Does collector feedback bias work like this in real life?

Question

I am reading about collector feedback biasing in BJT transistors.

Here is how it works:

Now suppose for some reason, a temperature change perhaps, β increases. This should cause an increase in IC. An increase in IC, though, would cause an increase in the drop across RC due to Ohm's law. This, in turn, would force VC to drop.

Here is the key: VC is also equal to the drop across RB plus the voltage VBE. The base-emitter potential is fixed at approximately 0.7 volts so any decrease in VC is reflected as a decrease in voltage across RB. By Ohm's law, that means that IB must decrease by a similar proportion. This decrease tends to offset the initial tendency of the collector current to increase.

Does this circuit work sequentially in real life as above described above?

When beta increases the collector current, the base current, Vce changes instantly so I can't imagine how it follows as above.

There is one error in your "sequence of events": It is not beta which is temperature sensitive. It is the collector current Ic - better: the saturation current Is in the equation Ic=Is[exp(Vbe/Vt)-1]. But the result is as described by you. Any temperature-caused Ic increase will reduce the collector potential which drives the curent Ib into the transistor (negative DC feedback). However - when used as a signal amplifier, the signal must be coupled to the base through a capacitor (and an additional resistor if signal feedback is desired). — LvW, Oct 03 '22 at 15:12
You've got the right idea - this feedback bias resistor \$R_B\$ does stabilize the DC operating point. But be aware that due to negative feedback of \$R_B\$, input impedance for AC signals between *base* and *grounded-emitter* is now lower than \$h_{ie}\$. — glen_geek, Oct 03 '22 at 17:06
Maybe it's obvious, but one thing to remember that sequential stories are just ways to help us understand behavior. Ultimately, as the beta changes with temperature, the circuit essentially solves the KVL and KCL equations at once. The sequence of events helps understand how those solutions would look like without solving the equations directly. But it all happens essentially at once. Same goes for feedback in Op Amps. Things can happen more sequentially with capacitors taking time to charge or discharge, but in such a case as the above it's neglegible.. — ee_student, Oct 03 '22 at 20:43
ee_student, study the slew rate effect for opamps and you will see that there is a clear sequence of events. — LvW, Oct 04 '22 at 10:16
Sure, and even without Op Amps transistors have small capacitance. But realistically, when you apply a DC voltage to an Op Amp, it's true that a 30pF capacitor doesn't become an open circuit instanteously, but in the same time would you agree that it's more helpful to look at it as an open circuit and assume the DC steady state is obtained in one go (barring large capacitors elsewhere). The story of the feedback is helpful to see the trend, but it's misleading to think that it's more than a story as the circuit assumes the KVL\KCL clearing solution essentially at once. — ee_student, Oct 04 '22 at 11:03
Sorry one more thought: in many circuit simulators transistor don't have parasitics, so if you build an Op Amp with negative feedback, it will clear in one time step of the simulator - at once. (Unless you add capacitors yourself). I see your point too, but from feedback stories, some people could literally think the circuit is doing a bunch of things, when all it ever really does is clear KVL and KCL. And without parasitics (or oscillation preventing caps due to parasitic phase shift), the clearing would be literally instantaneous. — ee_student, Oct 04 '22 at 11:08
All I wanted to say is that the slew rate effect is based on and can ONLY be explained using two timely succeeding effects (overdrive and delayed fedback). — LvW, Oct 04 '22 at 13:06

jonk · Answer 1 · 2022-10-04T08:06:39.450

Objective

Let's say that \$I_{_\text{C}}\$ is the DC operating point value you want to evaluate for its stability against certain changes. And let's say that temperature is one of these and that device variations are another of these.

Detailing the objective

The first thing you need to do is narrow down what temperature affects and, if possible, how it affects those things. Temperature will impact both \$\beta\$ and also \$V_{_\text{BE}}\$.

There are complex equations, depending upon something called the Boltzmann factor, that impact the saturation current model parameter, which itself impacts \$V_{_\text{BE}}\$. The range seems to be about from \$-1.8\:\text{m}\:\frac{\text{V}}{^\circ\text{C}}\$ to about \$-2.4\:\text{m}\:\frac{\text{V}}{^\circ\text{C}}\$ But in general, we can say that \$V_{_\text{BE}}\$ varies by about \$-2\:\text{m}\:\frac{\text{V}}{^\circ \text{C}}\$ near room temps. This works out to about a \$- 1\% \$ change for each \$+ 1\% \$ change in temperature.

