# Vbe/Ic equations

#### #12

Joined Nov 30, 2010
18,224
Annoyed by the inaccuracy of people saying, "It takes six tenths of a volt to turn a transistor on", I did a graph of a bipolar transistor (2N4250A) to see what happened. Some people say, "sixty millivolts per log" or, twenty some millivolts per doubling". One answer is a graph that shows an almost perfect straight line from 1na to 1 ma on a log/log graph. (I quit at 1 ma to avoid heating the transistor up.) I assume the line continues into the higher current ranges except for heat affecting the results.

Equations can be derived from this graph, but they won't be the same for every transistor in every range of current. They will be close for use as a relative calculation, as in, "new Vbe/ old Vbe = something something Ic2/Ic1".

The most interesting part, to me, is that the results go below a nanoamp while barely changing the slope of the line. Extrapolating would take the Vbe to zero when the current (Ic) is a tenth of a pico-amp. That isn't even as high as the advertised leakage for that transistor.

I had to attach the graph as a pdf because it takes over a megabyte to scan it well and there is an upload size limit on this site.

What say you as to the "standard" equations? Can you write them for me in your answer? I'm making a blog of this so I can point people to it when the subject comes up.

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#### t_n_k

Joined Mar 6, 2009
5,455
Not sure what you mean by "standard equation". Are you referring to the unapproximated Ebers-Moll equations which should explain your observations?

#### bountyhunter

Joined Sep 7, 2009
2,512
The standard equation from days of old is:

delta VBE = kT/q ln (I1/I2)

the term kt/q is usually approximated as .026.

The absolute value of a VBE at a given current varies based on many factors like doping, current density, temp, etc. But that equation approximates how the VBE changes with current.

#### #12

Joined Nov 30, 2010
18,224
Thanks. That's the one I was trying to remember.
Using .026 makes it come out as .05987 deltaV/log (base 10) of current change. Sixty millivolts is close enough.

The graph I made was actual measurements with a little bit of fudging when a single point didn't fall on the line due to measurement errors. You can see that this transistor, on this day, averaged .065 volts per log across 6 logs.

Any others I missed/forgot/never heard of?

#### WBahn

Joined Mar 31, 2012
28,182
I'm trying to figure out what it is you are annoyed about or what it is you would have people say differently instead.

In typical applications, would it be reasonable to assume that the collector current would likely be somewhere in the 100μA to 100mA range? If so, then that corresponds to a Vbe of 550mV and 750mV. So it would seem like saying that Vbe is about 0.6V or 0.7V when the transistor is "on" is not at all unreasonable.

The geereal Ebers-Moll equation is:

$$i_c\;=\;I_S \left( e^{\left( \frac{v_{be}}{\eta V_T} \right) } - 1 \right)$$

where V_T is the thermal voltage and is equal to

$$V_T=\frac{kT}{q}$$

where k is Boltzmann's constant, T is temperature, and q is the magnitude of the carrier charge.

The η is a "quality parameter" and is generally pretty close to 1, particularly in modern processes.

I_S is the "saturation current" and is typically in the picoamp range but is very seldom actually characterized.

But, like any such mathematical model, this is only an approximation of the actual BJT behavior. In most situations it is more than good enough. But if you are operating at a significantly different temperature or are specifically looking for differences due to temperature, then you need to account for the fact that the saturation current is a function of temperature.

More commonly, you need to account that a BJT has a finite output impedance that results in the collector current, at a constant Vbe, being a function of the collector-emitter voltage. In the active region this effect is generally fairly linear of reasonable collector currents and the slopes extrapolate back to roughly the same intercept on the Vce axis at the point -V_A where V_A is known as the Early voltage. You are possibly seeing this in your measurements because as you increase the collector current in your circuit the collector-emitter voltage is dropping and, as a result, it takes more base-emitter voltage to make up for dtop in collector current that would normally result. The higher the current, the larger the effect. So you expect to see a higher incremental voltage per decade at higher collector currents, which appears to be what you are seeing.

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#### #12

Joined Nov 30, 2010
18,224
I finished the blog. You can let this thread die now.

Thank you.

#### wayneh

Joined Sep 9, 2010
17,201
So it would seem like saying that Vbe is about 0.6V or 0.7V when the transistor is "on" is not at all unreasonable.
+1
Every rule has exceptions, but this rule is a pretty good one.

