# Emitter resistor and biasing voltage

In many articles regarding transistor biasing I see a circuit like below and I don't really get what is the point of biasing like this WITHOUT emitter resistor Re?

When emitter branch is connected straight to the ground, wouldn't the VB voltage always be around 0.7V no matter the voltage divider output? "The rest" of biasing voltage would only result in greater base current. In other words - it's like we tried to bias a diode.

With emitter resistor it's different story. VB would be then VB=0.7V+VRe - we get a steady VB voltage with any value we want (e.g. 3V). New contributor
Dawid W is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• You pose good arguments. But suppose a design requires collector voltage to swing from near zero to near +Vcc? An emitter resistor makes this impossible. – glen_geek Jan 13 at 18:37
• Please google "emitter degeneration." I believe it will answer all your questions. The emitter resistor makes the circuit better in almost every way. But it does reduce gain slightly. Since collector current depends on base current, it does not matter if the base voltage is constant. As long as base current varies, then output voltage (at collector) will also vary due to drop across load resistor. – mkeith Jan 13 at 21:27
• @glen_geek Nowadays we'd use rail-to-rail op-amps in such a situation instead, correct? Better stability. – Mast 2 days ago
• @Mast Likely so. R-to-R opamps very often use MOSfets instead of BJT's. Full output swing is an issue when supply voltages shrink - many analog circuits are now constrained by +Vcc of a few volts. – glen_geek 2 days ago

The point is to eliminate the need for one resistor. Having said that, I'm not going to argue that this is a good idea.

As you suspect, we assume that $$\V_{BE}\$$ is relatively constant, which implies that the current through R2 is relatively constant. If we know the current through R1 then we can use KCL to find the base current. Multiply by $$\h_{FE}\$$ and you have the collector current.

Using this kind of biasing requires very good knowledge of the actual value of $$\h_{FE}\$$ at the desired bias point....we usually don't have that knowledge in the real world, so adding an emitter resistor helps to compensate for changes in transistor parameters.

OP's circuit is susceptible to temperature variations, and very susceptible to transistor's current gain $$\ \beta \$$.

Below are shown two alternatives, which are somewhat less sensitive to transistor characteristics. Even so, transistors might be specified by binning $$\ \beta \$$. Doing so is usually considered not-worth-it.

Version (b) employs shunt feedback, which adds linearity, but reduces gain and reduces input impedance. It is biased more reliably than version (a). You might actually see such a circuit as version (b) in use, where maximum collector voltage-swing is an important design parameter: simulate this circuit – Schematic created using CircuitLab

• Some popular transistors are binned, eg. BC549A Hfe is 110~220, B is 200~450, and C is 420~800. – Bruce Abbott Jan 13 at 20:16
• Nice explanation... Version (b) resembles an "active Zener diode" where the Q2 collector is the "cathode" and the emitter is the "anode". Rshunt and Rc determine the "Zener voltage". It also resembles a transimpedance amplifier where the "virtual ground" appears at the base... – Circuit fantasist Jan 13 at 21:12
• So you can save the input blocking capacitor by replacing it with another resistor if the input AC source is galvanic. The network of two resistors will act as a resistor summer with weighted inputs that sums in a parallel manner VCC (VC for Version 2) and VIN... and applies the sum to the base. The price of this is less amplification. We can think of this arrangement as of a "transistor inverting amplifier" where the virtual ground appears at the base... – Circuit fantasist Jan 13 at 22:01

Just these are two different configurations - without and with negative feedback, each with its advantages and disadvantages...

The first is a special case of the common-emitter stage since its emitter is simply grounded (in the general case, the emitter of the common-emitter stage is "shifted" by another voltage). It has maximum gain and maximum output voltage swing (because a minimal voltage drop is lost on the transistor)... but it is affected by temperature variations and other interferences. Also, the voltage divider is heavily loaded by the low resistive base-emitter junction (diode).

Inserting a resistor into the emitter introduces a negative feedback (the so-called "emitter degeneration"). By passing its collector current through the resistor, the transistor itself "lifts" its emitter voltage... and it begins following the input bias voltage at the base. As a result of this opposition, the gain decreases and the output voltage swing decreases (because of the voltage drop across the transistor)... but the input resistance increases... and the voltage divider is less loaded.

The emitter resistor serves as a current-to-voltage converter for the collector current. Since the transistor keeps the voltage drop across it equal to the input bias voltage (VR2), the combination of these two elements acts as the opposite *voltage-to-current converter". It sinks a relatively constant current, which is slightly dependent on temperature changes, beta, etc...

When emitter branch is connected straight to the ground, wouldn't the VB voltage always be around 0.7V no matter the voltage divider output?

It won't get much above 0.7V, but as bias voltage is reduced the Base current decreases so it has less effect. If current going through the divider is much larger than the Base current it will set the bias voltage fairly accurately.

This is not a good biasing scheme for a linear amplifier because it lowers the source impedance which makes the relationship between Base current and source voltage more logarithmic than linear, and it makes the bias point more temperature sensitive.

However it is useful in situations where a well defined bias voltage is required, eg. in a switching circuit, or for peak detection of an AC waveform. It also makes the input less prone to 'blocking' caused by the Base rectifying large AC signals and building up a negative bias on the coupling capacitor.

it's like we tried to bias a diode.

Yes, but sometimes that's what you want.

• So, we can talk about a "stiff" and "soft" voltage divider... Besides the voltage divider viewpoint we can look at the input biasing circuit from the perspective of a "current divider" where R1 and VCC serve as a current source... and R2 and the base-emitter junction form the current divider. So R2 diverts the excessive base current. – Circuit fantasist Jan 13 at 21:30

R2 doesn't really do much of value in this circuit.

