View attachment diode.pdf
Hi.
Referring to the attached image, please help to advise on case c).

0V across diode may cause by an ideal diode that is turned ON or a practicle diode that are OPEN. But i cant continue on the analysis .
And i also do not understand when it says 'check the source voltage'. The the 5V considered as the source voltage?

Thanking you guys in advance

 

Justify your statement that 0V across the diode could indicate that a practical diode is OPEN. Then we'll go from there. Imagine the circuit with the diode removed and the lower end of the resistor hanging loose. What would your voltmeter show when connected to the lower end of the resistor and ground?

 

0V across the diode will be caused by an ideal turned ON diode (practical diode will have a drop of 0.3-0.7v depending of the material). It will also be caused if the diode is not connnected to anything.

However, there is still a voltage accross the branch between the voltage source (yes, the 5V) and ground. Think about this: when you have an open circuit, between the voltage source and ground you will read Vcc (the voltage source), but the elements within will not have any voltage.

 

Thanks BillB3857 & Varkatzas for the reply.

So, the question 'whats wrong with the circuit' can be answered as :
:The diode is OPEN which may due to it is not connected properly to the ground or to the resistor. No current flow to have the diode to turn ON and that is why Vdiode is reading 0V and the Vsource is 5V.

A question came up to my mind:-
1. If the diode is an ideal diode and turned ON, I will still have 0V , but will the Vsource reads 5V?
2. If the diode is a practical diode and connected properly, and it's broken (failed to operate as a practical diode, it will also be shorted - correct or not?) and with this, what will the Vsource measure? still 5V?
please help

 

In questions like this, Vsource is considered to be a solid, unlimited current rated voltage source. (Think BIG battery) To keep your responses in perspective, how did you answer each section (a, b, c) of the original question. Please respond as if the components were real, like you could buy. "Ideal" gets in the way of understanding what goes on in the "real" world.

 

My answers are ....
a) If the diode is ON, the voltage across the diode will depends on the material. If the diode is OFF (OPEN), there's no current flow neither through the resistor. the voltage measured across the diode are actually = 5V where it measured the value of the Vdc.
In this case, the diode is OPEN.

b) If R is shorted, the diode is still in ON state. The voltage is either 0.7V or 0.3V according to the material. It wont be 5V across the diode because the diode has a resistance which makes its forward bias to be either 0.7V or 0.3V.

(please correct me if i'm wrong)

 

BillB3857 , what do you mean by 'unlimited current rated voltage source' ?

 

Hello,

Have a look at this wikipage for the "ideal" voltage source:
http://en.wikipedia.org/wiki/Voltage_source

There are also "ideal" current sources:
http://en.wikipedia.org/wiki/Current_source

Bertus

 

Sorry for mixing "ideal" into the question by saying the power source had unlimited current ability. I should have said "well regulated and not affected by load." I would also inquire about your answer to "b". If R shorts, and you have a very high current capable voltage source, what will limit the current? How much current can the diode carry safely?

 

Also, how would you answer "c" of the question?

 

BillB3857, your 'ideal' was referring to the diode right?
Back to case b)
I am not sure what a maximum current can a diode carries, but there's a PIV value that shows the max allowable voltage. So,in this case, when R is short, nothing is limitting the current. An So.... is 5V equals to the PIV value in this case? hehehe... please help

 

Having the diode as practical diodes,
answering to c) will be as in the earlier post :-

:The diode is OPEN which may due to it is not connected properly to the ground or to the resistor. No current flow to have the diode to turn ON and that is why Vdiode is reading 0V and the Vsource is 5V.

A question came up to my mind:-
1. If the diode is an ideal diode and turned ON, I will still have 0V , but will the Vsource reads 5V? - this goes back to case b) and will wait for your reply
2. If the diode is a practical diode and connected properly, and it's broken (failed to operate as a practical diode, it will also be shorted - correct or not?) and with this, what will the Vsource measure? still 5V?
Yes, Vsource will still measure Vdc, but what happen to the diode is still a question.

 

Yes, I was mainly talking about the diode when saying that ideal can get in the way of an analysis. As for the power supply, most are designed to provide more current than would normally be expected while regulating the Voltage. Most devices have a voltage range that is acceptable to normal operation and then draw current as needed. Look at Personal Computer power supply. You will see multiple regulated Voltages, but a mention of current capability at each voltage. There are situations where current regulation is important and, as you will find, current is actually regulated by varying the voltage available.

You are correct in saying that diodes have a PIV rating, but they also have a maximum current rating. With an unlimited current source, what do you think happens to a diode such as a 1N4001 in your circuit?

 

After google-ing the datasheet of 1N4001 [http://www.diodes.com/datasheets/ds28002.pdf],
it mentioned the Surge Overload Rating to 30A Peak .
So i believe that's the max current the diode can have.

If the 5V can supply 30A and above (which i have no idea ) ,
it will failed the diode, and the diode will be shorted.

 

True, the diode WILL fail. It may short or it may let all the magic smoke out and become open. If the power source is not capable of supplying enough current to burn the critter to a crisp, then the supply voltage will tend to drop due to the source not being able to supply enough current and possible power supply damage would result. As they say, the devil is in the details but for a learning experience it is best to work with what is given in the question to obtain answers to that question. Too many assumed variations on what "could" be only confuse the issue.

