9
\$\begingroup\$

In the circuit below, why there is a resistor R2? What would be the difference if it wasn't there?

schematic

simulate this circuit – Schematic created using CircuitLab

I am not in school. This is not homework. The circuit comes from this source.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Do you know what a resistor divider is? It works in tandem with R1 as a divider (with some distortion since all that other stuff is also hanging off it). Not on its own. Unlike a divider the goal here is not necessarily a voltage but a current. \$\endgroup\$
    – DKNguyen
    Jan 8 at 16:04

7 Answers 7

1
\$\begingroup\$

Short answer

The problem is conceptual and not solely related to the operation of this particular transistor stage. So, it can be explained by simple equivalent electrical circuits.

The question is actually, "Can we set the voltage across a load (the input of the transistor stage here) with only one resistor connected in series to the power supply?”

The answer is, "Yes, it can but if the load has not very high (input) resistance". If its resistance is too high, then the resistance of our resistor must also be too high, and this is unacceptable (because of the leakages, parasitic capacitances, etc.)

Therefore, in such cases, we artificially reduce the load resistance by connecting a resistor in parallel with it; this allows us to reduce the resistance of our resistor. In other words, we make a voltage divider through the two resistors.

CircuitLab experiments

Let's now test these reasonings "in practice" through CircuitLab experiments. They trace the evolution of the idea step by step; that is why they are drawn in the same way. In order to simplify circuit diagrams, in some places I have combined resistors with meters ("visualized resistors"). I have assumed that the task is to provide 5 V across the load (half of the supply voltage). For convenience, I have tried to make the values of the quantities in these conceptual circuits to be multiples of 10.

1.Real high-resistance load

RL = infinity (open circuit): If there is no load connected, we cannot reduce the Vcc voltage, no matter what resistance Rb we set. The reason is clear - no current flows, so there is no voltage drop across the resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

RL = 100 MΩ: If the load is too high (eg 100 MΩ) we are forced to also increase Rb (to 100 MΩ) to decrease the load voltage to 5 V.

schematic

simulate this circuit

RL = 100 MΩ; Rleak = 100 MΩ to ground... However, if there is leakage to ground, the voltage drops by more than a volt.

schematic

simulate this circuit

... with Rleak to Vcc: If there is leakage to Vcc, the voltage rises by more than a volt.

schematic

simulate this circuit

Shunting resistor Rb2 (voltage divider)... Then let's artificially reduce the load resistance by connecting a 10 kΩ resistor Rb2 in parallel with it; this allows us to reduce the resistance Rb1. Thus we actually make a low-resistance voltage divider through Rb1 and Rb2.

schematic

simulate this circuit

... with Rleak to ground... Now the leakage to ground...

schematic

simulate this circuit

... with Rleak to Vcc ... or to Vcc does not affect the load voltage.

schematic

simulate this circuit

2.Virtual high-resistance load

The OP's circuit of common-emitter amplifier with emitter resistor has the same very high input resistance but its nature is different. This is not actually a resistance but an equivalent opposing voltage.

Behavioral voltage source: We can simulate it with a behavioral voltage source that follows the voltage on the left side of the ammeter. I have used a little trick by setting a small resistance (100 Ω) to the ammeter with the purpose to decouple the "copying" voltage source from the original voltage followed. As you can see, Rb has to be extremely high (1 GΩ) to get the 5 V desired voltage. By the way, this arrangement represents the famous "bootstrapping" technique for virtually decreasing resistance.

schematic

simulate this circuit

Simulated transistor: In the OP's stage, the transistor creates the "opposing voltage" by passing its collector current through the emitter resistor. Let's furst simulate the transistor by a voltage-controlled current source (transconductor) Ic. It is controlled by the voltage drop across the ammeter (the "base-emitter junction").

schematic

simulate this circuit

As you can see, the result is the same - need for a very high resistance Rb.

Real transistor... Finally, let's explore the real transistor stage. I have connected a voltmeter with 10 kΩ internal resistance as an emitter resistor Re.

... with base resistor... Now we need a 1.3 MΩ base resistor (still too high) to set a 5 V emitter voltage.

schematic

simulate this circuit

... with voltage divider: Here is a version with a voltage divider made up of moderately high resistance resistors.

schematic

simulate this circuit

The bottom line is that we need to drive high resistance loads through a low-resistance voltage divider.

