I have seen different common emitter amplifier designs with either a voltage divider or a simple resistor between the battery and the base.

What would make us choose a voltage divider over a simple resistor?

  • 2
    \$\begingroup\$ And, in those circuits have you also noticed an emitter resistor or not? \$\endgroup\$
    – Andy aka
    Commented Dec 23, 2016 at 18:54
  • \$\begingroup\$ If the current gain of the same transistor type varies from say 100 to 300 what would be more sensible? - Choose a bias circuit that depends on a specific gain (single resistor) or use a bias circuit that is (almost) independent of the gain (potential divider + emitter resistor) \$\endgroup\$ Commented Dec 23, 2016 at 19:01
  • \$\begingroup\$ That's still a mystery to me. We set a bias voltage with R1 and R2 (voltage divider) and R1 (the one between base and battery) should have been enough to keep the Base-Emitter current in control (no "thermal run away"). But still we add an emitter resistor. And we also don't need an emitter resistor in addition to the collector resistor to control the collector current. So there is something critical that I'm missing about the role of emitter resistor here. \$\endgroup\$
    – Xynon
    Commented Dec 23, 2016 at 19:15
  • \$\begingroup\$ Think about the common emitter amplifier's gain formula first. If you don't place an emitter resistor which is larger enough to neglect re (you should know what this is) then the gain formula will be dependent on transistor parameter(s). So, if you want the amplifier's gain to be constant and independent from transistor's parameters then you should set the base voltage --and so the collector (bias) current constant. That's why a resistor divider takes place on the base of transistor. \$\endgroup\$ Commented Dec 23, 2016 at 19:38
  • \$\begingroup\$ The beta variations are due to thermal voltage. The more current in the emitter the higher the heat dissipation and thus internal emitter resistance is 26/Ie ohms. A resistor between base and battery would have no effect on this "thermal increase in conductivity" because it is not from the battery. Therefore, without the emitter resistor, we cannot have a stable current gain. I see that now. But how that makes a voltage divider necessary rather than a simple resistor, I can't see yet. \$\endgroup\$
    – Xynon
    Commented Dec 23, 2016 at 19:57

4 Answers 4


The use of a biasing pair, as opposed to a single resistor at the base, provides another degree of freedom. If you just use a single resistor, then the source voltage on the other side of the resistor (if cheaply done) has to be one of the existing power supply rails. There are only a few of these, or just one. But if you use a biasing pair, then you can construct any reasonably desired Thevenin source voltage as well as choosing the equivalent Thevenin source resistance, too. So you get two degrees of freedom instead of one. This is actually pretty important. And all it costs is a resistor (and some operating power.)

The emitter resistor serves a variety of purposes. (But all of them amount to yet another degree of freedom for the system.)

Together with the biasing pair's Thevenin equivalent, it can be used to set the quiescent current for the system to be relatively independent of the BJT (\$V_{BE}\$ variation still has an impact, usually, with \$\beta\$ variation also still having a [still smaller] impact. But it becomes manageable.)

Without it and without some other form of negative feedback in hand, the gain is \$\frac{R_C}{r_e}\$, where \$r_e=\frac{k T}{q I_C}\$. You can see that this gain will vary with the collector current and the collector current, obviously, varies with the signal driving it. It's also \$\beta\$-dependent. And it's also temperature dependent.

So the biasing pair provides a necessary (for cost, if no other reason) extra degree of freedom needed to perform a good design.

And coupled with the biasing pair, the emitter resistor provides a predictable quiescent current AND thermal stability. The open loop gain variations aren't quite as important, as there is more than one way of providing the negative feedback needed to bring that into control. So, the truly indispensable aspects of the emitter resistor are mostly about thermal stability and setting up a predictable quiescent operating current. The ills of open loop gain variations due to other causes can often be remedied with the use of negative feedback and this doesn't always have to be through the use of an emitter resistor. That's only one possible way of doing it.

(You will often see the BJT with an emitter resistor simply bypassed with an AC capacitor. But in these cases, you will also see negative feedback used to bring things back into control.)

In short, the larger the voltage difference allowed across \$R_E\$, the better will be the thermal control and the quiescent current stability and the use of a biasing pair provides an important additional degree of freedom for the design. These are DC considerations, not AC.


The impedance looking into the base of a BJT transistor in a common emitter amplifier is approximately B*Re.


Re = the emitter resistor.
B = the transistor current gain.

If we use just one resistor between the power supply and the base then the bias point for the base would be approximately...

Vb = Vbe + (Vcc - Vbe) * B * Re / (B*Re + Rb)


Vcc is the supply voltage.
Vbe is the base to emitter voltage (usually 0.7V)
Vb is the base voltage.
Rb is the base resistor.

As one can see the bias point depends heavily on the value of B. The problem is that B varies widely from part to part.

For example, let Vcc = 10V, Re = 100 ohm, and Rb = 10K.

For B = 100 the bias point is...

Vb = 0.7 + (10V - 0.7V) * 100 * 100 ohms / (100 * 100 ohms + 10k) = 5.35V.

But for B = 300 the bias point is...

Vb = 0.7 + (10V - 0.7V) * 300 * 100 ohms / (300 * 100 ohms + 10k) = 7.68V.

In extreme cases the bias point could shift over so far that your usable AC output signal range is too small to be usable.

Using a divider, where the divider output impedance is much smaller than the impedance looking into the base gives a much more stable bias point.

For example, suppose we have Vcc=10V and we use a divider consisting of two 1K resistors to set the bias point on the base at near mid supply...

The divider forms a thevenin equivalent circuit consisting of a 5V source in series with a 500 ohm resistor.

In this case the equation for the bias point of the base is the same as above, except that we use the 5V equivalent source voltage, and the 500 ohm divider output impedance for Rb.

Let us now analyze the bias point again for the two example cases above...

For B = 100 we have...

Vb = 0.7V + (5V- 0.7V) * 100 * 100 ohms / (100 * 100 ohms + 500) = 4.79V

For B = 300 we have...

Vb = 0.7V + (5V- 0.7V) * 300 * 100 ohms / (300 * 100 ohms + 500) = 4.93V

As you can see the divider configuration is much more stable. Even though B changed from 100 to 300 the bias point only changed by 0.14V. Whereas when using the single base resistor it changed by 2.33V



simulate this circuit – Schematic created using CircuitLab

Important design rules

  • ratio to Base bias to base input impedance from 10 to
  • Ratio of R Collector to emiter impedance for gain
  • Independance on hFE for Ve (emitter) and thus Ic
    • ( better to determine by R ratios )

First, just forget about the "necessity" of divider or resistor at base.

What do we expect from a common emitter amplifier in terms of performance?

  • Transistor-independence (for serial production, for example).
  • Temperature-independence.
  • Cheap and effective.

For 1st and 2nd, placing a resistance between emitter and GND (or negative supply line) is the way to go. I think that you got the idea behind this from other comments and answers.

We want some gain, bandwith, lower output impedance, higher input impedance, symmetrical clipping etc...

So, what do we need first? Biasing. Constant biasing: A constant quiescent collector current which is temperature- and transistor-independent.

And what do we need to constantly bias the transistor? I can hear that you're saying "A constant voltage at base".

So, what's the simplest and cheapest solution to provide a constant voltage at base? A separate supply? No! Placing a divider will do the job. Simple and effective.


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