According to Wikipedia, a common emitter amplifier doesn't have big stability without an emitter resistor. When putting an emitter resistor gain is reduced. Why don't we put a base resistor in order not to lose any gain but stabilize the amplifier as well?

  • 3
    \$\begingroup\$ Because (1) that doesn't achieve the desired result (something you'd understand if you write out the equations); and, (2) the emitter resistor also reduces distortion. If the equations are rigorously derived from the circumstances, everything you need to know will be easily seen there in the equations. Mathematics is your friend. \$\endgroup\$
    – jonk
    Jul 24 '20 at 17:10
  • 1
    \$\begingroup\$ @HelenaWells you may struggle with the more advanced subjects of EE if you dont like math \$\endgroup\$
    – BeB00
    Jul 24 '20 at 17:17
  • 2
    \$\begingroup\$ @HelenaWells That's going to be a problem because you won't be able to think for yourself. That doesn't mean you won't do well. I know people who have done very well in electronics without more math than a few basic bits of algebra. I was shocked, frankly. But they learned how to use Excel, well. They also had really, really good intuition and knew how to set things up in Excel. But they were never able to take on novel circumstances without wasting oodles of time experimenting, first. I think this is something you need to address, though I also don't want to poor cold water either. \$\endgroup\$
    – jonk
    Jul 24 '20 at 17:20
  • 1
    \$\begingroup\$ @HelenaWells So, are you visually familiar with the idea of a tangent line to a curve? Do you have gut feeling for it? \$\endgroup\$
    – jonk
    Jul 24 '20 at 17:22
  • 2
    \$\begingroup\$ @HelenaWells So that's going to make answering your question especially difficult. You need to acquire a few concepts in order to understand still more concepts. So you'd need to first be taught an intuition/visual approach of some basic ideas, followed by more, one step at a time to reach a good answer to your question. I don't think anyone here has the time to write out that much, with feedback from you at each step along the way to make sure you get each early bit before proceeding to the next. You are right, this can be acquired entirely visually and without math. But laborously, I fear. \$\endgroup\$
    – jonk
    Jul 24 '20 at 17:34

A transistor amplifies current: you send current into the base-emitter junction, and an amplified current tries to flow in the collector-emitter junction. Current develops voltage when traversing resistors. So if you put a resistor down the emitter, the following happens:

  • when no current in the base flows, no current flows in the emitter too, so the voltage at the emitter is zero

  • you make current flow in the base, and the amplified current flows in the emitter. But the resistor raises the voltage of the emitter, so the voltage between base and emitter reduces, and less current enters the base.

It is a negative feedback, and this is why it stabilizes the amplifier.

  • \$\begingroup\$ OK but if we don't put any base or emitter resistor doesn't this mean Ib becomes too big the junction can't handle it? \$\endgroup\$ Jul 24 '20 at 17:28
  • \$\begingroup\$ Yes, a resistor in base is to limit the base current, and to raise the input impedance; a resistor at emitter also does that, but it also stabilizes. If your transistor is a switch, you can perhaps avoid emitter resistor; but if your transistor has to be even a little linear, you must stabilize it. \$\endgroup\$ Jul 24 '20 at 17:31
  • 1
    \$\begingroup\$ @HelenaWells Oh, you definitely have it right that a resistor is needed somewhere inserted between voltage supply rails an the PN junction of base-emitter. Otherwise, just as you say, the full voltage will be applied to the base-emitter and the current would be exponentially large (and bad.) But that doesn't mean that the resistor placed on the base side does the same thing as a resistor on the emitter side, for an amplifier circuit. It does do the same thing in terms of limiting current through the junction, though. So you see part of the issue. Just not all of it. \$\endgroup\$
    – jonk
    Jul 24 '20 at 17:31
  • \$\begingroup\$ Quote: "Current develops voltage when traversing resistors". I agree that during circuit analysis we ASSUME that this happens. But in reality (and physically spoken) this is wrong. No current without driving voltage. It is the E-field within a resistor (caused by the applied voltage) that allows current trough the devivce - NOT VICE VERSA! More than that - try to rethink your understanding of the transistor principle...don`t overestimate the role of the base current... \$\endgroup\$
    – LvW
    Jul 24 '20 at 18:34
  • \$\begingroup\$ @LvW of course you are right or, maybe, not always (how about the BEMF of a DC motor?). Anyway, when concentrating on currents like in this case, or for example when a shunt resistor is used to read current, it is correct to think in terms of current developing voltage. \$\endgroup\$ Jul 26 '20 at 15:59

It is a great challenge to explain a circuit in today's mathematical world without any mathematics. Let's try then!

Putting a base resistor means that you have connected an attenuator before an amplifier without negative feedback. The result of this cascading is a decreased overall gain with respect only to the input signal; the disturbances will not be attenuated.

