Engineering Optimizing Diode Performance: Load Line Analysis

[ARoyC]

Homework Statement

I am asked to draw the I-V characteristic graph of a p-n junction diode, with 1000 ohms and 10000 ohms resistors. I am also asked to draw the load line.

Relevant Equations

E-IR-V_d = 0

I have drawn the I-V characteristic graph of the diode. I am facing problems with drawing the load line. For what value of E (Source Voltage), should I draw the Load Line? To get the I-V graph, I had to continuously change E.

 

[LvW]

When both characteristics (diode and R) are displayed in one diagram, you must express Ue through Ud,
Hence: Ur=E-Ud and I=(E-Ud)/R.

 

[ARoyC]

When both characteristics (diode and R) are displayed in one diagram, you must express Ue through Ud,
Hence: Ur=E-Ud and I=(E-Ud)/R.

Yes. But what will be the value of E?

 

[LvW]

Yes. But what will be the value of E?

I don`t know. E=U0 is the voltage of the source - so YOU must know,
The load line is a line (negative slope) crossing the Vd axis (for I=0) at Ud=Uo and the I axis (for Ud=0) at Uo/R

 

[ARoyC]

I don`t know. E=U0 is the voltage of the source - so YOU must know,
The load line is a line (negative slope) crossing the Vd axis (for I=0) at Ud=Uo and the I axis (for Ud=0) at Uo/R

To get the I-V characteristic graph, I had to continuously change the Voltage of the source. By increasing the voltage of the source, the voltage across the diode increased and hence the current. So, I have a range of values of E. Which one should I take?

 

[LvW]

To get the I-V characteristic graph, I had to continuously change the Voltage of the source. By increasing the voltage of the source, the voltage across the diode increased and hence the current. So, I have a range of values of E. Which one should I take?

You are using the I=f(Ud) axis combination.
* At first, you draw the diode characteristic - by doing this, you assume a variation of U0=Ud, right?
* Now you add a resistor and you must express the resistor characterstics I=f(Ur) through Ud.
Otherwise, you cannot have both curves in one common diagram.
And for this purpose you set Ur=Uo-Ud.
And the resistor curve (load line) now is I(=Ur/R)=(Uo-Ud)/R. This gives a negative slope for rising Ud.
* Of, assuming a variation of Ud (as you did without R) is accomplished with a Uo variation.
But this is not shown in the diagram because you have Ud at the horizontal axis.

 

[ARoyC]

You are using the I=f(Ud) axis combination.
* At first, you draw the diode characteristic - by doing this, you assume a variation of U0=Ud, right?
* Now you add a resistor and you must express the resistor characterstics I=f(Ur) through Ud.
Otherwise, you cannot have both curves in one common diagram.
And for this purpose you set Ur=Uo-Ud.
And the resistor curve (load line) now is I(=Ur/R)=(Uo-Ud)/R. This gives a negative slope for rising Ud.
* Of, assuming a variation of Ud (as you did without R) is accomplished with a Uo variation.
But this is not shown in the diagram because you have Ud at the horizontal axis.

Sorry, I did not get it.

 

[LvW]

Sorry, I did not get it.

You should try to ask a specific question - otherwise, I don`t know where your lack of understanding is.

 

[ARoyC]

The specific question is, if I have drawn the I-V characteristic graph, how to draw the load line? First I made a circuit with a 1000 ohm resistor and a diode and a dc voltage source. Now I have constantly increased the source voltage. The voltage drop across the diode increased and also the current in the circuit. Then, with that data, I drew the I-V_d graph for the diode. Now to draw the load line, the equation is E-V_d-IR=0 where V_d and I are the variables. I know the value of R. But what of E should I take in the equation to draw the load line?

 

[LvW]

Take any value for E=Uo as you like. But, of course, it must be larger than Ud.
I cannot understand your problem as I gave you the inputs for drawing the load line (I gave you the points where the load line crosses the horizontal as well as vertival axis.)

 

[ARoyC]

So, if I take any value of E (of course greater than the corresponding V_d), I will get one straight line with a negative slope. If I take another value of E, I will get another straight line (shifted, because the constant value in the equation is changing). So. please let me know if my interpretation is correct. A diode can have different load lines for different source voltages. A load line is associated with the diode, the resistance and the particular source voltage. Am I correct?

 

[LvW]

Yes - I think, now you have the correct understanding. It is the purpose of the load line to find the DC values for current and voltage when a diode (non-linear) is combined with a resistor. Two linear resistors in series can be handled applying Ohms law - however, when one part is non-linear such a graphical method is necessary.
For this purpose both characteristics must be drawn in one common diagram. Therefore, the voltage across the resistor must also be expressed by the diode voltage Ud using the simple expression I=(Uo-Ud)/R.

 

[ARoyC]

Yes - I think, now you have the correct understanding. It is the purpose of the load line to find the DC values for current and voltage when a diode (non-linear) is combined with a resistor. Two linear resistors in series can be handled applying Ohms law - however, when one part is non-linear such a graphical method is necessary.
For this purpose both characteristics must be drawn in one common diagram. Therefore, the voltage across the resistor must also be expressed by the diode voltage Ud using the simple expression I=(Uo-Ud)/R.

Got it. Thank you.