# Effect of Vbe on Ic for a BJT

I am trying to find out where I have gone wrong in my understanding of BJTs:

According to the Ebers-Moll equation:

$$I_C = I_S\cdot (e^{ \frac{V_{BE}}{V_T}}-1)$$

So if VBE decreases, then IC should follow suit.

However, when looking at the following circuit:

Since VB is held at a fixed voltage by the biasing divider, if VBE were to decrease, then VE will rise, meaning more current flows through RE. Since IE = IC + IB, and so IE ≈ IC, surely IC will increase?

This contradicts the expected decrease in IC by the Ebers-Moll equation. I would greatly apreciate it if you could help me figure out where I have gone wrong here.

EDIT:

Adding context from The Art of Electronics (2nd edition), from which this example was taken, which states that VB is held constant by the voltage divider:

• Decresed Vbe causes higher Ie and also higher Ic. Commented Feb 8 at 21:26
• You might be interested in this: electronics.stackexchange.com/questions/699105/… Commented Feb 8 at 21:33
• Xavier, Assuming there's no applied signal via your input capacitor and assuming you don't alter the circuit in any way, then $V_{_\text{BE}}$ and $I_{_\text{C}}$ are co-determined by the circuit and $V_{_\text{BE}}$ can only change due to some external effect such as noise or temp change. +1C change might reduce $V_{_\text{BE}}$ by 2 mV. But if $V_{_\text{E}}=500\:\text{mV}$ then this means about 0.4% increase in $I_{_\text{C}}$ matching your starting expectation. Ebers-Moll would counter with $100\:\mu\text{V}$ increase in $V_{_\text{BE}}$. Much smaller than the disturbance. Commented Feb 8 at 23:09
• Xavier, Your just-added edit only emphasizes the point I just made. The higher the voltage drop across $R_{_\text{E}}$ you allow in your design, the smaller the % change in $I_{_\text{C}}$ due to thermal changes in $V_{_\text{BE}}$. It's why that is important. Commented Feb 8 at 23:13
• @periblepsis I'm sorry but I do not fully understand the point you are making. You are saying that the increase in Ve due to temperature (2mV) causes a negligable increase to Ic, but I don't get where the 100uV increase to Vbe has come from, or how it explains why the ebers-moll equation is condridicting what I am saying. Commented Feb 8 at 23:40

Including a resistance in the emitter path is called "emitter degeneration". Consider these scenarios, with different resistances at the emitter:

simulate this circuit – Schematic created using CircuitLab

Note that $$\R_{E1} = 0\Omega\$$, effectively tying node E1 to ground, and fixing its potential at 0V.

Sources V1, V2 and V3 are chosen to bias the base-emitter junctions to pass exactly 1mA in each case. Otherwise they can be disregarded, as all we are really interested in is how emitter current $$\I_E\$$ (and consequently collector current) changes as a result of changes in potential $$\V_{B}\$$, as opposed to any absolute potential there. That is, we want to see how transconductance $$\\frac{\Delta I_E}{\Delta V_B}\$$ is affected by the presence of a resistance $$\R_E\$$ in the emitter path.

Base potential may rise and fall, but the amount by which emitter potential $$\V_E\$$ rises and falls may or may not be the same.

On the left, where the emitter is connected directly to ground, of course we have:

$$V_{BE} = V_B - V_E = V_B - 0 = V_B$$

The Ebers-Moll equation is at work here:

$$I_C = I_S \left( e^{ \frac{V_{BE}}{V_T} } - 1 \right)$$

With emitter potential fixed at 0V, so that $$\V_{BE}=V_B\$$, this equation needs no "tweaking". However, the second and third circuits, B and C, do not behave the same way.

Circuits B and C permit the emitter to rise in potential, above 0V. The amount by which they rise is determined by Ohm's law. For any given emitter current $$\I_E\$$, $$\V_E\$$ will be:

$$V_E = I_ER_E$$

Now we need a new expression for $$\V_{BE}\$$ as a function of emitter current $$\I_E\$$:

\begin{aligned} V_{BE} &= V_B - V_E \\ \\ &= V_B - I_ER_E \\ \\ \end{aligned}

As you pointed out, $$\I_C \approx I_E\$$, and we can plug in this expression for $$\V_{BE}\$$ into the Ebers-Moll equation:

$$I_E \approx I_S \left( e^{ \frac{V_B-I_ER_E}{V_T} } - 1 \right)$$

Put another way, the presence of $$\R_E\$$ is reducing the change in emitter current $$\I_E\$$ that would result from any given change in base potential $$\V_B\$$.