 

Too many assumed variations on what "could" be only confuse the issue.

...agree

So I have the answer to (b) Thanks BillB3875! and my answer on (a) is correct right? and for case (c) are as below:- Are below answers explaining every possible situation?

(1) Considering that it is a practical diode, the diode is OPEN giving its voltage read as 0V. the diode is OPEN which may due to --> it is not connected properly to the ground, or to the resistor. No current flow to have the diode to turn ON and that is why Vdiode is reading 0V and the Vsource is 5V.

Another case which may make the diode voltage to be 0V is when:-
(2) Considering that it is a practical diode, and connected properly, and it's broken (failed to operate ) it will also be shorted and measured as 0V, Vsource will still measure 5V but the resistor will in the end get killed due to high current flow.

* what ever case it is, When measuring Vsource , it is always measuring the Vdc = 5V right?

 

0V across the diode will be caused by an ideal turned ON diode (practical diode will have a drop of 0.3-0.7v depending of the material). It will also be caused if the diode is not connnected to anything.

However, there is still a voltage accross the branch between the voltage source (yes, the 5V) and ground. Think about this: when you have an open circuit, between the voltage source and ground you will read Vcc (the voltage source), but the elements within will not have any voltage.

So the voltage measured across the diode is the 5V V source right? and not 0V right?

 

Here i conclude on case c), please help to clarify if this is correct :-

c) 0V across the diode will be caused by:-
(i) An ideal diode that is turned ON (because if practical diode turned ON, it will have a drop of 0.3- 0.7v depending of the material).
(ii) A practical diode is malfunction – broken and fail.
(iii) A diode (ideal or practical) is not connected properly – causing open circuit.
(iv) A practical diode that is turned OFF (OPEN circuit).

….. Considering that it is a practical diode,
(i) The diode is connected properly, and it's broken (failed to operate ) it will be shorted and measured as 0V, Vsource will still measure 5V but the resistor will in the end get killed due to high current flow.
Another case which may make the diode voltage to be 0V is when:-
(ii) The diode is OPEN giving its voltage read as 0V. The diode is OPEN which may due to --> it is not connected properly to the ground, or to the resistor. No current flow to have the diode to turn ON and that is why Vdiode is reading 0V and the Vsource is 5V. – Vdiode measured is 5V and Vsource measured is also 5V. [ This is not the case]

So answering to “what’s wrong with the circuit” is that the diode is broken and gets shorted as explained in (i).
Sooner…the resistor will get killed and also the diode. And there comes the smoke from the components

 

Here i conclude on case c), please help to clarify if this is correct :-

c) 0V across the diode will be caused by:-
(i) An ideal diode that is turned ON (because if practical diode turned ON, it will have a drop of 0.3- 0.7v depending of the material). True but there is no ideal diode for practical use.
(ii) A practical diode is malfunction – broken and fail. True even for a real world diode
(iii) A diode (ideal or practical) is not connected properly – causing open circuit. True statement.
(iv) A practical diode that is turned OFF (OPEN circuit). True. Open circuit at any point in series circuit would give this result.


….. Considering that it is a practical diode,
(i) The diode is connected properly, and it's broken (failed to operate ) it will be shorted and measured as 0V, Vsource will still measure 5V but the resistor will in the end get killed due to high current flow. Depending upon resistor value, it may or may not be killed. Ohms law would dictate how much current and power could then be calculated. Since resistor is not killed with a good diode, it probably would not be killed with a shorted diode, either. It would only see a voltage drope increased by between 0.3 to 0.7 volts. A small percentage of change.
Another case which may make the diode voltage to be 0V is when:-
(ii) The diode is OPEN giving its voltage read as 0V. The diode is OPEN which may due to --> it is not connected properly to the ground, or to the resistor. No current flow to have the diode to turn ON and that is why Vdiode is reading 0V and the Vsource is 5V. – Vdiode measured is 5V and Vsource measured is also 5V. [ This is not the case] The underlined would be better stated as the CIRCUIT IS OPEN rather than saying the DIODE is open. When you say the DIODE is open, that would indicate an internal failure of the component.

So answering to “what’s wrong with the circuit” is that the diode is broken and gets shorted as explained in (i).
Sooner…the resistor will get killed and also the diode. And there comes the smoke from the components

Again, don't try to over think the questions at this point. Plenty of time for that after you get a solid hold on the basics.

 

(ii) The diode is OPEN giving its voltage read as 0V. The diode is OPEN which may due to --> it is not connected properly to the ground, or to the resistor. No current flow to have the diode to turn ON and that is why Vdiode is reading 0V and the Vsource is 5V. – Vdiode measured is 5V and Vsource measured is also 5V. [ This is not the case] The underlined would be better stated as the CIRCUIT IS OPEN rather than saying the DIODE is open. When you say the DIODE is open, that would indicate an internal failure of the component.

In this case, when the diode become OPEN CIRCUIT. And the Vdiode measured will be 5V which is actually measuring the Vdc or the Vsource. But basically the diode is burnt , so … no diode – no voltage - This statement is correct?