bitrex said:
Someting I don't understand is, since Ies is such a small number, how does the Ebers Moll model accurately describe the emitter current? I mean, it's possible to have transistors with many milliamperes or amps flowing through them, and I don't see how you get those numbers when you're multiplying everything by 10^-16 or some such.
Edit: Also, if you have a chance - how would the above equations be different when the transistor is not in the active region, but in saturation with both the base to emitter and base to collector diode forward biased?
Good questions. Regarding the Ies value of 1e-16 to 1e-11, here it is. For a signal device, if Ies = 1e-15 (1.0 femtoamp), with Vbe = 0.65V, and T (temperature = 298 K (25 C), then Vt = nkT/q, where n = 1.0 in this region of operation, k = Boltzmann's constant = 1.3806e-23 joule/Kelvin, & q = 1.602e-19 C (charge on 1 electron), so that Vt = 0.0257 V, or 25.7 mV.
So if alpha = 0.99 typ, then Ic = 97.19 uA. If Vbe = 0.60V, then Ic = 13.87 uA. For a heftier device, if Ies = 1e-12 (1.0 pA), then at Vbe = 0.60V, Ic = 13.87 mA, etc.
The Ies values given in the fA & pA range are accurate in the operating region of uA and mA of collector current. The reason is as follows. The thermal voltage, Vt = nkT/q, is proportional to absolute temp, and "n", a factor which varies as current changes.
When Ic is in the uA to mA range, n is very close to 1. But at low values of current, below a uA, n increases, eventually reaching 2 in the nA region. This is explained in semiconductor physics. In this region, the E-M equations are not as accurate.
The exponential/logarithmic relation between I & V in a p-n junction, holds for about 5, or maybe 6 decades of current for a bjt, and up to around 10 decades for certain diodes. With a bjt there are 2 phenomena affecting the relation.
The first was just mentioned, the variation of "n" at low current values. Another is the c-b junction leakage current. Let's say that Ics is 1.0 nA. Even though the b-e jcn value of leakage, Ies, is 1.0 pA, the Ic minimum value is 1.0 nA. Even though the b-e junction could be at 0 V, there is always the leakage from c to b due to Vcb & Ics. This spoils the log-linear relation between Ic & Vbe, as well as Ie & Vbe. The E-M equations account for the c-b leakage Ics, with the 2nd term. This is why the log-linear function for I vs. V does not extend down into the pA or fA region.
But, if we hold the Ic value in the uA to mA region, or even amps for a power device, using an Ies value in the fA to pA range, produces good results. The slope of the I-V curve is linear from let's say 100 nA to 10 mA of Ic, or 5 decades of current. If Ies is 1.0 fA, that is way below the straight line region of the I-V curve. Below 100 nA of Ic, the curve flattens out. It is not log-linear way down to 1.0 fA at all. So why can we use the 1.0 fA value if the Ic-Vbe graph does not linearly extend down that low?
The answer is that a line drawn from Ies = 1.0 fA upward extends into the nA, uA, & mA region with good accuracy. From Ic = 100 nA until 10 mA, the lines nearly coincide. Below 100 nA, the actual Ic curve is greater than that predicted by E-M based on Ies = 1.0 fA.
But we seldom operate a bjt w/ fA or pA of Ic. So the error is not a big deal. Spice models for the bjt utilize Ies values in the fA to pA range with good results.
The Ies value of 1.0 fA or pA does not represent an actual physical current associated with the device. It is a scaling factor derived from doping density, junction geometry, and other intrinsic material properties. If the I-V curve was log-linear all the way down to the Ies value, then Ies is the forward current when the (exp(Vbe/Vt) - 1) factor is equal to 1. That occurs when exp(Vbe/Vt) = 2, so subtracting 1 leaves 1. Multiply this factor by Ies, and we get Ies. That is the physical meaning of Ies.
Ies is the current which would flow at Vbe = Vt*ln2 = 17.8 mV,
if the device was true log-linear down to this value of I/V.
As far as operation in the saturated region goes, yes, E-M covers that.
Let's say that we use Vbe = 0.60V again, w/ Ies = 1.0 pA, but we are in the active region. Since Vbc < 0, the 2nd term in E-M can be ignored. The exp(Vbc/Vt) is nearly 0. For example, if the collector is at 1.0V above the base, Vbc = -1.0V, and exp(Vbc/Vt) =
1.23e-17! This is virtually 0, and exp() - 1 =
-1. So, the 2nd term is simply -(-Ics), or Ics, the c-b jcn leakage current.
Ic is then alpha_n*Ies*(exp(Vbe/Vt) - 1), which equals 13.87 mA. But if we increase the collector resistor so that saturation takes place, we now must deal with that 2nd term.
Let Vce = 0.10V, the device is fully on hard. Vbc = Vbe - Vce = 0.60V - 0.10V = 0.50V. Let's say that the Ics value is 20 pA. The 2nd term of E-M gives
Ics*(exp(Vbc/Vt) - 1) = 5.71 mA.
But the E-M eqn has the 2nd term
subtracted from the 1st term. Hence Ic = 13.87 - 5.71 = 8.16 mA.
The actual Ic value is less in saturation, and E-M predicts the same. So E-M covers both active and saturated modes of operation. In active regions, the 2nd term is not needed, as it is small. But in saturation, we must use both terms.
Does this make sense?
Claude