A simulation will demonstrate this nicely. In my schematic, Source $$\V_{IN}\$$ is a 1mV amplitude sinusoid. Here's a plot of how emitter currents vary as $$\V_B\$$ is rising and falling by 1mV:

The blue trace is $$\I_{E1}\$$, where there is no emitter resistance. It clearly has greater amplitude than the others, telling us that transconductance decreases with increasing $$\R_E\$$.

I suspect that you have the right idea, but you failed to account for the biasing of the transistor, which has been carefully engineered in my examples to recreate the exact same quiescent emitter current in each case. In other words, there is indeed a decrease in collector and emitter current due to the presence of $$\R_E\$$, all other things being equal.

Strictly speaking, there is a decrease in effective transconductance, but that's relating change in collector/emitter current to change in base potential, and is highly dependent on absolute quiescent values everywhere. You must be operating in the same portion of the $$\V_{BE}\$$ vs. $$\I_C\$$ curve (according to the Ebers-Moll model), in order to make valid quantitative claims regarding $$\V_B\$$ vs. $$\I_C\$$, otherwise you have two changing independent variables, instead of just one, $$\V_B\$$.

From an intuitive perspective, since emitter degeneration permits $$\V_E\$$ to vary, any change in $$\V_B\$$ is necessarily going to result in a smaller variation in $$\V_{BE}\$$ than the scenario with no emitter resistance at all. Of course, effective transconductance will fall. My modification of the $$\V_{BE}\$$ term in the Ebers-Moll equation more formally reflects that.

In case this isn't yet clear enough, I'll say it with even more different words: The Ebers-Moll equation is always true, with or without emitter degeneration; a certain change in $$\V_{BE}\$$ will always result in the same change in $$\I_C\$$ (assuming you are operating in the same region of the $$\V_{BE}\$$ vs. $$\I_C\$$ curve). In the context of your question, though, it is $$\V_B\$$ that you are controlling, not $$\V_{BE}\$$. The resulting change in $$\V_{BE}\$$ will depend on $$\R_E\$$, and is therefore different in each case.

@periblepsis yes I would be grateful if you could elaborate on your point of view –
Xavier

@periblepsis I'm still interested in your point of view. –
internet


I'll try to get straight to the point. I can always expand, later. So I'll just focus on the point being made in the following section of the 3rd edition of The Art of Electronics:

The text above discusses a temperature change leading to a change in $$\V_{_\text{BE}}\$$. This occurs for several different physical and competing reasons. But the more important and dominant one is that the saturation current [a theoretical y-axis intercept value] shifts $$\\propto T^3\$$.

Their exact choices of $$\I_{_\text{C}}\$$, $$\V_{_\text{BE}}\$$, and $$\\Delta\,V_{_\text{BE}}\$$ are arbitrarily made for education purposes. (For example, for a device similar to the 2N2222 model in LTspice I'd need to use $$\I_{_\text{C}}\approx 125\mu\text{A}\$$ to get $$\V_{_\text{BE}}=600\:\text{mV}\$$ at $$\27.5^\circ\text{C}\$$. And in that case, I'd see $$\-1.9536\:\text{mV}\$$ change in $$\V_{_\text{BE}}\$$ for a $$\+1^\circ\text{C}\$$ change in temperature, from $$\27^\circ\text{C}\$$ to $$\28^\circ\text{C}\$$. The text is tossing out some broadly representative values. Not gospel!)

They've chosen a specific example requiring $$\A_v=50\$$ because this greatly exaggerates (or emphasizes) their point. A very large voltage gain necessarily also requires a very small voltage difference across the emitter resistor. (Discounting $$\r_e^{\:'}\$$ in their circuit -- which they didn't ignore, by the way -- says that $$\A_v\approx\frac{V_{_\text{CC}}-V_{_\text{C}}}{V_{_\text{E}}}\$$. So a high gain requires a small value for $$\V_{_\text{E}}\$$.) The choice of $$\A_v=50\$$ is no accident. It's pure malicious intent on their part.

They assumed, for purposes of their discussion, that the biasing resistor pair is arbitrarily stiff. (Writing, "since the base is held at constant voltage".) This means that they assume that the base voltage won't change, at all, and that any change to $$\V_{_\text{BE}}\$$ due to temperature variation causes a change to $$\V_{_\text{E}}\$$ and only to $$\V_{_\text{E}}\$$. (Reality, taking into account practical biasing, would involve changes to both $$\V_{_\text{B}}\$$ and $$\V_{_\text{E}}\$$.)