\$\beta\$ will be a different thing. It increases with increases in temperature. But let's just assume for now that \$\beta\$ changes by \$+ 1\% \$ for each \$+ 1\% \$ change in temperature.

The above doesn't account for changes in the Early Effect and a variety of other impacts. But we can assume those away, for now.

Capturing the boundaries

So, for a \$\pm 30\:^\circ\text{C}\$ temperature range we can expect to see \$V_{_\text{BE}}\$ varies about \$\pm 10\% \$, but inversely so, and \$\beta\$ varies also about \$\pm 10\% \$, but in the same direction.

As regards device variations, let's say that \$V_{_\text{BE}}\$ varies about \$\pm 10\% \$ and that \$\beta\$ varies about \$\pm 50\% \$.

Let's assume all these elements are orthogonal to each other and apply RMS to them to get a \$V_{_\text{BE}}\$ variation of about \$\pm 15\% \$ and a \$\beta\$ variation of about \$\pm 50\% \$.

This gives us some magnitudes to work with for the circuit.

Quantitative circuit analysis: \$I_{_\text{C}}\$

Oh. Almost forgot. What about the circuit? I guess we need some formulas!

If you analyze your circuit using KCL and KVL you will find that:

$$I_{_\text{C}}=\beta\cdot\frac{V_{_\text{CC}}-V_{_\text{BE}}}{R_{_\text{B}}+\left(\beta+1\right)R_{_\text{C}}}$$

To see how, read the following SymPy/Python code:

var( 'rb rc vb vc ib ic beta ie vbe vcc' )
eq1 = Eq( vc/rc + vc/rb + ic, vb/rb + vcc/rc )    # KCL collector node
eq2 = Eq( vb/rb + ib, vc/rb )                     # KCL base node
eq3 = Eq( vcc - rc*(ic+ib) - rb*ib - vbe, 0 )     # KVL through resistors to ground
eq4 = Eq( ic, beta*ib )                           # BJT
eq5 = Eq( ie, ic + ib )                           # BJT
ans = solve( [ eq1, eq2, eq3, eq4, eq5 ], [ vc, vb, ic, ib, ie ] )

ans[ic]
-beta*(vbe - vcc)/(beta*rc + rb + rc)

pprint( ans[ic] )
-β⋅(vbe - vcc)
───────────────
 β⋅rc + rb + rc

Sensitivity

A sensitivity equation is used to relate the %-change of one thing to the %-change of another thing. This looks like: \$\frac{\frac{\text{d}\, y}{y}}{\frac{\text{d}\, x}{x}}=\frac{\text{d}\, y}{\text{d}\, x}\cdot\frac{x}{y}\$, which gives you the %-change of y as compared to the %-change of x.

In this case, I find:

$$\begin{align*} \left|\%\,I_{_\text{C}}\right|&=\%\,V_{_\text{BE}}\cdot\frac{V_{_\text{BE}}}{V_{_\text{CC}}-V_{_\text{BE}}} \\\\ \left|\%\,I_{_\text{C}}\right|&=\%\,\beta\cdot\frac{R_{_\text{B}}+R_{_\text{C}}}{R_{_\text{B}}+\left(\beta+1\right)R_{_\text{C}}} \end{align*}$$

The SymPy code looks like this:

simplify( derivative( ans[ic], vbe ) * vbe / ans[ic] )
vbe/(vbe - vcc)

simplify( derivative( ans[ic], beta ) * beta / ans[ic] )
(rb + rc)/(beta*rc + rb + rc)

Getting more specific

So, let's assume \$I_{_\text{C}}=2\:\text{mA}\$ and \$V_{_\text{CC}}=10\:\text{V}\$. Assuming also a design value of \$\beta=200\$ and \$V_{_\text{BE}}=700\:\text{mV}\$ and a quiescent collector operating point near \$3.4\:\text{V}\$, we might then specify \$R_{_\text{C}}=3.3\:\text{k}\Omega\$ and \$R_{_\text{B}}=270\:\text{k}\Omega\$. (Recomputing, we'd then find \$I_{_\text{C}}=1.99\:\text{mA}\$. Close enough.)

From these values and the above %-change estimates, we'd say that the collector current might change by about \$\pm\,14.6\%\$ over part and temperature variations of \$\beta\$ and by about \$\pm\,1.1\%\$ over part and temperature variations of \$V_{_\text{BE}}\$.

