# How to calculate voltage divider transistor bias?

So I've been trying to learn how to calculate transistor bias, but I can't seem to find any appropriate equations. Let's say I have this (right) circuit which I can transform to the left circuit.

I know that Ucc = 10V, therefore Uce = 5V.

I chose Ic=Ie = 100mA and h21 = 100.

I can calculate Ib = 1mA (Ic/h21).

I want Ube to be 0,7V (so transistor openes).

Now if I want to get value of Uc (Rc voltage drop) to calculate Rc I would need Uc = Uce - Ue but I do not know Ue. If I try to calculate Ue = Ic * Re that doesn't work since I don't know Re.

Ue = Ub2 - Ube is the last option then.

Now, Ub2 = Rb2 * (Ib1 - Ib) - Ib1 is current at the beggining of the branch, Ib current that went into base.

The only thing left for me is to transform the 2 resistors into one, getting their resistance value together.

If I do that, I can get another equation Rb = Rb1 * Rb2 / Rb1 + Rb2 but that's it, I can't get those 2 out of it.

I heard that approximately Ue = 0,2 * Ucc and approximately Rb2 = Ube / 10 * Ib

but I would like to know how to properly calculate it with what I've been given, or how to get these approximations or at least where did these numbers came from so I can understand it and not just throw random values in.

• Why are you choosing $I_\text{C}=100\:\text{mA}$? That's rarely done for this kind of CE stage (I've never done it.) Also, this site (EESE) is packed with answers to questions just like this one, though I admit there are several approaches floating around. I think your problem regarding the search is more about not being able to recognize the answer that fits your case when you see it. Finally, it looks like you are in the "follow some recipe" phase and don't understand WHY any of the choices are made so that you can think on your own. This isn't a hard circuit, so you should learn why. – jonk Mar 8 '18 at 22:31
• Is this purely an educational question? Or do you have a known input source and a known output this circuit must drive? (Are you treating it as completely unloaded at the output and with a very low impedance input source?) – jonk Mar 8 '18 at 22:34
• My own writing, and I'm just a mere hobbyist, includes: Variation of BJT parameters on CE amplifier Q-point, Design steps for CE amplifier, Design steps for similar CE amplifier, and ... – jonk Mar 8 '18 at 22:44
• – jonk Mar 8 '18 at 22:45
• "I want Uce to be 0,7V (so transistor openes)." - do you mean Ube? (You already stated that 'therefore' Uce = 5V) – Bruce Abbott Mar 8 '18 at 23:11

Usually, you know what $V_\text{CC}$ is and you do in this case. It's $V_\text{CC}=10\:\text{V}$.

You also usually decide on the quiescent current, $I_{\text{C}_Q}$, for the CE amplifier as a matter of knowing what you have to drive with it (it rarely stands alone) and knowing something from the datasheet for the BJT. For example, let's look at this datasheet: PN2222A.

This sets a limit based on the worst case $V_\text{CE}$, which in this case might be perhaps a couple of volts less than our $V_\text{CC}$. If we want to limit the temperature rise of the BJT to a worst case of $30^\circ\text{C}$, then this means $I_{\text{C}_\text{MAX}}\approx 18 \:\text{mA}$. Looking at the following curves:

We can see a very convenient (and reasonable, perhaps) curve at $I_{\text{C}_Q}=10\:\text{mA}$. Personally, I think this is the maximum I'd accept for this device. It's enough underneath the temperature rise limitations to make me feel better and it's got a nice curve behavior in the chart, too. So let's call it that way.

(The upshot of this is that I cannot get to $I_{\text{C}_Q}=100\:\text{mA}$ with the PN2222A in a TO-92 package. However, you are perfectly free to select a different device with a different datasheet and curves and capabilities, if you feel the need for a quiescent current as high as you indicated.)

I generally want the DC point for the emitter to be at least $1\:\text{V}$ above ground, to aid temperature stability due to the emitter's AC $r_e=\frac{k\cdot T}{q\cdot I_\text{C}}$ dependence on T. More is better for this purpose. But more also cuts into the available collector voltage swing.

This is where some judgment comes into play. Requiring a lot of collector swing also implies a lot of variation in the collector current itself. And this impacts $r_e$, too. (Not just temperature, but also the collector current.) Here again, the more voltage you can afford to give to the DC quiescent point of the emitter, the better. Yes, it cuts into the collector swing... but this also means less collector current variation and therefore... more stability.

You've said nothing about the load that has to be driven here. So I'm just going to assume that there is no load, at all. (Or, not much.) So we'll just design it for the maximum swing possible given some other decisions that we've yet to conclude.

In addition to the DC emitter voltage value, there is also the minimum $V_\text{CE}$ for the BJT. I like to see at least $2\:\text{V}$ reserved here (partly because of the Early Effect and partly because it keeps the device well out of saturation.) However, how much you are willing to sacrifice depends also on what your $V_\text{CE}$ happens to be. I think I'll stick with $V_{\text{CE}_\text{MIN}}=2\:\text{V}$.

Now I can work out that the minimum collector voltage is $V_{\text{C}_\text{MIN}}=1\:\text{V}+2\:\text{V}=3\:\text{V}$. This means I have a total of $V_\text{CC}-V_{\text{C}_\text{MIN}}=7\:\text{V}$ to work with at the collector. The mid-point is half that, so the quiescent collector voltage should be set at $V_{\text{C}_\text{Q}}=1\:\text{V}+2\:\text{V}+\frac{7}{2}\:\text{V}=6.5\:\text{V}$. This gives a maximum swing of $\pm\: 3.5\:\text{V}$ at the collector.