Putting an emitter resistor means that you have introduced a negative feedback (the so-called "emitter degeneration"). The result of this connection between the output and input is also a decreased overall gain but with respect both to the input signal and any kind of disturbances (both they will be attenuated).

The clever trick is to amplify the input voltage without any attenuation but to attenuate only the disturbances. Such a distinction can be made on the basis of AC/DC by connecting a capacitor in parallel to the emitter resistor in the circuit of an AC amplifier. As a result, only the useful AC input voltage is amplified.

Another distinction can be made on the basis of common/differential in the circuit of a differential amplifier. As a result, only the useful differential input voltage is amplified.

A little more estravagant distinction can be made on the basis of current/voltage in the so-called "current interface". As a result, only the useful input current (proportional to the input voltage) is amplified; the voltage disturbances are not amplified.

As you can see, I was able to explain the phenomenon without using any mathematical symbols. Obviously, at this stage of understanding the basic idea, mathematics is not vital...


Try this approach.

Place 0.5 volts on the transistor base, you likely see about 0.1 milliamp.

Place 0.6 volts on the transistor base, you likely see about 1mA.

Place 0.7 volts on the transistor base, you likely see about 10 mA.

Notice every 0.1 volts increase causes about 10X more current.


NOW Insert 100 ohms between emitter and ground.

Place 0.5 volts on the transistor base, you likely see about 0.1 milliamps.

Place 0.6 volts on the transistor base, you likely see about 0.5 milliAmps.

Place 0.7 volts on the transistor base, you likely see about 1 milliAmps.

Place 0.8 volts on the transistor base, you likely see about 2 milliAmps.

Place 1.0 volts on the transistor base, you likely see about 4 milliAmps.

Notice we have linearized the behavior ---- 2X the voltage gives 2x the current, once we have about 0.05 or 0.1 volts across the resistor.

Get theee to a white_board, a transistor, and 5 volt supply, and a 100 ohm resistor. And some DVMs to measure.

Do some tinkering, and measure. Then you may be motivated to explore the use of some algebra, some trig, some other maths.


TL;DR answer:

A resistor in front of the base (as you suggested) does compensate for drifts in the applied base voltage.

However, the typically used emitter feedback resistor compensates for current changes due to varying base-emitter voltage of the transistor, which in general is the thing you want to compensate. A series resitor in the base will not provide this compensation!

Also the emitter current is much higher, in fact an amplified amout of the base current, and therefore creates much stronger feedback.

Furthermore, a high series resistor in the base path would significantly worsen the gain of the amplifier, due to the Miller effect at high frequencies.


Compensation for varying bias voltage

Consider the follwoing two configurations: Transistor schmeatics

The right version is the one you proposed. Let's run the numbers: Assumptions:

  • Constant base emitter voltage of 700 mV
  • Resistor 100 Ohm for the right schematic and 1 Ohm for the left schematic
  • Applied base voltage 0.8 V
  • base current amplification: 100

Right schematic

I'm sure you can calculate for yourself, that under the above assumptions, the base current is 1 mA and therefore the collector current is

$$I_C\bigg|_{V_{B} = 800~\mathrm{mV}} = 100~\mathrm{mA}$$.

Now let's assume a a base voltage increase of 10 mV: The base current will now be 1.1 mA which results in a collector current of:

$$I_C\bigg|_{V_{B} = 810~\mathrm{mV}} = 110~\mathrm{mA}$$.

Left Schematic

Assuming the same values for the transistor, the voltage over the resistor will be 0.1 Volt, resulting in a collector current of

$$I_C\bigg|_{V_{B}=800mV} = 100~\mathrm{mA} $$.

The same increase of 10 mV applied base voltage results in:

$$I_C\bigg|_{V_{B}=810mV} = 110~\mathrm{mA} $$.

As you can see: The effect of the resistors is the same in this case, however the one needed in the emitter path must only be one hundredth the size of the one in the base path for the same result. (the factor 1/100 comes from the current amplification of the transistor, which I assumed to be 100).

However, this is not the only instability cause, which has to be considered:

Compensation for thermal effects of the transistor

The following figure is mostly safe to assume, or at least in the right order of magnitude:

The base-emitter voltage for a given collector current drops by approximately 2 mV per Kelvin temperature rise.

2mV change in base-emitter voltage can lead to tremendous current fluctuations. Considering the right side schematic above:

Your proposed circuit (right schematic)

If the base-emitter diode becomes more conductive, the base node voltage will drop, essentially drawing more base current through the resistor and therefore also increase the collector current. This behavior is unwanted!

Working compensation (left schematic)

Consider the left schematic: If the base emitter diode becomes more conductive, the base current will raise, so will the collector current and the emitter current. This creates a higher voltage drop over the resistor, which is raising the emitter voltage of the transitor, and therefore reducing the base-emitter-voltage, which will give negative feedback to the base current and therefore limit the current through the transistor. The circuit is now compensated and will have a much lower temperature dependency than the other one.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.