3.Real low-resistance load

We only use one resistor Rb when the load has a low resistance (RL=100 Ω), and we actually control the current through it.

schematic

simulate this circuit

4.Virtual low-resistance load

So we can control a common-emitter stage in this way if the emitter is "fixed", eg connected to ground. The input base-emitter resistance is another example of "virtual resistance" but its nature is different. This is dynamically reduced resistance without the aid of a voltage source - when the voltage increases, it decreases its resistance and conversely, when the voltage decreases, it increases its resistance. Thus the voltage across this "dynamic resistor" is constant and the current is not affected.

schematic

simulate this circuit

The bottom line here is that we need to drive low resistance loads through a high-resistance resistor.

\$\endgroup\$
11
\$\begingroup\$

Here's all you need to know at your stage, the beginning hobby stage, of understanding transistors. There's plenty here that's not strictly correct, but you need to be an electronic engineer and designing professional equipment to worry about the detail.

  • VBE is about 0.6 or 0.7 V, for hobby-useful values of collector current, like the 1 uA to 10 mA sort of range for 'small signal' transistors.
  • The base current is 1% or less of the collector current.
  • R1/2 form a potential divider, to provide a fixed voltage at the base.
  • The emitter voltage will be 650 mV below the base voltage.
  • RE is only 1k. As it has roughly the same voltage across it as R2 has, this means the R2 current is roughly 1/10th of the collector/emitter current.
  • As the current needed in the base it at most 1/100th of the collector current, the base is drawing negligible current from the R1/2 potential divider, and so not significantly disturbing its output voltage.
  • The supply voltage is now dropped 1/3rd on RE, so 1/3rd on RC, so leaving 1/3rd across the transistor.

This is the classic low power 'linear amplifier' circuit, where now coupling an AC signal to the base via a capacitor results in an amplified version appearing at the collector. In this case, the amplification gain is -1 (-RC/RE). Typically, RC would be bigger than RE. Alternatively (and more commonly) part of RE would be bypassed by another resistor in series with a capacitor, so its AC value was much lower than RC, while the DC value stayed high for stable DC bias.

Without R2, R1 would be a current source into the base

  • The transistor would turn on, dropping half the supply voltage across RE, and half across RC, leaving only VCEsat (typically 100-400 mV) across the transistor.
\$\endgroup\$
2
  • \$\begingroup\$ Does second bullet mean: (a) The base current is at most (a factor of 100 less than the collector current) or (b) The base current is (a factor which is at most 100) less than the collector current? \$\endgroup\$
    – Ben Voigt
    Jan 9 at 21:26
  • \$\begingroup\$ @BenVoigt reworded, it was clumsy. \$\endgroup\$
    – Neil_UK
    Jan 10 at 6:52
7
\$\begingroup\$

The resistors on the left form a voltage divider used to bias the BJT in the desired operating point: they set the base voltage to provide the required base current. In analog applications, it is usually desirable to put the device in the forward-active region (which in this case requires that the base voltage be greater than ~0.7 V and less than the collector voltage), but this depends on what you are trying to accomplish.

Assuming the value of \$R_1\$ stated in the diagram, if you replace \$R_2\$ by an open circuit (i.e. set \$R_2 = \infty\$), the base voltage gets pulled high enough to put the BJT in the saturation region. If you replace \$R_2\$ by a short to ground (set \$R_2 = 0\$), the base-emitter junction is no longer forward-biased so the BJT is turned off.