#### #12

Joined Nov 30, 2010
18,224
and yet, I did an answer just a couple of days ago where the Vbe was .9 volts.

I'm just writing a blog that explains that there is no on/off point for a bipolar transistor. There is no sudden change in Ic at a particular Vbe. There is no (practical) lower limit where the Ic/Vbe is not proportional. If you need to work below 10ua, the transistor will work just the same as it does above a milliamp. This removes an area where doubt might happen to a beginner that has been taught to expect .6Vbe to .7Vbe for all circuits.

You can argue for and against this all you want. It won't change the facts of how a transistor works and it won't change the fact that most educators do not explain this properly to their students. If the question comes up, I can point them to the answer.

#### WBahn

Joined Mar 31, 2012
28,182
I'm not aware of any region of operation in whch Ic is proportional to Vbe.

Like any rule of thumb, the rule that Vbe for a silicon BJT in the active region is a constant is an approximation that is useful over a wide range of applications that cover a large fraction of the total usage. Also like any rule of thumb, it is not universal or absolute and plenty of exceptions can be found. I've worked on CMOS IC designs in which the design fundamentally depended on operating transistors in the deep subthreshold region.

I absolutely agree that students, particularly engineering students, should be expected to understand and be able to work with the basic constitutive relations for diodes, transistors, and other nonlinear devices. But the emphasis does need to be proportionate to the utility. Do you really want an engineer or technician that is NOT going to rely on the constant diode voltage drop model the overwhelming majority of the time? If anything, that would be worse, from the standpoint of actually getting anything done, than the engineer that doesn't comprehend that you can't always neglect the fact that base current is actually non-zero.

While there certainly are educators that don't explain things like this properly (or perhaps even at all), I believe it is more commonly the case, at least in engineering education, that the material IS covered, but that it tends to be covered as part of a theoretical background and/or introduction and students are only required to actually work with concepts at this level for a relatively short amount of time. As a result, the knowledge doesn't get engrained and gets purged.

There are those that would argue, with at least some merit, that time spent teaching something only to a level that is going to result in it getting purged is time wasted that could have been spent teaching more "useful" things. On issues of fundamental concepts, my attitude is that, when that is the case, what we need to do is figure out ways to reinforce those concepts so that they aren't purged but yet so that the amount of time spent on them is not increased significantly. One simple way to do this is to salt the curriculum with occasional problems that require the application of the fundamental concepts. Students don't have to revisit a concept very often in order for them to realize that it is worth understanding.

#### wayneh

Joined Sep 9, 2010
17,201
Wisdom is knowing both the rule and the exceptions where the rule doesn't work.

Learning only the rules reminds me of grade school education, whereas a decent college or graduate education should include more about what's going on at the exceptions. I learned the Arrhenius equation in high school but later learned how to derive it in college. Knowledge of the exceptions to a rule gives a much deeper understanding, closer to what a practitioner in the field would have learned by experience.

With any tool, knowing when it's the right tool for the job is what makes you the artisan.

#### wayneh

Joined Sep 9, 2010
17,201
If you need to work below 10ua, the transistor will work just the same as it does above a milliamp. This removes an area where doubt might happen to a beginner that has been taught to expect .6Vbe to .7Vbe for all circuits.

#### bountyhunter

Joined Sep 7, 2009
2,512
If you need to work below 10ua, the transistor will work just the same as it does above a milliamp.
Not sure if that's true. 10uA is actually what some transistors have for rated leakage current, especially at hotter temps. There are a lot of intrinsic gotchas that bite at very low currents, so the operation gets distorted.

#### tindel

Joined Sep 16, 2012
909
Interesting discussion.

Thanks for reminding me of the deltaVbe equation - I had forgotten about the equation. At my work we have to worst case these things over temperature, end of life, and initial tolerance at a certain operating point. Thankfully, we have a group that spends their entire career determining these values so that I don't have to (tough job IMHO) - interestingly enough it should come as no surprise that over worst case many bjt's saturated Vbe can range from as low as .2V to 2V! Many aren't that bad, but when I'm analyzing my circuits this is usually the ball-park that I try to analyze to for an initial ballpark so I don't have to worry too much about worst case. Of course as you push more current through the Vbe junction the voltage goes up - it's pretty intuitive when you think about a diode IV curve.

