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Solve: $$U_1=?\ ,U_2=-U_3=?,U=?$$

I tried to solve this circuit with Shockley Diode Equation : $$I_{\mathrm{D}}=I_{\mathrm{SAT}}\left(e^{\frac{U_{\mathrm{D}}}{\eta^{U_{\mathrm{T}}}}}-1\right)$$

and also with equation for dynamic resistance(for this example its conductance, inverse of dynamic resistance) : $$G=\frac{I_D+I_S}{I_S}$$

Correct me if I'm wrong:

$$from \ picture \ we \ see \ that \ : I_{D1}=I_{D2}+I_{D3} $$ $$from \ picture \ we \ see \ that \ : U_2=U_3$$

enter image description here

Results are: $$U_1=0.182(V)\ ,U_2=-U_3=0.2(V),U=-0.382(V)$$

How can be this circuit solved?

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  • \$\begingroup\$ All the currents go from + to - so \$ I_{D1} = I_{D2} + I_{D3} \$. Since D2 and D3 are sharing common nodes then \$ U_2 = U_3 \$. \$\endgroup\$
    – Transistor
    Commented Jun 9, 2020 at 3:53
  • \$\begingroup\$ @Transistor I corrected it, still cant get results. \$\endgroup\$
    – Hury H
    Commented Jun 9, 2020 at 4:30
  • \$\begingroup\$ Assuming you ground (make it \$0\:\text{V}\$) the (-) end, the voltage in the middle will be \$V_T\cdot\ln\left(\frac12\left[1+\sqrt{5}\,\right]\right)\$. It's pretty much the only possible value, independent of the voltage at (+). You should be able to work this out entirely from the Shockley equation. This is because this is the only possible value, \$x\$, where \$e^{^\frac{x}{V_T}}=1+e^{^\frac{-x}{V_T}}\$. \$\endgroup\$
    – jonk
    Commented Jun 9, 2020 at 5:42

1 Answer 1

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You are always allowed to select a node and call it \$0\:\text{V}\$. So, assuming you ground (make it \$0\:\text{V}\$) the (-) end, as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

Then \$V_X=\eta\,V_T\cdot\ln\left(\frac12\left[1+\sqrt{5}\,\right]\right)\$. It's pretty much the only possible value, unless the (+) voltage at the top is significantly less than \$10\cdot V_T\$. So, for any applied voltage, \$V\ge 10\cdot V_T\$ (about \$250\:\text{mV}\$ or more), you are just solving \$x\$ where \$e^{^\frac{x}{V_T}}=1+e^{^\frac{-x}{V_T}}\$.

You can work this out entirely from the Shockley equation.

$$\begin{align*} I_{_{D_1}}&=I_{_{D_2}}+I_{_{D_3}}\\\\ -I_{_\text{SAT}}\left(e^{^\frac{-\left(V-V_X\right)}{\eta\,V_T}}-1\right)&=I_{_\text{SAT}}\left(e^{^\frac{V_X}{\eta\,V_T}}-1\right)-I_{_\text{SAT}}\left(e^{^\frac{-V_X}{\eta\,V_T}}-1\right)\\\\ -e^{^\frac{-\left(V-V_X\right)}{\eta\,V_T}}+1&=e^{^\frac{V_X}{\eta\,V_T}}-e^{^\frac{-V_X}{\eta\,V_T}}\\\\ -e^{^\frac{-V}{\eta\,V_T}}e^{^\frac{V_X}{\eta\,V_T}}-e^{^\frac{V_X}{\eta\,V_T}}+e^{^\frac{-V_X}{\eta\,V_T}}+1&=0\\\\ -\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)e^{^\frac{V_X}{\eta\,V_T}}+e^{^\frac{-V_X}{\eta\,V_T}}+1&=0\\\\ -\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)\left[e^{^\frac{V_X}{\eta\,V_T}}\right]^2+e^{^\frac{V_X}{\eta\,V_T}}+1&=0\\\\\hline\\ \text{set }y&=e^{^\frac{V_X}{\eta\,V_T}}\\\\ \text{quadratic solution:}\quad y &=e^{^\frac{V_X}{\eta\,V_T}}=\frac{1\mp\sqrt{4\,e^{^\frac{-V}{\eta\,V_T}}+5}}{2\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)}\\\\\therefore V_X&=\eta\,V_T\,\ln\left(\frac{1\mp\sqrt{4\,e^{^\frac{-V}{\eta\,V_T}}+5}}{2\left(e^{^\frac{-V}{\eta\,V_T}}+1\right)}\right)\\\\\hline\\\text{for }V\ge 10 \cdot V_T\text{, }&e^{^\frac{-V}{\eta\,V_T}}\to 0\text{, and }V_X\to \eta\,V_T\,\ln\left(\frac{1+\sqrt{5}}{2}\right) \end{align*}$$

So, using \$V_T=26\:\text{mV}\$, this is \$V_X\approx \eta\cdot 12.51151\:\text{mV}\$. LTspice shows \$\eta=1.752\$ for the 1N4148 model. This computes out to \$V_X\approx 21.9\:\text{mV}\$. LTspice computes \$V_X=21.886\:\text{mV}\$ using a \$+10\:\text{V}\$ rail. Theory gets you close enough.

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