\$\endgroup\$
12
  • 1
    \$\begingroup\$ Does that mean the two resistors together "slow down" the voltage or current enough so things work? \$\endgroup\$
    – johnny
    Jan 7 at 21:10
  • \$\begingroup\$ @johnny I would advise against referring to a "speed" of voltage or current, but it is fair to say that \$R_1\$ and \$R_2\$ work together to provide the desired base voltage/current for correct operation. \$\endgroup\$
    – Puk
    Jan 7 at 21:15
  • \$\begingroup\$ If I were trying to read this or design something, would this be where I would think I need x voltage for the device, so I need a voltage drop to be y across or between these two resistors? @Puk \$\endgroup\$
    – johnny
    Jan 7 at 21:50
  • \$\begingroup\$ An application will either dictate one or more of the resistor values, or will require you to choose their values to meet certain specifications. \$R_1\$ and \$R_2\$ influence the base voltage (\$R_E\$ has a weak effect too), but a design will generally have additional requirements to consider like currents, power consumption and sensitivity to BJT \$\beta\$, all of which depend on \$R_E\$ as well. \$\endgroup\$
    – Puk
    Jan 7 at 22:40
  • 1
    \$\begingroup\$ @Johnny The way I started thinking about things like this is that resistors between a point and ground push the voltage at that point away from ground, in the same direction as whatever voltage is "above" ground. In other words, since there's a positive voltage at the top of the diagram then R2 is pushing the voltage of the point between R2 and R1 to a voltage higher than ground. Not sure if that's helpful but it worked for me. \$\endgroup\$ Jan 8 at 5:50
4
\$\begingroup\$

R2 makes voltage divider with R1 and because of it we can provide a more stable input to the bjt at the desired operating point.

\$\endgroup\$
3
\$\begingroup\$

straight to the heart of the question

The biasing pair, \$R_1\$ and \$R_2\$, manages the quiescent operating point against expected variations of BJT \$\beta\$, by allowing an independent (derived) Thevenin voltage and Thevenin source resistance to be set for the BJT base. To understand this well, a sensitivity analysis must be performed for each of the two different circuits, one with and one without \$R_2\$.

schematic

simulate this circuit – Schematic created using CircuitLab

Let's make the above into a concrete example. \$V_{_\text{CC}}=5\:\text{V}\$, \$I_{_{\text{C}_\text{Q}}}=2\:\text{mA}\$, \$V_{_{\text{E}_\text{Q}}}=500\:\text{mV}\$, \$V_{_{\text{C}_\text{Q}}}=2\:\text{V}\$, and assume \$V_{_\text{BE}}=700\:\text{mV}\$ and \$\beta=150\$ for a small signal BJT. Given the \$\beta\$ and for simplicity's sake, let's assume \$I_{_{\text{E}_\text{Q}}}=I_{_{\text{C}_\text{Q}}}\$. Then \$R_{_\text{E}}=250\:\Omega\$, \$R_{_\text{C}}=1.5\:\text{k}\Omega\$.

With stiff biasing for the left, I'd get \$R_{_{\text{B}_1}}=17.8\:\text{k}\Omega\$ and \$R_{_{\text{B}_2}}=6\:\text{k}\Omega\$. (Here, \$V_{_\text{TH}}=1.2605\:\text{V}\$ and \$R_{_\text{TH}}=4.4874\:\text{k}\Omega\$.) For the right, I'd get \$R_{_\text{B}}=285\:\text{k}\Omega\$.

To find out how a %-change in one thing changes (\$y\$) with respect to a %-change in another thing (\$x\$), you use the sensitivity equation: \$S^y_x=\frac{\frac{\partial\,y}{y}}{\frac{\partial\,x}{x}}=\frac{\partial\,y}{\partial\,x}\cdot\frac{x}{y}\$. This can then be applied, as in: \$\frac{\partial\,y}{y}=S^y_x\cdot \frac{\partial\,x}{x}\$.

Let's compare:

$$\begin{align*} &\quad\quad\quad\text{LEFT} && \quad\quad\quad\text{RIGHT}\\\\ I_{_{\text{C}_\text{Q}}}&=\frac{V_{_\text{TH}}-V_{_\text{BE}}}{\frac1{\beta}\,\cdot\, R_{_\text{TH}}+\frac{\beta+1}{\beta}\,\cdot\,R_{_\text{E}}} & I_{_{\text{C}_\text{Q}}}&=\frac{V_{_\text{CC}}-V_{_\text{BE}}}{\frac1{\beta}\,\cdot\, R_{_\text{B}}+\frac{\beta+1}{\beta}\,\cdot\,R_{_\text{E}}}\tag{DC} \\\\ S^{I_{_{\text{C}_\text{Q}}}}_{\beta}&=\frac{R_{_\text{TH}}+R_{_\text{E}}}{R_{_\text{TH}}+\left(\beta+1\right)\,\cdot\,R_{_\text{E}}} & S^{I_{_{\text{C}_\text{Q}}}}_{\beta}&=\frac{R_{_\text{B}}+R_{_\text{E}}}{R_{_\text{B}}+\left(\beta+1\right)\,\cdot\,R_{_\text{E}}}\tag{%$\beta$} \end{align*}$$

Suppose that \$\beta\$ can vary by 50%, one to another, when pulled from a bag.