From Figure 11 on the datasheet indicated above, I can see that $V_{\text{BE}_\text{Q}}\approx 700\:\text{mV}$ at $I_{\text{C}_Q}=10\:\text{mA}$. So $V_{\text{B}_\text{Q}}\approx 1.7\:\text{V}$.

I'll want the resistor divider current to be enough that it is stiff relative to the required base current (since the base current will vary with the signal.) This tends to mean that it is about $\frac{1}{10}$th of $I_{\text{C}_Q}$, or about $1\:\text{mA}$. So, $R_\text{B1}=\frac{10\:\text{V}-1.7\:\text{V}}{1\:\text{mA}}=8.3\:\text{k}\Omega$ and $R_\text{B2}=\frac{1.7\:\text{V}}{1\:\text{mA}-100\:\mu\text{A}}=1.889\:\text{k}\Omega$. Rounding these to nearby values, I decide to use $R_\text{B1}=8.2\:\text{k}\Omega$ and $R_\text{B2}=1.8\:\text{k}\Omega$.

At this point I also can compute $R_\text{E}=\frac{1\:\text{V}}{10\:\text{mA}+100\:\mu\text{A}}\approx 99\:\Omega$. Clearly, the nearest value is $R_\text{E}=100\:\Omega$. Similarly, $R_\text{C}=\frac{10\:\text{V}-6.5\:\text{V}}{10\:\text{mA}}=350\:\Omega$. Since I know in advance that the base current will probably be a little lower than I estimated ($100\:\mu\text{A}$), the base voltage will be a little higher and so the quiescent current probably just a little higher and so I will choose a slightly lower value, selecting the standard value of $R_\text{C}=330\:\Omega$.

This suggests a gain of about $A_V=\frac{330\:\Omega}{100\:\Omega}\approx 3.3$. (It will be a little less, perhaps more like $3.2$, because of $r_e\approx 3\:\Omega$.) If more gain than this is needed then an AC gain leg must be added to the emitter (a series resistor and capacitor to ground from the emitter.) However, your circuit doesn't allow for that. So that's your gain estimate.

If you are curious, you can compute the estimated base current using the following equation (derived from the Thevenin voltage law moving from the Thevenin equivalent at the base through to ground via the emitter and emitter resistor):

$$I_{\text{B}_\text{Q}}=\frac{V_\text{TH}-V_{\text{BE}_\text{Q}}}{R_\text{TH}+\left(\beta+1\right)\cdot R_\text{E}}=\frac{1.8\:\text{V}-700\:\text{mV}}{1476\:\Omega+101\cdot 100\:\Omega}\approx 95\:\mu\text{A}$$

Now it is your task to select a BJT and associated datasheet that is appropriate for the quiescent collector current you've selected. Keep in mind the power dissipation involved -- you won't be able to use a TO-92 packaged device here. But you can follow through with the above process, making your own choices here and there as you go. Then work out the resistor values.

• BRILLIANT answer! – Duck Sep 22 '18 at 21:35

I heard that approximately Ue = 0,2 * Ucc and approximately Rb2 = Ube / 10 * Ib

Yes, that is a commonly used design 'rule of thumb'.

but I would like to know how to properly calculate it with what I've been given, or how to get these approximations or at least where did these numbers came from so I can understand it and not just throw random values in.

Assuming you only know that Ucc = 10V and 'therefore' Uce = 5V, you cannot 'properly' calculate any resistor values unless you make some other arbitrary choices. You decided that Ic will be 100mA, but this doesn't give you Ue so you don't know what Ub will be. Therefore you cannot determine the values of Rb1 and Rb2.

You have to decide what voltage you want for Ue. For maximum output voltage swing it should be as low as possible, but this compromises stability. For maximum stability it should be as high as possible, but this reduces output voltage swing. So you must compromise.

Ue = 0.2 * Ucc is one possible choice. In this case it is reasonable because it is much larger than the Base-Emitter voltage so it provides good stability, while it (hopefully) doesn't reduce output voltage swing too much.

However it may not be the best choice. This rule was developed back in the days of Germanium transistors which had imprecise characteristics and were prone to thermal runaway. Modern silicon transistors generally have tighter tolerances and better thermal stability, so they don't need as much Emitter degeneration.

• Good discussion, I think. I wish it had more about the why's, though. What does "stability" mean, for example? (I know at least some of it, but the OP probably doesn't know any.) – jonk Mar 9 '18 at 4:57
• @BruceAbbott Thank you. Is higher resistance (higher Ue) better for thermal stabilization because higher Ic will cause bigger voltage drop at Re and due to Kirchhoff's laws in that current loop it will decrease Ube more than if Re was lower? Also higher Re meaning bigger voltage drop Ue is bad for voltage swing because Re is in the series with output voltage? – Patrick Mar 9 '18 at 11:15
• "higher Ic will cause bigger voltage drop at Re and... decrease Ube more than if Re was lower?" - Yes. "bigger voltage drop Ue is bad for voltage swing because Re is in the series with output voltage?" - yes. – Bruce Abbott Mar 9 '18 at 15:32