The SymPy code looks like:

abs( simplify( derivative( ans[ic], beta ) * beta / ans[ic] ).subs(
    { beta:200, rb:270e3, rc:3.3e3 }
) * 50 )
14.6415943426551

abs( simplify( derivative( ans[ic], vbe ) * vbe / ans[ic] ).subs(
    { beta:200, rb:270e3, rc:3.3e3, vcc:10, vbe:.7 }
) * 15 )
1.12903225806452

If the argument so far has been soundly reasoned, then it's now obvious that \$\beta\$ variation swamps out \$V_{_\text{BE}}\$ variation and you should figure that there will be as much as \$\pm\,15\:\%\$ variation in the operating point over temperature and device variations.

(You could remove the orthogonality assumption I made and argue for still worse. But this gives an idea about an approach to quantify variations you care about.)

This particular circuit topology

This circuit topology is never used in isolation where a specific voltage gain is required, because the voltage gain is highly dependent upon the collector current and because it is also highly dependent upon the signal, itself. So this is used to get the highest gain possible from a BJT, with the expectation of very high distortion. It is therefore always wrapped with negative feedback by placing it within a larger circuit context to remove distortion and set the desired voltage gain.

The analysis so far tells us something useful about this topology.

To get the highest possible voltage gain from this circuit, \$R_{_\text{C}}\$ should be as large as possible without forcing the BJT into deep saturation. But we must allow for collector current variation of about \$15\,\%\$. We might have been tempted to set \$R_{_\text{C}}=3.9\:\text{k}\Omega\$ to get better voltage gain. But then we'd find a quiescent collector voltage that could be under one volt. And that's without any signal applied! So, with signal, we'd almost certainly be driving the BJT well into deep saturation. That's why this kind of analysis can be vital in deciding how much to compromise on the voltage gain in order to deal with circuit variations, even with NFB applied.

Engineering logbooks

As an aside, what you read above is something that should be placed into the engineering logbook you keep as a circuit designer. (A must-do, if you are a professional, and should-do, otherwise.) These kinds of details are important to record and to be able to refer back to, should circumstances require it, in order to show why you made the choices you made at the time. For professionals, it can save your life if things ever go to court. For professionals and hobbyists, it can save a lot of your time later on when you need to revisit a design.

In short, keep logbooks for the rest of your life!

Hi jonk You have mentioned that beta will rise with temperature. According to my knowledge (supported by some IEEE papers) the opposite is true (but in a negligible amount). I think, the most important temperature effect comes from the saturation current in Shockleys exponential equation. However, it should be clear that Vbe does not change (automatically) but MUST be reduced externally for rising temperatures. — LvW, Oct 04 '22 at 10:32
@LvW Are you saying that the beta does not increase with temperature? What about this figure 4 from the BC847 datasheet https://docs.rs-online.com/3bb0/A700000006782199.pdf And it is common to assume that with a temperature increase of 1 K, the current gain increases by about 0.5%...1%. — G36, Oct 04 '22 at 13:53
Yes - you are right. I think, I must qualify my statement regarding the temperature-dependent properties of factor B (or beta). The sensitivity is very low - but there are contradictory statements regarding the sign (probably depending on the power). However, I do not have any own findings. — LvW, Oct 04 '22 at 14:02
@LvW I understand the confusion. Resistance of metal conductors rises with increasing temperature and one might infer from this similar behavior for semiconductors. But the extra charge carrier production outweighs it in the case of semiconductors, I believe. I don't think any BJT has a beta that declines with increasing temperature. Theory says I shouldn't see one. See *"Introductory Electronics for Scientists and Engineers,"* 2nd ed, 1987, by Robert Simpson or *"Analogue and Digital Electronics,"* 1984, by H. Ahmed and P. J. Spreadbury. — jonk, Oct 04 '22 at 20:53
Hi jonk - thank you for this information and the links provided. — LvW, Oct 05 '22 at 08:50
@jonk Thank you . But I have a question , explanation of this circuit they explained in sequential events. Where does this explanation come ? Is it about block diagram? — Heroz, Oct 11 '22 at 05:44
@Heroz I've seen your recent question. And I've seen you write up an NFB diagram. No, the above description isn't about using an NFB model. It can be used. I just didn't use it, here. If you want to see it, then it looks like [this](https://i.stack.imgur.com/W8sk1.png). There will be a LambertW() function involved in a closed solution. — jonk, Oct 11 '22 at 06:38
@jonk thank you very much. In website they explain in sequential events so I doubt does in reality it work like that? And I think that why they wrote in sequential events maybe because NFB diagram. So I tried to create them .And now I think in order to understand this circuit I don't need to deeply understand to guanta level right? Do you have any advice to understand circuit? — Heroz, Oct 11 '22 at 06:57
Here explaination: Now suppose for some reason, a temperature change perhaps, β increases. This should cause an increase in IC. An increase in IC, though, would cause an increase in the drop across RC due to Ohm's law. This, in turn, would force VC to drop. — Heroz, Oct 11 '22 at 07:09
Sorry in your NFB diagram where did these two block from ? How did you find it? — Heroz, Oct 11 '22 at 07:24
@Heroz Regarding **sign**, I suppose I could have been clearer. But in my writing, I specifically was excluding the **sign** of the change and instead focusing on the **magnitude** of the change. So I allowed myself to change the order of the terms in the denominator of one of the expressions, without explaining that fact. If you want the sign, in that case, reverse their order. — jonk, Oct 11 '22 at 19:57
@Heroz Regarding the block diagram? I don't see any block diagram in this answer I gave. What do you mean? — jonk, Oct 11 '22 at 19:58

