Doubts on Diodes -- Why is voltage shifted by 0.7V?

[Bling Fizikst]

Homework Statement

refer to the image

Relevant Equations

refer to the image

I have just started diodes .
What's $V_{ON}$ here? Could anyone explain what's going on in the graph? It seems like the one from the definition of half wave rectifier but shifted below the x axis by $0.7$

 

[.Scott]

An ideal diode would begin passing current as soon as there was forward voltage.
However, semiconductor diodes don't begin passing current until the forward voltage exceeds Von.

 

[256bits]

Does this help.
A clearer description from where the output voltage is being measured would have been a nice thought from the book, or instructor.

 

[Steve4Physics]

@Bling Fizikst, this seems a bit unclear to me:
- is the required graph for $V_{DC} = 2V$ or for $V_{DC} = 0$?
- what does the y-axis represent? I suspect it is the voltage across the resistor (essentially the 'load'); this needs to stated explicitly or the question is ambiguous/incomplete.

EDIT: The y-axis is not the voltage across the resistor. See @256bits Posts #3 and #6.

More generally, it may help to note that there are 3 typical ways to treat diodes:

a) The simplest (and least accurate) way. The diode is ideal. It has infinite resistance for one polarity of applied voltage (‘reverse bias’) and zero resistance for the other polarity (forward bias).

b) The slightly more complicated (and slightly less inaccurate) way. The diode is 'semi-ideal' (my own terminology). It has infinite resistance when reverse biassed AND still has infinite resistance when forward biassed by less than its ‘threshold voltage', $V_{ON}$. If forward biassed with a voltage greater than $V_{ON}$, the diode has zero resistance.

For the most common type of diode (the silicon pn junction diode) $V_{ON} \approx 0.7$V. I guess that the Post #1 question assumes this type of diode. So, in this question, you would assume that the diode only has zero resistance when the applied voltage is greater than $0.7$V. Otherwise the diode is a pefect insulator. It's also important to note that $V_{ON}$ values are different for other types of diode.

c) The 'full' (most accurate) way. The diode's behaviour is determined from its 'characteristics' (as a graph or data-table or equation(s)) which gives the relationship between the applied voltage and the resulting current. This information would be supplied by the manufacturer in a data-sheet.

from https://eleobo.com/introduction-to-diode-and-types/

Edit - minor cosmetic changes.

 

[phyzguy]

An ideal diode would begin passing current as soon as there was forward voltage.
However, semiconductor diodes don't begin passing current until the forward voltage exceeds Von.

This is not true. Semiconductor diodes also begin passing current as soon as there is a forward voltage. However, the current is an exponential function of the voltage, so the current is quite small until the voltage gets large enough. The current as a function of voltage is $$ I = I_0 \exp{\frac{q V}{kT}} $$ Since kT/q is about .026V, the voltage needs to increase until approximately 0.7V until the current is significant. See the attached plots. The left one is on a log scale, the right one on a linear scale.

 

[256bits]

- what does the y-axis represent? I suspect it is the voltage across the resistor (essentially the 'load'); this needs to stated explicitly or the question is ambiguous/incomplete.

The resistor voltage is either 0 in the with the diode reverse biased, or v
_i - v
_diode when the diode is forward biased.
The graph does not represent the voltage across the resistor.

what does the y-axis represent? I suspect it is the voltage across the resistor (essentially the 'load'); this needs to stated explicitly or the question is ambiguous/incomplete.

As I said in my post. I do agree the graph is incomplete.
But refer to my post for the representation of output. ie diode output
On the graph, the black line represents the source voltage, with the red line representing the diode output ( I should have edited the picture with red marker rather than blue ).

Not sure why the course has even have a valve diode symbol along with a silicon diode symbol in the circuit the top picture - with V
_{dc = 2v}, and then eliminate the valve diode in the bottom circuit.
Would this be a common thing to do in a present day course explaining silicon diodes? hmmm!!!

 

[DaveE]

Why questions are kind of tough to answer because there's always another "why?" at a deeper level. I would spend a bit of time reading about P-N junctions in semiconductors, the archetypal version of diodes (there are other similar types). In particular the charge distribution and the depletion region, which is greatly affected by the applied voltage. There is a lot of information available on the web to read.

https://www.electronics-tutorials.ws/diode/diode_3.html

 

[Steve4Physics]

The resistor voltage is either 0 in the with the diode reverse biased, or v
_i - v
_diode when the diode is forward biased.
The graph does not represent the voltage across the resistor.

I agree. I hadn’t thought it through. I had the typical teaching-scenario in mind - a supply, diode and load where we are interested in the voltage across (and the current through) the load.