Answer:
[tex] v = \frac{2 V}{3}= 0.667 v[/tex]
Since we have identical diodes we can use the equation:
[tex] I_D =I= I_S e^{\frac{V_D}{V_T}}[/tex]
And replacing we have:[tex]I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA[/tex]
Since we know that 1 mA is drawn away from the output then the real value for I would be
[tex] I_D = I = 3.86 mA -1 mA= 2.86 mA[/tex]
And for this case the value for [tex] v_D[/tex] would be:
[tex] V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V[/tex]
And the output votage on this case would be:
[tex] V = 3 V_D = 3 *0.660 V = 1.98 V[/tex]
And the net change in the output voltage would be:
[tex] \Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V[/tex]
Explanation:
For this case we have the figure attached illustrating the problem
We know that the equation for the current in a diode id given by:
[tex] I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}[/tex]
For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode [tex] v_1 + v_2 + v_3= 2[/tex] and each voltage is the same v for each diode, so then:
[tex] v = \frac{2 V}{3}= 0.667 v[/tex]
Since we have identical diodes we can use the equation:
[tex] I_D =I= I_S e^{\frac{V_D}{V_T}}[/tex]
And replacing we have:
[tex]I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA[/tex]
Since we know that 1 mA is drawn away from the output then the real value for I would be
[tex] I_D = I = 3.86 mA -1 mA= 2.86 mA[/tex]
And for this case the value for [tex] v_D[/tex] would be:
[tex] V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V[/tex]
And the output votage on this case would be:
[tex] V = 3 V_D = 3 *0.660 V = 1.98 V[/tex]
And the net change in the output voltage would be:
[tex] \Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V[/tex]
