current through the base divider pair: \$I_{_\text{BIAS}}=\text{BIAS%}\cdot I_{_\text{Q}}\$. This sets a kind of stiffness.

If the base divider pair passes enough current through itself, ignoring the base current, and if the the base current of the BJT is small enough by comparison, then BJT variations (they all vary, one to another) won't impact the biasing much. That's the hope, anyway. So we want to establish a value for \$\text{BIAS%}\$ that is about 10 times larger than \$\frac1{\beta}\$. If it is that much larger, then slight variations in \$I_{_\text{B}}\$ won't significantly impact the quiescent current. And that's a useful goal.

In this case, \$\text{BIAS%}=10\cdot\frac1{\beta=100}=0.1\$ and therefore \$I_{_\text{BIAS}}=0.1\cdot I_{_\text{Q}}=100\:\mu\text{A}\$.

\$R_1\$ and \$R_2\$

We want \$R_2=\frac{V_{_\text{B}}}{I_{_\text{BIAS}}}\$. That means that when the BJT base is properly biased at \$V_{_\text{B}}\$ then the current in \$R_2\$ will be \$I_{_\text{BIAS}}\$, which we argue is stiff compared the BJT's base current.

We also want \$R_1=\frac{V_{_\text{CC}}-V_{_\text{B}}}{I_{_\text{BIAS}}+\frac{I_{_\text{Q}}}{\beta}}\$. Here, the current in the denominator includes not only the bias current but also the necessary base current, too, which will be removed by the BJT (before it gets to \$R_2\$.)

Returning to \$R_{_\text{TH}}\$

From those two, you can work out \$R_{_\text{TH}}=\frac{V_{_\text{CC}}-V_{_\text{B}}}{I_{_\text{Q}}\cdot\left(\text{BIAS%}\,\cdot\,\frac{V_{_\text{CC}}}{V_{_\text{B}}}+\frac1{\beta}\right)}\$.

If \$\frac1{\beta}\$ is small enough compared to \$\text{BIAS%}\cdot\frac{V_{_\text{CC}}}{V_{_\text{B}}}\$ (which it usually is) and if \$V_{_\text{B}}\$ can be neglected when compared to \$V_{_\text{CC}}\$ (which is a judgment call) then this can be reduced to \$R_{_\text{TH}}\approx \frac{V_{_\text{B}}}{I_{_\text{BIAS}}}\$.

Here, we'd guess that \$R_{_\text{TH}}\approx 14\:\text{k}\Omega\$.

And again, this is nothing like the \$33\:\text{k}\Omega\$ given in the example.

In their case, their choice of \$33\:\text{k}\Omega\$ is equivalent to selecting \$\text{BIAS%}\approx 0.03\$. Which is only \$3\times\$ the base current. That's not very stiff and it very well may lead to different biasing points when several copies of the same circuit are later built and tested.

Summary

Well, that's it. Well, except that they made a typo:

You will find the left side written earlier and the right side, later on. They don't match up. Just be aware that all of us make mistakes like this.

And I suppose there's another lesson of sorts in all this. Algebra is an important skill to own. Keep working to improve the skill as well as maintaining it.

(*) There is an Early Effect that argues the BJT collector is not an ideal current source. But this effect matters more for AC gain calculations (and even then, the effect is only 2nd order and can often be ignored.) For DC quiescent biasing analysis, this effect is so small that it can be almost completely ignored, due to the DC negative feedback (NFB) arriving from the emitter resistor working against the BJT's base-emitter Shockley equation. (So long as the emitter resistor drops some few tenths of a volt or more, anyway.)