Question

The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtain an output voltage VO = 2.0 V. If a current of 1 mA is drawn away from the output terminal by a load, what is the change in output voltage?

Answer 1

`v = (2 V)/(3)= 0.667 v`

Since we have identical diodes we can use the equation:

`I_D =I= I_S e^{(V_D)/(V_T)}`

And replacing we have:
`I = 10^(-14) A e^{(0.667)/(0.025)}= 3.86x10^(-3) A = 3.86 mA`

Since we know that 1 mA is drawn away from the output then the real value for I would be

`I_D = I = 3.86 mA -1 mA= 2.86 mA`

And for this case the value for
`v_D` would be:

`V_D = V_T ln ((I_D)/(I_T))= 0.025 ln ((0.0029)/(10^(-14)))= 0.660 V`

And the output votage on this case would be:

`V = 3 V_D = 3 *0.660 V = 1.98 V`

And the net change in the output voltage would be:

`\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V`

Step-by-step 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:

`I_D = I_s [e^{(V_D)/(V_T)} -1] \approx I_S e^{(V_D)/(V_T)}`

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
`v_1 + v_2 + v_3= 2` and each voltage is the same v for each diode, so then:

`v = (2 V)/(3)= 0.667 v`

Since we have identical diodes we can use the equation:

`I_D =I= I_S e^{(V_D)/(V_T)}`

And replacing we have:

`I = 10^(-14) A e^{(0.667)/(0.025)}= 3.86x10^(-3) A = 3.86 mA`

Since we know that 1 mA is drawn away from the output then the real value for I would be

`I_D = I = 3.86 mA -1 mA= 2.86 mA`

And for this case the value for
`v_D` would be:

`V_D = V_T ln ((I_D)/(I_T))= 0.025 ln ((0.0029)/(10^(-14)))= 0.660 V`

And the output votage on this case would be:

`V = 3 V_D = 3 *0.660 V = 1.98 V`

And the net change in the output voltage would be:

`\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V`