Yes. But what will be the value of E?LvW said: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.
I don`t know. E=U0 is the voltage of the source - so YOU must know,ARoyC said:Yes. But what will be the value of E?
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 said: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
You are using the I=f(Ud) axis combination.ARoyC said: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?
Sorry, I did not get it.LvW said: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.
You should try to ask a specific question - otherwise, I don`t know where your lack of understanding is.ARoyC said:Sorry, I did not get it.
Got it. Thank you.LvW said: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.
Load line analysis is a method used to optimize the performance of a diode by determining the maximum power output and the corresponding operating point on the diode's characteristic curve. It involves plotting the diode's forward voltage and current characteristics on a graph, and then finding the intersection of the load line with the diode's characteristic curve.
Load line analysis is important because it allows us to determine the maximum power output of a diode and the corresponding operating point. This information is crucial for ensuring that the diode is operating at its most efficient and effective point, which can improve its overall performance and longevity.
The load line for a diode is determined by using the diode's forward voltage and current characteristics, which can be found in its datasheet. These values are plotted on a graph, and the load line is then drawn by connecting the maximum power output point (Vmax, Imax) with the origin (0,0).
There are several factors that can affect the load line of a diode, including temperature, aging, and variations in manufacturing. These factors can cause changes in the diode's forward voltage and current characteristics, which can shift the load line and affect the diode's performance.
While load line analysis is a useful tool for optimizing diode performance, it does have some limitations. For example, it assumes that the diode's characteristics are constant, which may not always be the case due to factors such as temperature and aging. Additionally, load line analysis does not take into account other factors that may affect diode performance, such as parasitic capacitance and inductance.