If \$\beta\$ varies by 50% then the quiescent current for the left-hand circuit can vary by 5.61% while the right-hand circuit would vary by 44.2%. 44.2% isn't a good thing. On this \$5\:\text{V}\$ system that could mean that \$674\:\text{mV}\le V_{_{\text{C}_\text{Q}}}\le 3.326\:\text{V}\$, which is not well managed. 5% by comparison means \$1.83\:\text{V}\le V_{_{\text{C}_\text{Q}}}\le 2.17\:\text{V}\$ which, of course, is fine.

So that's why.

summary

Putting the above into words, the base current depends entirely upon the BJT's \$\beta\$. So if the \$\beta\$ varies a lot, then so does the base current, with it. (Obvious!) But the base current must arrive after coming through whatever source resistance is supplying it from an ideal voltage source. So variations in that base current, due to device variations, are multiplied by the source resistance.

If \$\beta\$ variation is a concern (and it is, not just for device variations but also for operating temperature variations on \$\beta\$) then one of the objectives is to lower this source resistance in order to mitigate the impact of \$\beta\$ variations on the circuit's operating point.

surrounding thoughts

Gaining some stability vs device and temperature variations has its costs.

The main cost is how the left-hand circuit will load down whatever drives it. The right-hand schematic presents less of a load, which is a good thing. But its quiescent operating point wiggles around a lot with varying BJT \$\beta\$. So a compromise is made.

Another cost is that the sensitivity with respect to \$V_{_\text{BE}}\$ will be higher in the left-hand schematic. This is because there is a greater voltage difference across the source resistance on the right side and that works in its favor. But \$V_{_\text{BE}}\$'s don't vary by 50%, either. Perhaps a tenth that much? So it's less of a concern.

One can have their cake and eat it, too, getting the best of all worlds. Not only taking advantage of the high input impedance of the right schematic but also becoming essentially immune to BJT variations with respect to both \$\beta\$ and \$V_{_\text{BE}}\$, while also perfecting the predictability of the resulting voltage gain of the circuit. And, to top it off, achieve immunity against operating temperature! All this and still more. But it requires more than one BJT to get there.

\$\endgroup\$
2
\$\begingroup\$

given that \$\beta_{dc} \$ is known then yes R2 is not needed, R1 could be increased to compensate.

In the real world \$\beta_{dc}\$ in generally not precisely known (and varies due to temperature and aging) and so a pair of resistors are used instead

\$\endgroup\$
1
\$\begingroup\$

Short answer:

The key to the answer is the fact that the BJT is a voltage-controlled device: Ic=f(Vbe). (For a current-controlled device we need a bias current and for a voltage controlled device we ned a bias voltage).

For the purpose of providing a relatively "stiff" voltage Vb to the base node we are using a voltage divider which allows a current Io through the grounded resistor R2 which is much larger than the expected base current Ib (which goes through R1 and, thus, has an influence on the voltage Vb)

This is important because the base current Ib has very large tolerances (due to beta uncertainties). But when the current Io through the divider chain is much larger (rule of thumb: At least factor 10), these tolerances have only a minor influence on the base voltage Vb.

Rough analysis for the shown circuit:

Current through R1-R2: Io=30/32k=0.94mA

Base voltage Vb=30(10/32)=9.4V and emitter voltage Ve=9.4-0.7)=8.7V

Emitter current: Ie=8.7/1k=8.7mA.

Estimated base current (beta=100): Ib=0.087mA=87µA.

Factor Io/Ib=0.94/0.087=10.8

Because the base current Ib is app. 10 times lower than the current Io through R2, any uncertainty/tolerance of beta (resp Ib) will have a minor influence on the DC bias conditions. With other words: The divider R1-R2 provides a "good/stiff" base voltage Vb - nearly independent on beta-variations.

Together with the emitter voltage Ve=IeRe this stable voltage Vb allows current-controlled voltage feedback which further stabilizes Vbe=Vb-Ve and, thus, the whole DC bias point.

(Sorry, my answer is not as short as announced).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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