0
\$\begingroup\$

enter image description hereenter image description here

In the following circuit Ic must be found.

I begun by open-circuiting the 3 capacitors, and finding the voltage at the base which is 2.5V (from the voltage divider). By assuming that Vbe=0.7V it's easy to progress, however no such thing is stated (the only data given is that β is very large), so i was wondering if there is a way to find Ic without that assumption.

If we do assume that Vbe=0.7V, then when is the Ic=Is*e(Vbe/Vt) formula used?

\$\endgroup\$
  • \$\begingroup\$ "however no such thing is stated" - what do you mean? \$\endgroup\$ – Andy aka Jun 3 '18 at 18:46
  • \$\begingroup\$ The data given for the exercise dont include that info. \$\endgroup\$ – Manouil Jun 3 '18 at 18:50
  • \$\begingroup\$ It might be clearer if you provided the full detail of the exercise rather than feeding snippets. \$\endgroup\$ – Andy aka Jun 3 '18 at 18:51
  • 2
    \$\begingroup\$ Read this electronics.stackexchange.com/questions/337010/… \$\endgroup\$ – G36 Jun 3 '18 at 19:01
  • 1
    \$\begingroup\$ Yes - assuming Ib=0 you can use the simple voltage divider formula. Hence, the voltage at the base node is simply 2.5 volts. As shown in my detailed answer, such a simplification is allowed only because of the negative feedback effect of RE. Such negative feedback always reduces the sensitivity of gain stages against simplifications (in your case: Ib=0 and Vbe=0,7V). Because of these simplifications you do not need the exponential relationship between Ic and Vbe which would require the knowledge of the exact VBE value, \$\endgroup\$ – LvW Jun 4 '18 at 6:54
3
\$\begingroup\$

In the enclosed figure (first diagram) it is shown how an emitter resistor RE stabilizes the operating point against uncertainties of VBE. Hence, it is common practice to assume for VBE a suitable value between 0.6 and 0.7 volts because such a relatively large VBE uncertainty results only in small Ic variations (for sufficient RE-feedback).

The dotted line (vertical) shows how RE=0 (no feedback) would result in an unacceptable large Ic variation (uncerainty).

enter image description here

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

By assuming that Vbe=0.7V it's easy to progress, however no such thing is stated (the only data given is that β is very large), so i was wondering if there is a way to find Ic without that assumption.

Sure. Let's solve it for the general case, assuming active mode operation and ignoring the Early effect:

schematic

simulate this circuit – Schematic created using CircuitLab

From KVL, you have:

$$V_\text{TH}-I_\text{B}\:R_\text{TH}-V_\text{BE}-I_\text{E}\:R_\text{E}=0\:\text{V}$$

Where, of course, \$R_\text{TH}=\frac{R_1\:R_2}{R_1+R_2}\$ and \$V_\text{TH}=V_\text{CC}\:\frac{R_2}{R_1+R_2}\$.

From \$I_\text{E}=\left(\beta+1\right)\:I_\text{B}\$, you can substitute in and solve for \$I_\text{B}\$:

$$I_\text{B}=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)\:R_\text{E}}$$

Now substitute from your nice formula and solve for \$V_\text{BE}\$:

$$\begin{align*} \frac{I_\text{SAT}}{\beta}\:e^{\frac{V_\text{BE}}{V_T}}&=\frac{V_\text{TH}-V_\text{BE}}{R_\text{TH}+\left(\beta+1\right)\:R_\text{E}}\\\\ \frac{V_\text{TH}-V_\text{BE}}{V_T}\:e^{\frac{-V_\text{BE}}{V_T}}&=\frac{I_\text{SAT}\left[R_\text{TH}+\left(\beta+1\right)\:R_\text{E}\right]}{\beta\:V_T}\\\\ \frac{V_\text{TH}-V_\text{BE}}{V_T}\:e^{\frac{V_\text{TH}-V_\text{BE}}{V_T}}&=\frac{I_\text{SAT}\left[R_\text{TH}+\left(\beta+1\right)\:R_\text{E}\right]}{\beta\:V_T}\:e^\frac{V_\text{TH}}{V_T}\\\\ \frac{V_\text{TH}-V_\text{BE}}{V_T}&=\mathcal{LambertW}\left(\frac{I_\text{SAT}\left[R_\text{TH}+\left(\beta+1\right)\:R_\text{E}\right]}{\beta\:V_T}\:e^\frac{V_\text{TH}}{V_T}\right)\\\\ V_\text{BE}&=V_\text{TH}-V_T\:\mathcal{LambertW}\left(\frac{I_\text{SAT}\left[R_\text{TH}+\left(\beta+1\right)\:R_\text{E}\right]}{\beta\:V_T}\:e^\frac{V_\text{TH}}{V_T}\right) \end{align*}$$

Note that no assumptions were made about \$V_\text{BE}\$ above. None are needed. Just as you thought might be the case! (Also note when \$u\:e^u=z\$ then \$u=\mathcal{LambertW}\left[z\right]\$. See Lambert W Function.)


This technique works over 3-5 orders of magnitude, with modest error bounds over part variations over that range; takes into account circuit details in the process for a direct solution; and it's not complicated to develop a practical value for the saturation current by looking at a datasheet; as I demonstrate here. The general solution here simply works better than the assumption and gets you a more accurate answer, and with far less effort, than alternatives.

\$\endgroup\$
  • \$\begingroup\$ Jonk, I think, you only have shifted the problem from Vbe (assumption necessary) to Isat (assumption necessary). Do you really think that a guess of Isat is easier than a guess of Vbe ? For my opinion, a calculation of Ic starting with an assumption for Vbe (instead of Isat=....) is (a) much more simpler and (b) exact enough because of strong DC negative feedback which makes the calculation rather insensitive to assumption uncertainties. \$\endgroup\$ – LvW Jun 4 '18 at 6:36
  • \$\begingroup\$ @LvW If you test the function with a range of values (and you can get the value pretty easily from most datasheets -- I've explained how, recently, on a different post) you'll find it is better than the assumption and far more widely useful (much larger dynamic range) this way. So yes, I think it improves the situation. \$\endgroup\$ – jonk Jun 4 '18 at 13:06
  • \$\begingroup\$ as far as I have understood, the questioner needs the value of Ic for a given gain stage. However, you have proposed a method which needs prior knowledge of Ic (see the graph in your link as given with "different post") - as an input for the given formula Vbe=f(Isat). More than that, this formula contains the current gain beta which has very large tolerances. So - I cannot see how this method "improves he situation". Did I misunderstand something? \$\endgroup\$ – LvW Jun 4 '18 at 15:05

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.