# BJT Veb matter & use of the Ic-Veb Exponential Relationship

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?

• "however no such thing is stated" - what do you mean? – Andy aka Jun 3 '18 at 18:46
• The data given for the exercise dont include that info. – Manouil Jun 3 '18 at 18:50
• It might be clearer if you provided the full detail of the exercise rather than feeding snippets. – Andy aka Jun 3 '18 at 18:51
• – G36 Jun 3 '18 at 19:01
• 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, – LvW Jun 4 '18 at 6:54

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). 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: 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.

• 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. – LvW Jun 4 '18 at 6:36
• @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. – jonk Jun 4 '18 at 13:06
• 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? – LvW Jun 4 '18 at 15:05