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1. First example:

$$U=68mV\\ U_1=?\\ U_2=?\\ I_{s1}=I_{s2}=10pA\\ I_{s3}=20pA\\ U_T=25mV $$ Is=reverse current, and UT=termic tension. For this one I thought that because the D3 diode is reverse biased no current flows but then there's that Is3. If the problem wasn't there I would use the formula:

$$I_d=I_s(e^\left(\frac{U}{UT}\right) -1)$$

And find the Id from there. But how can I find the tension for D1 and D2, and since D2 is parallel to D3 is their tension the same. I don't know if I'm wrong but that's what I'm concluding from this one.

2. Second example:

The diodes have different characteristics. These three are known: $$I_{s1}\,I_{s2}\,I_{in} $$

Find: $$I_{D1}=f(I_{in}\,I_{s1}\,I_{s2})\\ I_{D2}=f(I_{in}\,I_{s1}\,I_{s2})$$ I'm currently solving some examples from the book but these two are very different from the others that I did.

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  1. The correction helped a ton a simplifies the problem though complicates the math a bit. So for simplification \$I_{s3} = 2I_{s1} = 2I_{s1} = 2I_{s}\$. Considering \$D_3\$ to reverse biased then the KCL at junction of D1, D2 and D3 is given by $$ I_{D1} = I_{D2} + I_{D3} $$ and yes the voltage across D2 and D3 are same hence we know $$ U = U_1 + U_2 = 68mV$$ or $$ U2 = 68mV - U_1$$

Therefore the current equation can be re-written using the diode current equation:

$$ I_{s}(e^{\frac{U_1}{U_T}}-1) = I_{s}(e^{\frac{U_2}{U_T}}-1) +2I_{s}(e^{\frac{U_2}{U_T}}-1) $$

divide by \$I_s\$

$$ (e^{\frac{U_1}{U_T}}-1) = (e^{\frac{U_2}{U_T}}-1) + 2 (e^{\frac{U_2}{U_T}}-1) $$ or,

$$ (e^{\frac{U_1}{U_T}}-1) = 3e^{\frac{U_2}{U_T}}-3 $$

substituting \$ U_2 = 68mV-U_1\$;

$$ e^{\frac{U_1}{U_T}}-1 = 3e^{\frac{68m-U_1}{U_T}}-3$$

$$ e^{\frac{U_1}{U_T}} - 3e^{\frac{68m-U_1}{U_T}} + 2 = 0 $$

substituting \$ e^{\frac{U_1}{U_T}} = x \$;

$$ x^2-3e^{\frac{68m}{U_T}} + 2x = 0 $$

\$ e^{\frac{68m}{U_T}} = 15.180 \$;

$$ x^2 + 2x - 45.54= 0 $$

Therefore the roots are \$ x = 5.822 \$ or \$ x = -7.822 \$ (not possible because \$ x\$ is an exponential and can not take negative value) ;

therefore $$ x = e^{\frac{U_1}{U_T}} = 5.822 $$ or

$$ U_1 = U_Tln(5.822) = 44mV $$ hence $$ U_2 = 68mV - U_1 = 68mV-44mV = 24mV $$

  1. If you know \$I_{in}\$ then you can calculate the voltage across diode as $$U_{D1}=U_{D2} = U_D = V - I_{in}*R_1$$

once you know the voltage across the diodes then its easy to calculate the currents from equation: $$I_{D1} = I_{s1}(e^{\frac{U_D}{U_T}}-1)$$ $$I_{D2} = I_{s2}(e^{\frac{U_D}{U_T}}-1)$$ with \$U_D = V-I_{in}*R_1\$

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  • \$\begingroup\$ Sorry I made a mistake while writing, it isn't 10A, it is 10pA. \$\endgroup\$ – nor Nov 8 '17 at 21:51
  • \$\begingroup\$ Shouldn't the current be a function of the three values, in the second example? \$\endgroup\$ – nor Nov 8 '17 at 21:54
  • \$\begingroup\$ replace \$U_D = V-I_{in}R\$, this should give you function of those three variables \$\endgroup\$ – rsg1710 Nov 8 '17 at 21:56

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