# Circuits with diodes examples!

1.

$$U=68mV\\ U_1=?\\ U_2=?\\ I_{s1}=I_{s2}=10pA\\ I_{s3}=20pA\\ U_T=25mV$$ 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.

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.

## 1 Answer

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 $U_D = V-I_{in}*R_1$

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