Why the Q-point of an amplifier with collector-feedback resistor are stable and not randomly moves on output BJT-curves when the power supply in applied to the circuit? The schematic is below:

enter image description here

  • \$\begingroup\$ what is negative feedback? think. What is gain ratio of feedback? who does that reduce error? \$\endgroup\$ Nov 17 '17 at 16:01
  • \$\begingroup\$ The question was updated to clarify the situation \$\endgroup\$
    – MaxMil
    Nov 17 '17 at 16:07
  • \$\begingroup\$ To help you think about it: why would it move "randomly about"? I see a clear feedback in the circuit which counteracts (works against) changes in the DC voltage on the collector. Do you see it as well? \$\endgroup\$ Nov 17 '17 at 16:17
  • \$\begingroup\$ Yes, I see. But I found that all circuits like this are analyzed with KVL equations first, for example for my: Vcc - Vbe - betta * Ib * (Rc + Re) - Ib * Rf = 0, i. e. 10 - 0.7 - 90*Ib*(4.7k+1.2k) - Ib * 250k = 0. From here I can found Ib. But why that value of Ib in reality must happen? Why when the 10V is applied to schematic the Ib and Ic may not be much smaller (to satisfy Ohms law of course) and stay there? \$\endgroup\$
    – MaxMil
    Nov 17 '17 at 16:26

There are a couple of ways of approaching such a question. One is to think from a reductionist approach and see where that takes you. The other is to just go look at the equations using Thevenin, solve them completely, and see if you can gain additional insight from those.

From a reductionist point of view \$I_Q=I_E=I_C=\beta\cdot I_B\$ and \$V_{BE}=700\:\textrm{mV}\$, so therefore it follows that \$V_{CE}=V_{CC}-I_Q\cdot\left(R_C+R_E\right)\$, with the value of \$V_{BC}\$ being slightly smaller by exactly one \$V_{BE}\$. So, assuming active mode, then:


The collector voltage is then:

$$\begin{align*} V_C&=V_{CC}-I_Q\cdot R_C\\\\& =V_{CC}-\left(V_{CC}-V_{BE}\right)\frac{R_C}{R_C + R_E + \frac{R_B}{\beta}} \end{align*}$$

And from here you can see that this equation turns on how the last term varies with changes in \$\beta\$. But it also appears that the value of \$R_B\$ is also important, when compared to the other resistor values, and that its importance increases with smaller values of \$\beta\$.

There are a couple of mental directions to go from here.

  1. How good are the above reductions (simplifications) and should we consider doing a somewhat more detailed analysis to figure out how much error there is in making those reductions? The answer to this lies in the assumptions themselves. The active mode assumption we simply have to accept for an amplifier that operates well. So there was no risk from that. The \$V_{BE}\$ assumption varies only by about \$60\:\textrm{mV}\$ per decade change in collector current. So that isn't a biggy, either. So most of the answer here turns on \$\beta\$. If we are using "larger" values of \$\beta\$ then our reductionist approach is pretty good. But for "smaller" values, not so good. So the sum-up here is that in the case where \$\beta\$ is larger, then \$V_C\$ is very stable and also our reduction is sound; and in the case where \$\beta\$ is smaller, then \$V_C\$ is less stable and our reduction is no longer as sound. Keeping in mind that we may still ignore the Early Effect, and temperature, and variation of \$V_{BE}\$ with collector current, it may still be worthwhile doing some extra math (full KCL, with solutions) to see how that changes the situation (and if so, by how much.)
  2. Another is to ask ourselves, "What is the real question we are asking?" Is it, "How much does \$V_C\$ vary with a change of '1' in \$\beta\$?" Or is it instead, "How much does \$V_C\$ vary with a change of '1%' in \$\beta\$?" These are actually different questions, despite seeming similar. But what do we really want to know? Chances are, it's more about the percent-variation. We probably are not as interested in comparing what happens to \$V_C\$ when \$\beta\$ changes from 300 to 301 vs changes from 90 to 91 (by 1 in each case), as much as we are interested in comparing what happens when there is a 10% change, regardless of starting value. And we actually haven't even expressed how to address either question, yet.