On a somewhat related topic (I'll start a new thread, if you wish, #12 as to not hi-jack yours):
What I'm having trouble with these days is how low of a voltage is low enough to guarantee that your transistor turns off. Does the base voltage really need to be below the emitter voltage? I would think not as most of the time when saturating a transistor as a switch the emitter is tied to ground and the input can only go as low as ground... but the collector emitter cutoff current (Icbo, usually) can be quite high over temperature. So if your drive resistor is high (100k) so that you consume little energy when on - if your collector-base leakage is 1uA worst case, you can have 0.1V on the input of your transistor - is that low enough to know it will remain off?

#### WBahn

Joined Mar 31, 2012
28,182
This "Δvbe" equation isn't something that people should be trying to remember. Of course, if you use it all the time then you will remember it, but it is trivially derivable from the diode equation. But for most things all you need to recall is that it is about 60mV/decade at room temperature and that it is proportional to temperature.

$$i_d\;=\;I_S e^{\frac{v_d}{\eta V_T}}$$

The "-1" has been neglected since it is negligable at anything above about v_d=0.1V.

So pick two currents, I_1 and I_2, corresponding to V_1 and V_2, respectively, and define ΔV_d to be V_2-V_1:

$$I_1\;=\;I_S e^{\frac{V_1}{\eta V_T}} I_2\;=\;I_S e^{\frac{V_2}{\eta V_T}} \Delta V_d\;=\;V_2 - V_1$$

Now take the ratio:

$$\frac{I_2}{I_1}\;=\;\frac{I_S e^{\frac{V_2}{\eta V_T}}}{I_S e^{\frac{V_1}{\eta V_T}}} \frac{I_2}{I_1}\;=\;e^{\frac{\Delta V_d}{\eta V_T}}$$

The big thing to note here is that the saturation current drops out, so as long as I_1 and I_2 are at the same temperature, the fact that the saturation current is quite temperature-dependent is immaterial.

Finally, take the ln() of both sides:

$$\ln \left( \frac{I_2}{I_1} \right) \;=\;\frac{\Delta V_d}{\eta V_T} \Delta V_d\;=\;\eta V_T \ln \left( \frac{I_2}{I_1} \right)$$

The ideality (or quality) factor, η, is usually assumed to be close enough to 1 to ignore the difference.

At room temp (300K, which is "room temp" for most device characterization), the thermal voltage is 25.85mV. So if you want a decade change, then I_2 = 10*I_1 and you have:

$$\Delta V_d/\text{decade}\;=\;25.85mV \cdot \ln(10)\;=\; 59.52mV/\text{decade}$$

Given the approximation with the ideality factor, it makes no sense to use this value to any greater precision than is afforded by 60mV/decade (which would correspond to an ideality factor of 1.01).

If you want the voltage change per doubling (an octave), the just change the ln(10) to ln(2) and you get 17.92mV/octave. Again, you can use 18mV/octave but 20mV/octave is completely reasonable since this would, in effect, just be assuming that the ideality factor is 1.12.

By understanding where these relationships come from you can very easily get the result for a different temperature without plugging recalculating the thermal voltage at the new temperature because the result is directly proportional to the thermal voltage which, in turn, is directly proportional to the temperature. So if you wanted the result at T=-10°C, you just scale it appropriately:

$$\Delta V_d/\text{decade}\;=\;\left( 60mV/\text{decade} \right) \cdot \frac{263.15K}{300K}=52.63mV/\text{decade}$$

This property is used to make pn-junction thermometers (or diode thermometers). You simple run some reference current through one diode and some multiple of that current and then amplify the voltage difference between them. One of the beautiful things about this is that it doesn't matter if the reference current changes with temperature, which it almost certainly will.

The current mirror multiplier doesn't have to be 10. If you use 3.19 then you get a differential voltage of 100μV/K. Since you want to have the scaling constant be rational so that you can use identical transistors in your mirror, you could use 100 transistors for the reference side and 319 on the output side. This may sound like a lot of transistors, but for critical bias generators and other things where matching is important, using dozens if not hundreds (or even thousands) is not that uncommon as it allows you to significantly reduce the impact of short-reach process variations by means of a host of layout tricks.