For it to act as an amplifier, the base is biased with a current from the supply to result in the transistor being in the active mode. So if you have a transistor like the common 2SC2712Y you know the hFE is between 120 and 240 @25°C so you can calculate the required resistor for a given load resistance. BCX70H is even tighter specified (180-310 @2mA 25°C).

If you use the latter, with a 10V supply, R2 = open and RL = 2.5K then R1 would be around 1.1M. Worst case (at 25°C) the DC collector voltage would be between 6.15V and 3.37V, so it could work with output amplitudes up to around 6Vp-p before clipping. Less than that over a range of temperatures and taking tolerances of resistors into account.

R2 diverts some of that base current to ground and adds more uncertainty to the resulting base current, as well as reducing the input impedance, none of which you probably want.

Others have made relevant comments about the advisability of including a bypassed emitter resistor to add negative feedback to the DC operating point, which are valid enough where that is necessary (very often it is, especially if you have high output levels, wide temperature range). When you do that, you usually want R2 to better control the voltage on the base, and the resulting emitter current. Of course you lose the voltage across the emitter resistor/capacitor (maybe a volt or two) from the maximum output swing so it's not all gain- in fact it may be worse at room temperature, but you gain temperature stability at the expense of complexity and parts cost. Another approach is to derive base current from the collector voltage and lose the emitter resistor (again, you need a bypass capacitor to go along with the resistors or you excessively degrade your AC gain).

Not shown in your circuits is the input coupling capacitor.

• Just a simple idea - you can insert the input AC (galvanic) source between the lower end of R2 and ground. Then R1 and R2 will act as a resistor summer with weighted inputs that sums in a parallel manner VCC (through R1) and VIN (through R2)... and applies the sum to the base. The price of this is less amplification. We can apply this trick to the Version (b) of @glen_geek above. – Circuit fantasist Jan 13 at 21:42

Vbe isn't actually a constant 0.7V. It's just mostly constant. It varies very slightly with current, just like a diode does, and this is fundamental to its operation (as in it actually needs this behaviour to operate the way you expect it to). As an approximation to make certain forms of analysis easier, we say 0.7V is constant with respect to base current, but since that is an approximation it doesn't always hold up such as in the case of your reasoning.

• Can you explain your statement "it actually needs this behaviour to operate..."? – Elliot Alderson Jan 13 at 21:22
• @ElliotAlderson I don't know enough to really get into it except just to say that the relationship between the relationship between the behaviour of the collector-emitter and the base-emitter of the BJT depends on these small voltages changes. It's not just some parasitic behaviour where if you somehow got rid of it the BJT would still continue to function. – DKNguyen Jan 13 at 21:42
• So we can think of diodes as of a kind of "dynamic resistors"... "voltage-stabilizing nonlinear resistors"... or simply "voltage stabilizers"... – Circuit fantasist Jan 13 at 22:09
• @Circuitfantasist I wouldn't try to make that relationship simply because it's too easy to be led astray. It's simple enough to think of diodes as what they are without throwing the term "resistor" into the mix: components with a nonlinear voltage to current relationship. – DKNguyen Jan 13 at 22:10
• @ DKNguyen, Then why they model the forward-biased diode by a voltage source? Or the transistor by a current source? Is this more true? I think it is more correct to define at the very beginning these two types of elements... to add them to the library of elements... and then model the diodes and transistors through them... rather than through sources... what they are not... – Circuit fantasist Jan 13 at 22:22

I've read all the answers (surprised to receive so many, thank You all) and I feel like I need to specify my way of thinking.

In my understanding, biasing allows us - among others - to amplify an AC wave by setting the transistor at a certain operating point.

Suppose we have an AC input signal with maximal value 1Vpp (+-1V). Now, to avoid clipping we need to set the VB at around 1.8V - 2V in order to prevent VB dropping below ~0.7V. We will not achieve this WITHOUT RE resistor.

In other words, circuit I presented in the first post may serve ONLY for really small signals (maybe dozens of mV). In case of bigger amplitude the voltage divider will do no good since it simply can't force Base Voltage high enough to cover the whole range of an input signal - it always be (the VB) at around 0.7V and all we will get clipped output.

Am I right?

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Dawid W is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Don't confuse $V_B$ (the voltage at the base with respect to ground) and $V_{BE}$ (the voltage from emitter to base). Without $R_E$ they are the same but if you add $R_E$ then $V_B$ will be significantly greater than $V_{BE}$. It is $V_{BE}$, not $V_B$, that is going to be about 0.7V when the transistor is active. – Elliot Alderson 2 days ago
• "Don't confuse VB (the voltage at the base with respect to ground) and VBE (the voltage from emitter to base)." I do not, that's the whole point. Without Re they both the same which in particular cases means that voltage divider is useless and doesn't actually BIAS anything since VB (with respect to ground) still sits at ~0,7V. All we get is the logarithmic changes of Base Current. – Dawid W 2 days ago
• "All we get is...changes of Base Current." But that is exactly what we want! We want to set the base current at an appropriate bias point, and if we know the transistor parameters, and those parameters are relatively constant, then we can do that without an emitter resistor. – Elliot Alderson 2 days ago
• You said "we need to set the VB at around 1.8V - 2V in order to prevent VB dropping below ~0.7V". That doesn't make sense. If you are talking about a circuit without $R_E$ then you can not set VB to 1.8V. If you are talking about a circuit with $R_E$ then it is VBE that is 0.7V, not VB. – Elliot Alderson 2 days ago
• 'If you are talking about a circuit without RE then you can not set VB to 1.8V. " I wrote that exactly: "We will not achieve this WITHOUT RE resistor." Circuit without Re - VB=VBE; Circuit with Re - VB=VRe+VBE **** My final question - Is it true, that without Re we cannot amplify higher magnitudes? – Dawid W yesterday