Piecewise Linear Modeling (diode)

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 5K views
hogrampage
Messages
107
Reaction score
1

Homework Statement


A pn junction diode is connected to a Thevenin circuit consisting of a 0-to-1V, 1kHz square wave and 1kΩ resistor. Use piecewise linear modeling to estimate the diode voltage for the low (0V), and high (1V) states of the voltage source. Assume the diode to have an approximate turn-on voltage of 0.7V for small currents.


Homework Equations


rd = [itex]\frac{nV_{T}}{I_{D}}[/itex]
iD = [itex]\frac{V_{th}-V_{f}}{R_{th}+r_{d}}[/itex]
VD = Vf + iDrd


The Attempt at a Solution


I was thinking of just picking ID = 1mA, n = 1, and VT = 25mV so that:

rd = 40Ω

I don't know if I'm going about this right, or even where to go next.
 
Physics news on Phys.org
For the 1V case...

Are you familiar with the graphical method of finding the operating point? Plot a graph with two "curves" on it:

First one for the voltage source/resistor: Plot open circuit voltage, short circuit current and join with straight line.

Another for the diode, in particular the equation VD = iDrd + Vf is of the form y = mx + c ... a straight line.

Where they cross is your answer.
 
hogrampage said:

Homework Statement


A pn junction diode is connected to a Thevenin circuit consisting of a 0-to-1V, 1kHz square wave and 1kΩ resistor. Use piecewise linear modeling to estimate the diode voltage for the low (0V), and high (1V) states of the voltage source. Assume the diode to have an approximate turn-on voltage of 0.7V for small currents.

The Attempt at a Solution


I was thinking of just picking ID = 1mA, n = 1, and VT = 25mV so that:

rd = 40Ω

I don't know if I'm going about this right, or even where to go next.

Picking iD = 1 mA is definitely not a good idea.

You know VD will be close to 0.7V. So what is a logical choice for iD if the input to the 1K resistor is 1V?
 
With that info, I found iD to be 0.3mA. Then, rd is ~83Ω with n=1 and VT=25mV. So, when the voltage source is 1V:

VD=0.7+(0.0003)(83)=725mV
 
Assuming the piecewise-linear approximation they have in mind is a straight line along the V axis (iD=0) up to V = 0.7V and then a straight line of slope 1/rd up to the operating point iD ~ 0.3 mA, then that would be correct.
 
Alright, thanks :D.