score 2 · Accepted Answer · answered Oct 03 '22 at 14:50

2

Does collector feedback bias work like this in real life?

Yes it does; it creates negative feedback and, that concept is widespread in analogue electronics (particularly op-amps). Negative feedback is the "great stabilizer" of analogue circuits.

Fighting against the basic problems of a BJT circuit is something that has to be accommodated anyway (whether negative feedback is used or not).

answered Oct 03 '22 at 14:50

Andy aka

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I have a question. this is what I understand when beta increase Ic try to increase but when Ic increase it makes Vce drop and also Vbe. So Ib decrease and it tries to keep Ic current from rising right? – Heroz Oct 03 '22 at 16:15
Negative feedback is good at reducing the effects of things like this @Heroz – Andy aka Oct 03 '22 at 17:42
Thank you. In physical reality if I measure does it work sequentially as above? – Heroz Oct 04 '22 at 00:50
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In the analogue world, nothing works sequentially @Heroz <-- there are no discrete time steps. – Andy aka Oct 04 '22 at 08:59
Andy aka is negative feedback really a "great stabilizer"? For opamps negative feedback always reduces the stability margin (against dynamic stability). – LvW Oct 04 '22 at 10:10
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@LvW yes, I know what you mean; it's a double-edged sword. Maybe there's a better word than stabilizer..... – Andy aka Oct 04 '22 at 10:21
@Heroz I'm sorry but I didn't understand your questions in your comment. – Andy aka Oct 04 '22 at 11:26
I'm sorry . This is what I mean. Quote"In the analogue world, nothing works sequentially" , does that mean In physical reality when beta change the circuit essentially solves the KVL and KCL equations at once right ? It's not work sequentially and why they explained in sequential events ? – Heroz Oct 04 '22 at 12:10
1

No circuit solves any of Kirchhoff's equations; the equations are just mathematical concepts to help us predict currents and voltages @Heroz (sorry if that disappoints) – Andy aka Oct 04 '22 at 12:20
I have one doubt Quote"In the analogue world, nothing works sequentially" but why they explained in sequentially events. When beta change circuit change instantaneously right ? – Heroz Oct 04 '22 at 16:58
Nothing changes instantaneously either and I cannot tell you why they used a sequential kind of explanation without some overriding words that explain that sequential explanations are useful but things don't happen in quanta in the analogue world. – Andy aka Oct 04 '22 at 17:41
@Andyaka Sorry for ask you many question . Could you explain me how Rb( feedback resistor ) stabilize this circuit? – Heroz Oct 11 '22 at 08:27
@Heroz initially you accepted an answer I gave on a different question then, yesterday you deselected it. How do you think this motivates me to help you? The explanation of stabilization using a feedback resistor is not trivial and therefore is deserving of a new question. – Andy aka Oct 11 '22 at 08:32
@Andyaka... I'm sorry . I did not mean to deselected it to motivate you answer me but I'm not sure in answer because I still not understand how it works . I think maybe it's as you said " in analogue world things don't happen in quanta " . I'm dig too deep it doesn't matter in analogue world right? Do you have any advice to understand circuit? thank you. – Heroz Oct 11 '22 at 09:04
OK, that's fine. Do you know anything much about negative feedback? The problem of trying to explain how this circuit works is that I could waste pages of explanation on stuff you already know but, you could say to me "assume I know nothing" and then I'll have to do pages and pages of stuff on ohms law and other basic transistor circuits before moving to the one above. Pitching this just right is the difficult thing hence, why I suggested you raise a new question. – Andy aka Oct 11 '22 at 09:16
You might also be interested in [this answer](https://electronics.stackexchange.com/questions/556899/what-is-the-purpose-of-the-emitter-resistor-in-a-common-emitter-amplifier/556905#556905) that clearly shows why emitter degeneration (adding an emitter resistor) is a pretty useful thing to consider as the best method of stabilizing a BJT circuit @Heroz – Andy aka Oct 11 '22 at 13:02