So in their example, since $$\V_{_\text{BE}}\$$ is assumed to decline by $$\-2.1\:\text{mV}\$$ per Kelvin degree increase, this means that $$\V_{_\text{E}}\$$ rises by the same amount. That leads to a necessary increase in the emitter current and associated increase in collector current.

Their example says $$\\Delta T=20^\circ\text{C}\$$, so this adds $$\\frac{20\,\cdot\,2.1\:\text{mV}}{175\:\Omega}=20\cdot 12\:\mu\text{A}=240\:\mu\text{A}\$$ to $$\I_{_\text{C}}\$$.

And that makes their point as they write a "nearly 25%" increase in the collector current.

Ebers-Moll, though, provides some negative feedback to this change. In their example, they used $$\V_T=25\:\text{mV}\$$. Ebers-Moll says that the transistor responds to this change such that $$\\Delta V_{_\text{BE}}=V_T\cdot\ln\left(1+\frac{240\:\mu\text{A}}{1\:\text{mA}}\right)\approx 5.4\:\text{mV}\$$.

So start with the $$\20\cdot -2.1\:\text{mV}=-42\:\text{mV}\$$ change due to temperature, suggesting that previously computed $$\240\:\mu\text{A}\$$ increase in $$\I_{_\text{C}}\$$. But this is countered by an increase in $$\V_{_\text{BE}}\$$ due to that increased collector current, which increases $$\V_{_\text{BE}}\$$ and therefore reduces $$\V_{_\text{E}}\$$. So instead of a $$\42\:\text{mV}\$$ increase, there's really more like $$\36.6\:\text{mV}\$$ change, instead. And this means not quite so much of a collector current change -- perhaps 20%-21% instead of 25%.

(It doesn't stop there because this Ebers-Moll effect reduces $$\I_{_\text{C}}\$$, which means the previously computed alteration of $$\5.4\:\text{mV}\$$ that assumed no Ebers-Moll effect is also wrong. Since the new current is less, the $$\5.4\:\text{mV}\$$ is a little less, which means $$\I_{_\text{C}}\$$ is a little higher. Etc. You can iterate this over and over. Or you can compute it once using an equation involving the product-log/LambertW function. But it's really not worth all that effort. One step of iteration gets close enough. And the authors of The Art of Electronics felt that even this first step of correction due to Ebers-Moll wasn't worth applying -- for the very reason of just how much I just wrote here to even discuss it.)

While Ebers-Moll helps to reduce the impact of temperature changes, in this case to perhaps about 80% of what the authors of The Art of Electronics suggested in the text, it doesn't really change their main point. The high gain requires a low voltage drop across the emitter resistor, which makes temperature changes to $$\V_{_\text{BE}}\$$ more pronounced by comparison, which leads to greater changes in the quiescent operating point of the circuit due to operating temperature differences. That point remains.

Obviously, reducing the voltage gain would help because this would allow a larger voltage difference across the emitter resistor and then temperature changes to $$\V_{_\text{BE}}\$$ would have less impact on the quiescent state of the stage.

The authors then go on to suggest two other roughly equivalent ways to get a higher AC voltage gain while still keeping a substantial voltage difference across the emitter leg. These examples are in Fig. 2.50 and Fig. 2.51.

If Vbe reduces, it's because Vb has reduced and Ve has reduced slightly less.

The voltage at the base is the driving force.

• Ok so the general consus seems to be that when Vbe reduces (e.g. due to a temperature increase), Vb decreases rather than Ve increasing. However this seems to condradict what The Art of Electronics says about this example: "The problem is that the emitter voltage of only 0.175 volt will vary significantly as the -0.6 volt of base-emitter drop varies with temperature (-2.lmV per degree, approximately), since the base is held at constant voltage by R1 and R2" Commented Feb 8 at 23:16
• @Xavier No - your statement in the 1st two lines is not correct. See my comments to some other contributions.
– LvW
Commented Feb 9 at 12:54

Think about why Vbe would decrease. If the temperature is fixed then Vbe is constant, all other things being equal.

So to decrease Vbe we could reduce R2 or increase the 175Ω resistor, or maybe increase R1 or decrease the 10kΩ resistor.

