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VT = 25.8649606e-3                              # measured VT
ISAT = 10e-15                                   # model IS
def A(b): return 1/(1+1/b)                      # alpha factor
def Exp(v,vt): return exp(v/vt)                 # ~ exponential factor
def IC(vbe,vce,vt,isat,bf,br): return isat*Exp(vbe,vt)*(1-Exp(-vce,vt)/A(br))
def IE(vbe,vce,vt,isat,bf,br): return isat*Exp(vbe,vt)*(Exp(-vce,vt)-1/A(bf))
def IB(vbe,vce,vt,isat,bf,br): return isat*Exp(vbe,vt)*(1/bf+Exp(-vce,vt)/br)

sci( IC(.65,.1,VT,ISAT,200,20), 4 )
'802.4051E-06'

sci( IC(.65,0,VT,ISAT,200,20), 4 )
'-41.0221E-06'

sci( IC(.65,0.00126195567390989,VT,ISAT,200,20), 4 )
'-182.1746E-21'

VT = 25.8649606e-3                              # measured VT
ISAT = 10e-15                                   # model IS
def A(b): return 1/(1+1/b)                      # alpha factor
def IC(vbe,vce,vt,isat,br): return isat*exp(vbe/vt)*(1-exp(-vce/vt)/A(br))
def IE(vbe,vce,vt,isat,bf): return isat*exp(vbe/vt)*(exp(-vce/vt)-1/A(bf))
def IB(vbe,vce,vt,isat,bf,br): return isat*exp(vbe/vt)*(1/bf+exp(-vce/vt)/br)

sci( IC(.65,.1,VT,ISAT,20), 4 )
'802.4051E-06'

sci( IC(.65,0,VT,ISAT,20), 4 )
'-41.0221E-06'

sci( IC(.65,0.00126195567390989,VT,ISAT,20), 4 )
'-182.1746E-21'

There's another parlor trick, of sorts. Saturation can only be handled by the full 4-quadrant model, unlike active mode which only requires half of it.

Suppose \$V_{_\text{CC}}=3.3\:\text{V}\$ for an MCU and you need to ensure at or below \$V_{_\text{CE}}=100\:\text{mV}\$ with a collector load of \$160\:\text{mA}\$ that will be activated by an I/O pin and an NPN transistor. You can solve for the required \$V_{_\text{BE}}\$:

def VBE(ic,vce,vt,isat,br): return vt*ln(ic/isat/(1-exp(-vce/vt)/A(br)))
float( sci( VBE(0.16,0.1,VT,ISAT,20), 4 ) )
0.7869631         # required VBE
                  # note: bipolars are voltage-driven, not current driven
float( sci( IB(0.7869631,0.1,VT,ISAT,200,20), 4 ) )
0.0009892429      # required base current

(This would be \$\beta\approx 162\$.)

Remember, the expectation is no more than \$V_{_\text{CE}}=100\:\text{mV}\$, or \$\ge 3.2\:\text{V}\$ across the load at \$160\:\text{mA}\$. So the collector load can be treated as a \$\frac{3.2\:\text{V}}{160\:\text{mA}}=20\:\Omega\$ resistance.

Assume that the I/O pin will yield exactly \$3.3\:\text{V}\$ at \$0\:\text{mA}\$ and has a FET output resistance, when HI, of \$\approx 40\:\Omega\$.

Then it follows that the base resistor should be \$\frac{3.3\:\text{V}-786.9631\:\text{mV}}{989.2429\:\mu\text{A}}-40\:\Omega\approx 2500.364\:\Omega\$. Round that to \$2.5\:\text{k}\Omega\$ and the result should be very close.

I'll run both the exact and approximate values:

Again, quantitatively predictable.

Take note here. Normally, using a bipolar transistor as a switch (in deep saturation) isn't treated as predictable. Base current is usually just taken to be about \$\frac1{10}\$th of the collector current with a \$\beta\approx 10\$ assumption and the analysis stops there. It's easy.

But I want to point out that the full Ebers-Moll model is a powerful model and it just works right, even in saturation. It is only the full Ebers-Moll model that can handle saturation and active mode behavior (and reverse-active and cutoff modes, too), quantitatively. (It also, of course, clearly and precisely demonstrates that the bipolar transistor is a voltage-driven device. Not current-driven.)

It was and still remains a remarkable achievement.

Let's try an extreme test by setting \$V_{_\text{CE}}=0\:\text{V}\$:

The prediction is negative??? It's \$I_{_\text{C}}\approx -41.0221\:\mu\text{A}\$.

Let's try an extreme test by setting \$V_{_\text{CE}}=0\:\text{V}\$. In this case, the prediction is that the collector current will be negative. This is because the transistion from positive collector current to negative collector current (due to the differences in the two current generators) is at \$V_{_\text{CE}}=V_T\,\ln\left(1+\frac1{\beta_{\small{R}}}\right)=0.00126195567390989\approx 1.262\:\text{mV}\$. Below that value, expect a negative collector current:

The prediction is in fact negative. It's \$I_{_\text{C}}\approx -41.0221\:\mu\text{A}\$.

Lastly, just to verify that the collector current reaches zero at \$V_{_\text{CE}}=1.26195567390989\:\text{mV}\$:

sci( IC(.65,0.00126195567390989,VT,ISAT,200,20), 4 )
'-182.1746E-21'

Which is as close to zero as one may wish to see.

LTspice calculates Ic(Q1) = -4.09828e-011.

So this is also a match.

• \$I_{_\text{C}}=\alpha_{\small{F}}\,I_{\small{F}}-I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[1-\frac1{\alpha_{\small{R}}}\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)\right]\$
• \$I_{_\text{B}}=\left(1-\alpha_{\small{F}}\right)I_{\small{F}}+\left(1-\alpha_{\small{R}}\right)I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[\frac1{\beta_{\small{F}}}+\frac1{\beta_{\small{R}}}\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)\right]\$
• \$I_{_\text{E}}=\alpha_{\small{F}}\,I_{\small{F}}-I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)-\frac1{\alpha_{\small{F}}}\right]\$

• \$I_{_\text{C}}=\alpha_{\small{F}}\,I_{\small{F}}-I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[1-\frac1{\alpha_{\small{R}}}\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)\right]\$
• \$I_{_\text{B}}=\left(1-\alpha_{\small{F}}\right)I_{\small{F}}+\left(1-\alpha_{\small{R}}\right)I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[\frac1{\beta_{\small{F}}}+\frac1{\beta_{\small{R}}}\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)\right]\$
• \$I_{_\text{E}}=-I_{\small{F}}+\alpha_{\small{R}}\,I_{\small{R}}\approx I_{_\text{S}}\exp\left(\frac{V_{_\text{BE}}}{V_T}\right)\left[\exp\left(\frac{-V_{_\text{CE}}}{V_T}\right)-\frac1{\alpha_{\small{F}}}\right]\$