Solving Currents in BiCMOS Darlington Pair Using KVL and assuming that the BJT is in forward-active and that the MOSFET is under saturation and using the given, I was able to generate three equations with three unknowns:

1) $\frac{3V-2V}{1000Ω}=I_D+I_C$

2) $I_D=(10^{-3}\frac{A}{V^2})(V_{BE}-0.6V)^2$

3) $I_C=(0.1(10^{-15})A)(e^\frac{V_{BE}}{0.026V}-1)$

. The problem is that, based on my calculations, the above system yields:$V_{BE}=1.6V, I_D=0.001A, I_C=-4.8658(10^{-17})A$. Does this really mean that $I_C$ is negative and that the npn BJT $Q_2$ is actually not in forward active mode or is there something wrong with my method/equations?

• $I_D=I_C/\beta+V_{BE}/R_e ~~,~~ and ~~assume~~ V_{BE}=0.6V @Ib~10uA$ – Sunnyskyguy EE75 Feb 18 '17 at 18:21
• So what part is wrong with my equations? My equations require no assumptions. – John Smith Feb 18 '17 at 18:22
• 2) should it be? $I_D = k*(V_{GS} − V_{th} )* V_{DS}$ – Sunnyskyguy EE75 Feb 18 '17 at 18:30
• No, it's technically $I_D = k*(V_{GS}-V{TH})^2 (1+\lambda V_{DS})$ if operated in saturation mode (which is the usual transistor operation for MOSFETS) but $\lambda$ in this case is zero. – John Smith Feb 18 '17 at 18:33
• OK TY but Vbe MUST be <=0.6V at this low current so this affects everything else. – Sunnyskyguy EE75 Feb 18 '17 at 19:04

Well for this circuit we have $I_D + I_C = \frac{V_{CC}-V_{Out}}{R_L} = 1mA$ Additional we knows that $I_S=I_D=\frac{V_{BE}}{R_B}$ So we we assume $V_{BE} = 0.6V$ we have $I_D= 0.6mA$ and $I_C=0.3mA$ therefore $V_{BE} = V_T*ln\left(\frac{Ic}{Is}\right)= 0.7469V$ (I assume Vt = 26mV).

So we have a new value for Vbe, so, the new value for Id and Ic is:

$I_D = \frac{0.7469V}{1k} = 0.7469mA$

$I_C = 0.253mA$

so again we can find new value for $Vbe$

$Vbe =V_T*ln\left(\frac{Ic}{Is}\right)= 26mV *ln(\frac{0.253mA}{0.1fA}) = 0.74254V$ and the new $I_D = 0.74254mA ;I_C = 0.25746mA$ values.

And once more I repeat the iteration $Vbe = 26mV *ln(\frac{0.25746mA}{0.1fA})=0.742995V$

$I_D = 0.742995mA ;I_C = 0.257005mA$

The new $Vbe$ value is $Vbe = 0.742949V$

At this step, I end the iteration process and conclude that $Vbe = 0.7429V$.

And $I_D = 0.7429mA$ and $I_C=0.2571mA$

Since we know the MOS drain current $I_D$ we can find $Vgs$

$V_{gs} = V_{th}+\sqrt{\frac{I_D}{0.5k}} = 0.6V+\sqrt{\frac{0.7429mA}{0.5m}} =1.81893V$

And finally $V_{BIAS} = V_{BE}+V_{gs} =0.7429V+1.81893V = 2.56183V$

In all this, calculations I ignore the BJT base current.

EDIT

To get exact solution you need to solve this:

$$I_C = 1mA - \left(\frac{I_C}{\beta}+\frac{Vbe}{1k}\right);I_C = 1*10^{-16}*e^{\frac{Vbe}{V_T}}$$

And if I plug this into computer I get $V_{BE} =0.742718V; I_C=0.254735mA$

• To neglect Ib when it is only 2.5uA seems OK but i would expect Vbe to then be around 0.55V to 0.6V from experience and never 0.74V. Why? – Sunnyskyguy EE75 Feb 18 '17 at 18:49
• @TonyStewart.EEsince'75 Normally for Ic<1mA most BJT will have Vbe around 0.6V this is why. – G36 Feb 18 '17 at 19:21
• Not 0.74 and actually closer to 0.55V at Ib= 2.5uA , why 0.74? – Sunnyskyguy EE75 Feb 18 '17 at 19:56
• I guess the unusually large Vbe is due to a very tiny junction with an unusually small reverse saturation current , $I_S= 0.1fA$ rather than typical ranges for discrete BJT's of 1 fA~10pA So 0.74V is theoretically correct – Sunnyskyguy EE75 Feb 18 '17 at 22:14
• Shouldn't the equation be $V_{gs} = V_{th}+\sqrt{\frac{I_D}{k}}$? Why do you use $0.5k$ in the denominator? Maybe your definition of k is actually twice my definition of k. Here's a reference, check equation 6: leachlegacy.ece.gatech.edu/ece3050/notes/mosfet/mosfet2Rev.pdf – John Smith Feb 19 '17 at 5:17