score 1 · Answer 3 · 2022-10-03T19:29:34.047

There's a balance where the voltage at the collector Vce has just the exact value needed to feed just enough base current to maintain the collector current what's needed to get the voltage at the collector to be dropped to Vce from value Vcc. It can be coarsely calculated with Vbe and transistor's Hfe (see NOTE1). Or it can be found by simulating or by making a test circuit.

That biasing solution has one advantage: It reduces the dependency on transistor's Hfe which vary easily -50...+100% between the individuals of the same type. Also the collector leakage current doesn't affect as much the operating point as it would do in case Rb was connected between Vcc and the base.

Rb between the collector and the base essentially causes a negative feedback, which as a side effect reduces the gain in multistage transistor amplifiers. But also that can be considered as an advantage: The non-linearity of the transistor causes less distortion in a class-A amplifier stage.

NOTE1: To get the operating point (=idle state Ic and Vce) one must solve the next equation group

Ic = Hfe(Vce - 0,7V)/Rb

Vce = Vcc - Rc(1 + 1/Hfe)Ic

The first one says that Ic = Hfe * Ib

The second one says the voltage drop in Rc is the collector current plus the base current multiplied by Rc.

It can be shown that an emitter resistor - together with a resistive voltage divider - has better properties as far as stability of the DC operating pont against BJT parameter variations is concerned. — LvW, Oct 04 '22 at 10:12
The resistor between b and c can be good enough in many applications and it needs fewer parts. — , Oct 04 '22 at 10:19

score 1 · Answer 4 · answered Oct 03 '22 at 19:01

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The sequential reasoning you described is used quite frequently to explain the operation of many circuits, especially those in which feedback is employed.

However in the physical reality there is no such clear-cut sequence.

That sequence is an easy way of analyzing the influence of a parameter variation in a qualitative way, but that is just a simplification. With that "sequencing" you assume a certain chain of cause-effect steps whose actual justification lies in how the circuit equations works.

In particular, that reasoning is valid as long as you implicitly assume a small variation relative to a quiescent point. This is necessary because this method can be formally justified only if the circuit is linear or, as in your case, can be linearized (that small variation assumption is needed for the linearization to be performed).

This reasoning can be formally justified by writing the non-linear (maybe differential) equations system describing the circuit and applying a linearization technique. This involves things like multidimensional Taylor expansion, which in turn is based on partial derivatives and multidimensional function theory. Tough stuff for a beginner, but it is something that many electrical engineers may have to grok during their engineering university education.

However it is not something that you need to understand in order to apply that sequential method to understand how the circuit behaves, as long as you assume small variations and the circuit can be linearized around the chosen quiescent point.

