# General approach to solving base resistor for a CE amplifier circuit that has no voltage divider at the base

Here's the general DC equivalent for the CE amplifier I am working on:

I have looked at different CE amplifier circuit problems to help me solve this one and almost all of them have the voltage divider configuration at the base side of the transistor circuit. With this circuit, RC, Vcc, Beta, and RE have been given. RB is the resistor I need to solve for. My approach at the moment is using load line analysis. However, I am not sure if it is the correct way to solving this circuit.

The full circuit looks similar to this:

Please note that I have presented this in a general form because I just want to be steered to the right path of doing the calculations for this particular setup. I am not seeking a full answer, just a guide :)

• Homework? Given what you know of resistors and transistors, can you tell us what sets the emitter current? – TimWescott Aug 13 '19 at 14:46
• The base current sets it? – AugieJavax98 Aug 13 '19 at 14:55
• Why not use two resistors - you get a much stabler operation. – Andy aka Aug 13 '19 at 14:58
• Yes, the base current sets the emitter current. So if you know the beta, and the supply voltage, how do you set the base current? – TimWescott Aug 13 '19 at 14:58
• Try calculating the base current you need to get the $V_{out}$ you want. Then calculate the base resistor you need. Then (and this should be the point of the lesson, although it may not come out until a later class) change the beta by a factor of two and see if your circuit still works. – TimWescott Aug 13 '19 at 16:01

One of the problems with BJTs, even within a single family of parts, is that their $$\\beta\$$ may vary widely between each other and also their datasheet values. Another is that their $$\V_\text{BE}\$$ can be somewhat different, as well (variations of their saturation current.) And that's true even assuming only a single temperature of operation. Add in ambient and operating temperature vagaries, and these variations are wider still. A lot of experience and skill can go into making good choices.

Apparently, you are given $$\V_\text{CC}\$$, $$\R_\text{C}\$$, $$\\beta\$$, and the total of $$\R_\text{E}=R_{\text{E}_1}+R_{\text{E}_2}\$$. You are not told the voltage gain, $$\A_v\$$, so you don't need to worry about the details in the second schematic. That's what I understand, so far.

Since you only have to solve for a DC operating point and do NOT have to deal with any AC behavior (voltage gain, frequency range of operation, variation of the dynamic resistance ($$\r_e\$$), etc.) or input or output loading questions, then it is probably better to solve it with an eye towards better toleration of BJT and thermal variations.

With the above in mind, then:

1. Compute the largest achievable (impractical, though) value for the collector current as $$\I_{\text{C}_\text{MAX}}=\frac{V_\text{CC}}{R_\text{C}+R_\text{E}}\$$. This will help you select a BJT that is designed for collector currents at or below this value. For example, let's say that this value works out to $$\I_{\text{C}_\text{MAX}}=10\:\text{mA}\$$. Then this means that you can probably select a so-called "small signal BJT." On the other hand, if you find that $$\I_{\text{C}_\text{MAX}}=5\:\text{A}\$$, then that would be an entirely different matter and you'd need to find a very different BJT. For discussion purposes, let's assume that you can use a small signal BJT and that $$\I_{\text{C}_\text{MAX}}\le 10\:\text{mA}\$$.
2. Unless your $$\V_\text{CC}\$$ is very small, variations in $$\V_\text{BE}\$$ will be less of a problem than variations in $$\\beta\$$. So you may want to focus more on tolerating $$\\beta\$$ variations. Select a small-signal BJT part and grab its datasheet and look it over. For example, the OnSemi 2N2222A. (You are free to select something else, of course.)
3. Decide upon a temperature range you want to tolerate. Let's say $$\-10^\circ\:\text{C}\le T\le 55^\circ\:\text{C}\$$. Now look at Figure 1 in the datasheet and "guesstimate" that $$\170 \le \beta\le 270\$$. (You can round those down and up to widen the range, if you like.) I'd probably take the geometric mean of $$\\beta\approx 214\$$ for design purposes on a homework problem like this. This allows the value to rise or fall in equal proportions. In this case, about $$\\pm 25\$$%.
4. Looking now at Figure 4 in the datasheet we can see that this temperature range also means that $$\600\:\text{mV} \le V_\text{BE}\le 800\:\text{mV}\$$. Let's use the average for design purposes, or $$\V_{\text{BE}_\text{Q}}=700\:\text{mV}\$$.
5. At this point, you need to decide upon $$\V_{\text{CE}_\text{Q}}\$$ or else $$\V_{\text{E}_\text{Q}}\$$. It's a good idea to set $$\500\:\text{mV} \le V_{\text{E}_\text{Q}}\le 2\:\text{V}\$$, if possible, for thermal reasons. Larger is better. Smaller is worse. So pick a value and then compute $$\V_{\text{CE}_\text{Q}}=V_\text{CC}-V_{\text{E}_\text{Q}}\cdot\left(1+\frac{R_\text{C}}{R_\text{E}}\right)\$$ and then check to make sure that $$\V_{\text{CE}_\text{Q}}\ge 4\:\text{V}\$$. (If this value is negative, nothing will work at all. And if $$\V_\text{CC}\$$ is very low, then you may have to accept smaller values for $$\V_{\text{CE}_\text{Q}}\$$.)
6. At this point, just compute $$\R_\text{B}=R_\text{E}\cdot\left(\beta+1\right)\cdot\left[\frac{V_\text{CC}-V_{\text{BE}_\text{Q}}}{V_{\text{E}_\text{Q}}}-1\right]\$$. (And, of course, verify that it seems reasonable -- especially, not negative!)