Thevenin expansion of question (1) above follows:

$$\begin{align*} \frac{V_C}{R_C}+\frac{V_C}{R_B}+I_C &= \frac{V_{CC}}{R_C}+\frac{V_B}{R_B}\\\\ \frac{V_B}{R_B}+\frac{I_C}{\beta}&=\frac{V_C}{R_B}\\\\ \frac{V_B-V_{BE}}{R_E}&=\frac{\beta+1}{\beta}\:I_C\\\\ \therefore\\\\ V_C&=V_{CC}-I_Q\cdot R_C\\\\& =V_{CC}-\left(V_{CC}-V_{BE}\right)\frac{R_C}{\left(R_C + R_E\right)\frac{\beta}{\beta+1} + \frac{R_B}{\beta+1}} \end{align*}$$

This seems to be hardly any difference. But it provides us with a quick look at the modifications, too. Given the relatively minor differences, it seems like the earlier, easier, and less complex equation wasn't too bad, after all.

Question (2) above really wants us to ask about the sensitivity of \$V_C\$ to percent changes in \$\beta\$. We can answer the question about the sensitivity of \$V_C\$ to incremental changes in \$\beta\$ by solving for \$\frac{\textrm{d} V_C}{\textrm{d} \beta}\$. But what if that's not as interesting as asking about percent changes?

Thinking just a little deeper, we find a curious answer we may want to find is:

$$\begin{align*} \frac{\left[\frac{\textrm{d} V_C}{V_C}\right]}{\left[\frac{\textrm{d} \beta}{\beta}\right]}&=\frac{\beta}{V_C}\cdot\frac{\textrm{d} V_C}{\textrm{d} \beta}\\\\ &=-\frac{R_B \: R_C\:\left(V_{CC}-V_{BE}\right)}{\left(\frac{\beta+1}{\beta}\left(R_C+R_E\right)+\frac{R_B}{\beta}\right)\left(\frac{\beta+1}{\beta}\left(V_{CC} R_E+V_{BE} R_C\right)+V_{CC}\frac{R_B}{\beta}\right)} \end{align*}$$

Note that there is factor inserted here that scales the differential equation that would apply for a "change by 1" to make this a unitless comparison of % changes, instead. This form of equation answers the following question: "What percent change in \$V_C\$ would occur for a given percent change in \$\beta\$?"

And that's actually a decent question.

And with the values you provided with the schematic, I find that the answer is approximately -0.32. Which means that you'd get about a 10% downward change in \$V_C\$ for a 31% upward change in \$\beta\$. Since one BJT might vary that much in \$\beta\$, from one to another, this suggests that you probably should not expect to see more than about a 10% change in \$V_C\$ when swapping BJTs into the socket from a bag of them (discounting temperature effects, the Early Effect, and variations in \$V_{BE}\$.)

None of the above deals with overall DC operating point design questions. By this, I mean for example, "If we know \$V_{CC}\$ and \$I_Q\$ as design inputs, and either \$R_C\$ or \$V_C\$ also as a design input, then how would \$R_B\$ and \$R_E\$ depend on them (or other design inputs as yet unmentioned?)" How might asking this question lead to more interesting additional questions about the BJTs used in this topology, generally?

I'll just drop these follow-ups out there for now.

From the above, you may also see that you have to refine your questions, too. The question starts out being, perhaps, "How does \$V_C\$ vary with a change in \$\beta\$?" But to get perhaps a more interesting and useful answer, you may need to refine your question further and ask a perhaps better question, instead.

Which leads us to the first of my two laws:

  • The question matters as much as the answer

The second being:

  • An equal right to an opinion isn't a right to an equal opinion

In your collector Feedback circuit, Base resistor RB is connected between base and collector. Hence collector voltage relations are,

Vc = Vcc - IcRc Vc = 0.7 + Ib*Rb + IeRe

Now, suppose collector current increases for some reason like beta change, voltage drop across collector decreases (eq. 1) and also it in turn automatically reduces the base current. Thus collector current is reduced or adjusted back. Thus it keeps transistors at a fixed Q-point. Rb provides that negative feed back to achieve this goal.

  • \$\begingroup\$ It's what I know. I need to understand how much value of Ic the BJT decide to use for Ib value? And that will influence the Q-point position. \$\endgroup\$
    – MaxMil
    Nov 17 '17 at 17:09

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.