answered Oct 03 '22 at 19:01

LorenzoDonati4Ukraine-OnStrike

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Quote:" ...in the physical reality there is no such clear-cut sequence....". I think, such a general statement is not correct. Counter example: The slew rate of an opamp is caused by a short overdrive of the first stage (now operating as a current source) - until the feedback (with a slight delay) comes into play and brings the first stage back to linear operation. – LvW Oct 04 '22 at 10:07
@LvW I was stating a general statement on purpose. These are not digital system, where you can clearly mark a discrete event and their consequences at subsequent time instants. In the analog domain it's all about small (ideally infinitesimal) increments around the feedback loop. In general you could "cut" the loop in any point and start applying small increments from that point and follow the evolution around the loop. – LorenzoDonati4Ukraine-OnStrike Oct 04 '22 at 12:09
@LvW Of course if you *know* that in a *specific* instance the perturbation of the system happens in a *specific* point in the loop, that's another issue. Feedback (and the kind of analysis I outlined) works anyway. – LorenzoDonati4Ukraine-OnStrike Oct 04 '22 at 12:09
@LvW For example, take the usual linear power supply topology employing a pass transistor. You can analyze the system considering what happens when, for example, the perturbation is an increase in temperature of the pass transistor, or an increase in input voltage, or a variation of load current. It doesn't matter which perturbation actually happens, the feedback loop works the same and the same kind of analysis can be applied. – LorenzoDonati4Ukraine-OnStrike Oct 04 '22 at 12:15
@LvW However, if you are *not* in a situation where small perturbations can be assumed, then that kind of analysis may well fail brutally, e.g. when applying a big/quick enough perturbation pulse. In this case other factors you didn't initially model could play a crucial role (e.g. parasitic capacitance or wire inductance) and the system may, for example, oscillate between two operating points. In system theory this is a problem of stability of the equilibrium points of a non-linear system. – LorenzoDonati4Ukraine-OnStrike Oct 04 '22 at 12:19
I did not refer to digital systems - perhaps I did not express myself clear enough. The effect of slew rate (opamp with feedback) is caused by a timely sequence of two "events". And it can only be explained, verified and quantified using these effects: (1) Overdrive of the first stage and (2) delayed action of negative feedback. That`s all I wanted to say. I am sure you will agree. – LvW Oct 04 '22 at 13:15
@LvW Yes, I got it. I cited digital systems because digital systems are, like the example you made, cases where you don't work with small increments around an operating point, but you drive the opamp in saturation (heavily in your case). This means that you start in a state where the operating point is in the linear region to another operating point where the system behaves in a completely different way (saturation), so this behavior cannot be explained by a linear mathematical model, such as that of "small increments" I referred to. – LorenzoDonati4Ukraine-OnStrike Oct 04 '22 at 15:41
@LvW My explanation above was intentionally general purpose. As I said in a comment, when you work with linear (or linearized) models, it doesn't matter where in your circuit you start applying a perturbation *for analyizing the behavior of the circuit*. I can assume there is a small variation of Ic that, because of feedback, causes a change in Vbe. Or I can assume there is a change in Vbe that in turn causes a change in Ic. This doesn't mean that, *in some specific case* you know the initial perturbation is either in Ic or in Vbe. ... – LorenzoDonati4Ukraine-OnStrike Oct 04 '22 at 15:48
@LvW ... This works out because the model is linearized around a specific quiescent point (operating point, equilibrium point - call it whatever you like). If you jump between operating points with your "perturbation", the linearized model fails. It's like trying to use a 1st order Taylor expansion when the increment is too big. The error increase more and more as you get farther from the point where you performed the expansion. The more you get away from that point, the more higher Taylor terms get important. – LorenzoDonati4Ukraine-OnStrike Oct 04 '22 at 15:51
I agree that my comment does not apply to linear systems. That´s clear. – LvW Oct 04 '22 at 20:20
@ Lorenzo Donati support Ukraine . Thank you , but which equation that this circuit explanation come from ? – Heroz Oct 07 '22 at 10:01
@Heroz The equations for this circuit were given by Jonk in his answer. What I was trying to tell you, without repeating what other had already said, is that the rationale behind *any* circuit analyzed with that "sequential" approach lies in some very difficult math, such as [this](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor's_theorem_for_multivariate_functions). Since you didn't state which level of expertise you have, I gave an explanation suitable for a beginner, without entering in further details. – LorenzoDonati4Ukraine-OnStrike Oct 07 '22 at 15:40
@ Lorenzo Donati support Ukraine So sequential events can write in block diagram right? For example this link https://www.circuitbread.com/tutorials/block-diagrams-1-4 RLC circuit they write it in block diagram form. – Heroz Oct 08 '22 at 03:50

LvW · Answer 5 · 2022-10-04T14:01:36.547

For my feeling, some comments are in order:

The most important temperature influence on the DC bias point comes from the positive temparature sensitivity of the current Is in the well-known equation Ic=Is[exp(Vbe/Vt)-1)]
This increase can be compensated when Vbe is externally reduced (negative DC feedback using an emitter resistor Re or Rb as shown) by a app. -2mV/K. In the latter case (Rb stabilization) it is the voltage division between Rb and the DC input resistance at the B-E junction (Rbe=Vbe/Ib) that is responsible for the Vbe reduction).
In contrast to this sensitivity of Is, the sensitivity of the current gain B (resp.beta) is comparably small.