An interesting value to compute is the sensitivity of this design. For example, what percent change in $$\V_{\text{C}_\text{Q}}\$$ would you expect for some given percent change in $$\\beta\$$ (which was estimated above as $$\\pm 25\$$%?) Or, alternatively, what percent change in $$\V_{\text{CE}_\text{Q}}\$$ would you get, then? This requires calculus, though. And so I'll avoid that bit. Still, once you have your design, you should try out the minimum and maximum estimates for $$\\beta\$$ and recompute the quiescent values for the collector and emitter voltages to make sure that it works well enough to handle those changes. You could go a little further and plug in not only those changes but also the $$\\pm\:100\:\text{mV}\$$ variation in $$\V_\text{BE}\$$ that we estimated from the datasheet, as well. If you balanced out the design sufficiently well, then you shouldn't find impossible or unworkable results.

If you need any help in developing the above equations, I'd be happy to answer specific questions there. (Also, who knows? I may have made my own mistakes above.)

## Equation Development Appendix

Terms first:

• $$\V_\text{CC}\$$ -- Single-supply's positive supply voltage. A given quantity in this problem.
• $$\R_\text{B}\$$ -- DC base resistance. The value to be computed in this problem.
• $$\R_\text{E}\$$ -- DC emitter resistance. This may either be a single
• $$\R_\text{C}\$$ -- DC collector resistance. A given quantity in this problem. resistor in the emitter leg or else it may be $$\R_\text{E}=R_{\text{E}_1}+R_{\text{E}_2}\$$. The total is a given quantity in this problem.
• $$\\beta\$$ -- The ratio of the collector current to the base current. In this problem, a value that assumes active mode operation. A given quantity in this problem.
• $$\I_{\text{C}_\text{MAX}}\$$ -- A value that specifies the maximum hypothetical collector current, given the circuit elements and topology.
• $$\I_\text{C}\$$ -- The collector current under discussion. See context for its meaning at that point in the discussion. (See also definition for $$\V_\text{BE}\$$ that follows.)
• $$\V_\text{BE}\$$ -- The base-to-emitter voltage under discussion. For an NPN, this is a positive value. This value either determines $$\I_\text{C}\$$ or else it is determined by $$\I_\text{C}\$$, depending upon the context. The relationship between the two is taken from the Shockley equation for an active mode BJT. For more details on this relationship, see: Three Equivalent Ebers-Moll Models of the BJT.
• $$\V_\text{CE}\$$ -- The collector-to-emitter voltage under discussion. For an NPN, this is a positive value.
• $$\T\$$ -- Operating temperature of the device(s). The meaning is taken from context. But for a discussion on BJTs, this is usually taken to be the operating temperature of the physical BJT (assumed uniform across it.) This temperature affects the BJT $$\\beta\$$ and also the BJT $$\V_\text{BE}\$$, among other things.
• $$\V_{\text{B}_\text{Q}}\$$ -- Quiescent base voltage (relative to ground.)
• $$\V_{\text{E}_\text{Q}}\$$ -- Quiescent emitter voltage (relative to ground.)
• $$\V_{\text{C}_\text{Q}}\$$ -- Quiescent collector voltage (relative to ground.)
• $$\V_{\text{BE}_\text{Q}}\$$ -- Quiescent base-to-emitter voltage. Or, put another way, "the quiescent base voltage relative to the quiescent emitter voltage."
• $$\V_{\text{CE}_\text{Q}}\$$ -- Quiescent collector-to-emitter voltage. Or, put another way, "the quiescent collector voltage relative to the quiescent emitter voltage."

