CALCULATING THE INTERNAL BASE-EMITTER RESISTANCE OF A TRANSISTOR.

by David Cline

In the field of electronics engineering, the internal base-emitter resistance of a transistor is usually not significant enough to insert into design calculations. However, in some cases this small amount of resistance can have a noticeable affect on circuit operation - as is the case with a Common Collector amplifier, or unity gain amplifier. As we shall see, the internal base-emitter resistance plays a key role in determining the actual gain of a Common Collector amplifier. Before we go into base-emitter resistance calculations, I want to examine the impact that the base-emitter resistance will have on the Common Collector Amplifier.

For a unity gain amplifier, you would expect the voltage gain to be "one" as calculated from the output voltage divided by input voltage or the emitter voltage divided by the base voltage.

AV = Vout / Vin = 1

However, in reality, the voltage gain is slightly less than one (typically around 95%). At first I thought that this was attributed to the base-emitter threshold voltage (typically around .7 volts between the emitter and base) which does cause a voltage offset between emitter and base. Then I realized that the voltage offset simply means that the base voltage will just ride a little higher than the emitter voltage. Variations in the output emitter voltage should still precisely follow, in the same magnitude, as variations in the input voltage at the base, no matter what their respective quiescent voltages are. For instance, if the input is increased by 1 volt, then the output should be increased by 1 volt. However, the output will always swing slightly less than the input. What is really going on is that part of the unity gain is being eaten up by some intrinsic resistance within the transistor. This resistance is called the base-emitter resistance, which we shall designate as
Rbe.
The added resistance of Rbe prevents the load resistor RL from inheriting the full voltage drop. In other words, Rbe steals a tiny bit of the voltage drop and reduces the unity gain to slightly less than one. Since RL would have received the full voltage drop (full unity) if it weren't for that pesky Rbe, it's reasonable to assume that we can calculate the actual gain by dividing the "almost full" voltage drop experienced by RL by the full voltage drop experienced by the sum of RL and Rbe.

AV = VRL / VRL + V
Rbe

Problem is, we can't know the voltage drop of
Rbe quite yet. However, because Ohm's Law (I = V/R) is a linear equation, we can calculate the gain by using the resistance ratio instead of the voltage ratio.

AV = RL / RL +
Rbe

But wait, how do we know what the value of
Rbe is? Well, just go ask any Electronics Engineer. They, having been endowed with the magical powers of mystical insight, will always tell you that the internal base-emitter resistance is about 25 or 26 ohms at 1 ma of current. It doesn't seem to matter whether you're talking about the small signal 2N2222 or the high-powered 2N3055 transistor, the base emitter resistance is always 26 ohms (under certain conditions). Why is that? Even though there may not be much practical use in answering this question, the answer to this question is still very interesting. Here's what I found:

The base-emitter resistance is determined by a principle called "thermal voltage". Apparently this voltage is produced within the PN junction (base-emitter junction) due to thermal energy acting on the energy states of electrons. This means that thermal voltage is determined soley by the ambient temperature. Its value is not specific to any type or brand of transistor. We can even verify this with a few calculations. There is a way to calculate the thermal voltage at any given temperature. Once we do that, we can calculate the base-emitter resistance. Here are all the required steps:

We first have to figure out how much thermal energy (in Joules) is contained in a single atom per degrees Kelvin.

Once we know the thermal energy exhibited by a single atom per Kelvin, we have to translate this value from thermal Joules to "electrical" Joules. This will tell us the thermal voltage per particle per Kelvin.

After we have translated the energy from "heat" to "volts", we want to know what this thermal voltage is going to be at room temperature.


Once we have determined what the exact thermal voltage is, we can then use ohm's law to calculate the value of the base-emitter resistance at any given ambient temperature.


In order to figure out how much energy is contained in a single atom at a certain temperature we first have to observe the energy content in a much larger mass of matter. It doesn't matter what kind of matter (did I just make a pun?). It only matters that we know how many atoms are in this mass. We know that a mole of matter contains 6.02214179(30) x 10^23 atoms, also known as the Avogadro Constant (the number in parentheses represents the uncertainty in the last two digits of the measured value). It has been measured that one mole of matter (at the ideal gas constant, which we won't go into) contains 8.314472(15) Joules per Kelvin. This is referred to as the Gas Constant "R". It means that for each degree Kelvin, there is about 8 Joules of energy. So a mole of matter at two degrees Kelvin (that is dang cold!) contains about 16 and a half Joules. To figure out how much energy there is in a single atom from a mole of matter we have to divide the Gas Constant "R" by the Avogadro Constant "NA". In other words, we have to divide the total energy content by the number of particles that contribute to this energy. So here we go:

R / NA = 8.314472(15) / 6.02214179(30) x 1023 = 1.3806504 x 10^-23 Joules per Kelvin

This number, in fact, is called the Boltzmann's Constant, (usually designated as "k"). It basically states how much thermal energy (in Joules) a single particle (atom) contains at a given temperature in Kelvins.
Now that we know the thermal energy per particle per Kelvin, we would like to translate this in terms of electrical energy. Well, how do we do that? We first have to define what a Joule is. A Joule is simply a unit of energy - it can be any kind of energy. One Joule can be: 

0.24 food calories, which is a tiny pinch of white granulated sugar (.016 teaspoon to be exact) or even a little tiny doughnut crumb.

The amount of energy it takes to lift one apple 3 feet straight up into the air from the ground (Say what! A tiny pinch of sugar can do that?).

The amount of energy needed to raise the temperature of one gram of water one degree Celsius.

The amount of energy that is equivalent to 6.2415097410^18 electron volts.

Only the last example is relevant to our case. O.K. so there are lots of electron volts in one Joule, how many are there in a tiny fraction of a Joule, as expressed by Boltzmann's Constant. To find out, we just multiply Boltzmann's Constant by the number of electron volts in a Joule, which we shall designate as "E".

k x E = (1.3806504 x 10^-23) x (6.2415097410^18) = 8.617343(15) x 10^-5 Electron volts per Kelvin

Another way we could do this is to divide Boltzmann's Constant by the elementary charge of a single electron. This will give
us Boltzmann's Constant expressed in terms of electron volts instead of in Joules - same value of energy, just different expression. One electron exhibits 1.602176487 x 10^-19 volts. We'll designate this single electron as "e".

k / e = (1.3806504 x 10^-23 / 1.602176487 x 10^-19) = 8.617343(15) 10^-5 Electron volts per Kelvin.
See, we get the same answer as before. We'll designate the result as "kB"

Now that we have expressed Boltzmann's Constant as volts per Kelvin, all we have to do now is multiply the answer by the number of Kelvins (which is 300 at room temperature, i.e., 80 degrees F). We'll designate Kelvin as "T"

kB x T = (8.617343(15) x 10?5) x 300 = .025852029 Volts

We've finally solved the thermal voltage. We'll call it VT. Now we just use ohm's law to figure out what the base-emitter resistance is with an emitter current of 1ma.

Rbe = VT / IE = .026 / .001 = 26 ohms

Yay! We did it!
2011 David Cline. All Rights Reserved