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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?

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  • \$\begingroup\$ \$I_D=I_C/\beta+V_{BE}/R_e ~~,~~ and ~~assume~~ V_{BE}=0.6V @Ib~10uA\$ \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 '17 at 18:21
  • \$\begingroup\$ So what part is wrong with my equations? My equations require no assumptions. \$\endgroup\$ – John Smith Feb 18 '17 at 18:22
  • \$\begingroup\$ 2) should it be? \$I_D = k*(V_{GS} − V_{th} )* V_{DS}\$ \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 '17 at 18:30
  • \$\begingroup\$ 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. \$\endgroup\$ – John Smith Feb 18 '17 at 18:33
  • \$\begingroup\$ OK TY but Vbe MUST be <=0.6V at this low current so this affects everything else. \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 '17 at 19:04
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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 \$

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  • \$\begingroup\$ 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? \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 '17 at 18:49
  • \$\begingroup\$ @TonyStewart.EEsince'75 Normally for Ic<1mA most BJT will have Vbe around 0.6V this is why. \$\endgroup\$ – G36 Feb 18 '17 at 19:21
  • \$\begingroup\$ Not 0.74 and actually closer to 0.55V at Ib= 2.5uA , why 0.74? \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 '17 at 19:56
  • \$\begingroup\$ 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 \$\endgroup\$ – Sunnyskyguy EE75 Feb 18 '17 at 22:14
  • \$\begingroup\$ 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 \$\endgroup\$ – John Smith Feb 19 '17 at 5:17

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