From the above, you can see that where you see a subscripted $$\Q\$$, it refers to the quiescent value. The meaning of quiescent is "without applied signal that is actively changing circuit values." It's the DC "set-point," so to speak. (The place the circuit sits when you power it up without adding a signal.) So if I were to write, $$\I_{\text{B}_\text{Q}}\$$, even though I didn't provide a definition above for it you should know to read this as, "the quiescent base current." (But if I wrote, $$\R_{\text{C}_\text{Q}}\$$, then you should imagine that I'm joking around, or something. The collector resistor doesn't have a quiescent resistance that's any different from any of its AC operation resistance. So I would never need to write such a thing. I could. But there's never a need for it.)

Equation discussion:

• The first equation I present is from item #1 above: $$\I_{\text{C}_\text{MAX}}=\frac{V_\text{CC}}{R_\text{C}+R_\text{E}}\$$. You should be able to easily see why. Assume that the emitter-collector voltage is $$\V_\text{CE}=0\:\text{V}\$$. (This might be approximated in the case where the BJT is no longer in active mode but is, instead, saturated and operating like a switch -- abnormal for an amplifier, obviously.) If so, then the entire $$\V_\text{CC}\$$ appears across exactly two resistors: $$\R_\text{C}\$$ and $$\R_\text{E}\$$. So the equation falls out of this reasoning as an absolute maximum possible collector current.

In reality, the emitter and collector currents would NOT be equal, since some of the emitter current must be coming via the base. But we are also in the context of something that might be operating close to active mode and hopefully, therefore, the base current is a tiny, ignorable addition. Of course, in saturation, that wouldn't be as true.

If you wanted be safer and assuming active mode only, then you'd need to assign something like $$\V_\text{CE}\ge 400\:\text{mV}\$$ (still in light saturation, but now the portion of the base current wasted on the forward-biased BC junction and not available for recombination is now low enough that it can also be ignored), then the equation might be modified as: $$\I_{\text{C}_\text{MAX}}=\frac{V_\text{CC}-400\:\text{mV}}{R_\text{C}+R_\text{E}}\$$, or something similar.

I didn't want to complicate that discussion, which is why I kept the equation someone simpler, here.

• The second equation I present is from item #5 above: $$\V_{\text{CE}_\text{Q}}=V_\text{CC}-V_{\text{E}_\text{Q}}\cdot\left(1+\frac{R_\text{C}}{R_\text{E}}\right)\$$. This follows readily from the fact that if you subtract $$\V_{\text{CE}_\text{Q}}\$$ from $$\V_\text{CC}\$$, then the remaining voltage (ignoring the base current contribution, as this is in active mode) sits across the sum of $$\R_\text{C}\$$ and $$\R_\text{E}\$$ and that this result can be taken as both the collector and emitter currents.

(Again, let me emphasize that I'm simplifying by treating $$\I_\text{C}=I_\text{E}\$$, which isn't strictly true but is usefully true for these purposes.)

So we can estimate that $$\I_{\text{E}_\text{Q}}=I_{\text{C}_\text{Q}}=\frac{V_\text{CC}-V_{\text{CE}_\text{Q}}}{R_\text{C}+R_\text{E}}\$$. If we now multiply this current by $$\R_\text{E}\$$, we should easily find $$\V_{\text{E}_\text{Q}}=I_{\text{E}_\text{Q}}\cdot R_\text{E}=\frac{V_\text{CC}-V_{\text{CE}_\text{Q}}}{R_\text{C}+R_\text{E}}\cdot R_\text{E}\$$. Solving that equation for $$\V_{\text{CE}_\text{Q}}\$$ gets you the equation I provided.