Or, we could capacitively couple a negative-going signal to the base, which is really where this circuit is performing a useful function. In that case, Vbe will decrease and Ic will decrease. The emitter voltage will decrease too, but by less than base voltage decreases. This emitter degeneration reduces the gain and makes it more linear and less dependent on the transistor characteristics.

If the transistor was 'ideal' the gain with signal taken from the collector would be just the negative of the ratio of the collector to emitter resistors or about -57, and the emitter would have the same voltage change as the base. With a real transistor the numbers would be more like -8mV change on Ve for a -10mV step at the input, and about +0.5V change at the collector, so the -2mV Vbe change is what does the work.

• So you are saying that Vbe decreasing will decrease Vb, rather than increasing Ve? Does this mean that the passage from The Art of Electronics I have added in the edit is incorrect? Commented Feb 8 at 23:32
• Win Hill doesn't make many mistakes IME, and I don't disagree with the passage. If Vbe changes due to temperature (the dominant effect being the temperature dependency of Is, not Vt, BTW) then Ic will increase with decrease of Vbe. Ve will increase, and Vb will decrease but not nearly as much as Vbe decreases. Maybe a few percent. So the simplification that Vb is constant with temperature changes that appears in AoE is substantially correct. Commented Feb 8 at 23:53
• Yeah, I understand that Ic will increase due Is increasing, I just can't wrap my head around how that is compatible with an increase in Ve. What is the wrong with me saying "Ve increases, so Ie increases, so Ic increases", if we both agree that Ve is increasing? Commented Feb 9 at 0:14
• Perhaps I can solve a misunderstanding: Vbe does NOT "automatically" change with temperature. Vbe is always generated externally. However, the current Is is very temperature sensitive and causes an increase of Ic with rising temepratures. And the tempco of -2mV/K says that Vbe must be reduced by 2mV/K for keeping Ic constant. And that`s what a resistor RE in the emitter leg can do (approximately) because of Vbe=Vb-Ve. Hence, Vb should remain fixed (voltage divider).
– LvW
Commented Feb 9 at 8:01
• The use of simpler separate models for large and small signal analysis does have advantages when it comes to gaining intuition. Commented Feb 9 at 14:45

Since Vb is held at a fixed voltage by the biasing divider,

The base-emitter branch loads the voltage divider. So Vb cannot be fixed by the voltage divider. If Vbe decreases (perhaps from a temperature increase), Vb will decrease, Ib will increase causing Ic and Ie to increase. Ve will increase slightly, reducing the effect of the Vbe decrease.

• I have added an excerpt from The Art of Electronics to the post, which seems to sugggest that Vb would remain constant, whilst Ve will vary due to temperature changes. This goes against what you are saying (that Vb will decrease). Also how does Ve increasing slightly reduce the effect of the Vbe decrease, surely Ve increasing only further reduces Vbe? Commented Feb 8 at 23:28
• You are right - it is the aim to keep Vb constant and let Ve=IeRe react upon unwanted temperature changes. Hence Vbe reduces as a consequence of an Ie increase - and thus can keep such an unwanted change very small.
– LvW
Commented Feb 9 at 12:38

In normal or typical operation as an inverting gain stage Vb is not held at a fixed voltage. The voltage at the base is perturbed by the input signal.

"Ve will rise, meaning more current flows through Re."- No- Decreasing Vb (from an external input signal coupled through the capacitor) means less Vbe and less voltage across Re (175 ohm). Lower Ic, and less voltage across Rc (10 k), and higher Vc.

• What about is Vbe is reduced due to an increase in temperature? The relevent excerpt from The Art of Electronics I have added in the edit seems to suggest that Ve will vary as I described? Commented Feb 8 at 23:23
• Quote: "if Vbe were to decrease, then Ve will rise, meaning more current flows through Re." No - exactly the other wy round: If Ve will rise (due to an temperature-induced increase of Ie ), the voltage Vbe will be reduced with the aim to compensate this unwanted increase.
– LvW
Commented Feb 9 at 12:30
• Perhaps another point of misunderstanding: In the copied text from AoE they speak about a quantity re. Please keep in mind that this quantity is NOT something like an emitter resistor (causing feedback). More than that - it is not a resistor at all! It is another way of writing for the inverse transconductance (re=1/gm), which appears in the gain formula. I never use such a nomenclature - it can be (and often is) a source of misinterpretations and misunderstandings.
– LvW
Commented Feb 9 at 12:34