• The third equation I present is from item #6 above: $$\R_\text{B}=R_\text{E}\cdot\left(\beta+1\right)\cdot\left[\frac{V_\text{CC}-V_{\text{BE}_\text{Q}}}{V_{\text{E}_\text{Q}}}-1\right]\$$. From KVL, you find $$\I_{\text{B}_\text{Q}}=\frac{V_\text{CC}-V_{\text{BE}_\text{Q}}}{R_\text{B}+R_\text{E}\cdot\left(\beta+1\right)}\$$. But also find $$\I_{\text{B}_\text{Q}}=\frac{V_\text{CC}-V_{\text{B}_\text{Q}}}{R_\text{B}}=\frac{V_\text{CC}-\left(V_{\text{E}_\text{Q}}+V_{\text{BE}_\text{Q}}\right)}{R_\text{B}}\$$. Setting these two equal to each other provides $$\\frac{V_\text{CC}-V_{\text{BE}_\text{Q}}}{R_\text{B}+R_\text{E}\cdot\left(\beta+1\right)}=\frac{V_\text{CC}-\left(V_{\text{E}_\text{Q}}+V_{\text{BE}_\text{Q}}\right)}{R_\text{B}}\$$. Solving that for $$\R_\text{B}\$$ gives the equation I provided here.

It's not always clear how to proceed. Sometimes, you have to just sit down and think about the circuit in different ways.

This isn't unlike the early problems you face with mesh (KVL) or nodal (KCL) analysis, where you have to figure out which loops are the important ones (mesh) or which nodes are unknown and whether or not some are inter-related (nodal.) There isn't a formulaic approach that always takes you efficiently from A to B. (Okay. That's technically wrong. There is a formulaic approach, though, and Spice uses it. But it can be quite complex to try and apply that method by hand.)

Usually, you supply your own creativity in simplifying what needs to be done. And this means also having an idea of what is important and what isn't. In the above, for example, I sometimes assumed that the base current was ignorable. But, of course, when it comes to working out the value of $$\R_\text{B}\$$, then I actually have to worry about $$\I_{\text{B}_\text{Q}}\$$. So it matters sometimes. Doesn't matter other times. You have to know when it is, and isn't, worth worrying over. That is, if you want to keep your analysis simple enough, but no simpler, in order to solve practical values for a problem.

• Thanks for this! I would love to see how you developed these equations. Oh and another thing, $$V_BEQ$$ is just $$V_(cc) - Transistor barrier voltage$$ right? – AugieJavax98 Aug 14 '19 at 4:25
• @AugieJavax98 I'll add the equation development and context when I get a chance (some hours from now.) And no. The $V_{\text{BE}_\text{Q}}$ is just the quiescent base-emitter voltage -- the voltage required for the quiescent collector current, $I_{\text{C}_\text{Q}}$. – jonk Aug 14 '19 at 6:44
• With this comment I like to underline the last paragraph of jonk´s very detailed answer. In electronics, no formula and no equation involving voltage, current and part parameters is correct by 100%. That means, we always have to deal with simplifications and neglects. And it is one of the most important challenges for an engineer to decide if a specific simplification for a specific task or application is allowed or not. – LvW Aug 15 '19 at 9:35

In this loop only base current is unknown. Find it. Next, apply kirchhoff's voltage law in this loop:

Use, Emitter Current=Collector Current=beta*base current. Find voltage between Collector and Emitter. If the final result satisfies that the transistor is in active region your answer is correct. Otherwise proceed with the assumption of Saturation or cutoff region!

Will you make only one circuit then customize the calculation of its base resistor and operate it at only one temperature? Never replace the transistor without re-calculating for its beta.

But if you make many of this circuit then you must measure the beta of each transistor and customize the appropriate base resistor for each one.

Or you can buy thousands of transistors, measure the beta of each one and select only the ones that have the appropriate beta.

Or use a voltage divider for the base like most circuits.

• This comment applies to the given configuration only!! And that is the big disadvantage of the configuration. In contrast, using a voltage base divider for producing the required voltage VBE (app 0.65..0.7 volts) the circuit is much...much less sensible to beta changes (nearly independent). – LvW Aug 15 '19 at 6:48