Huh? Why on earth would it have to have a rating of 5.5 MA?

And how do you justify changing the problem by putting in resistor that is nine orders of magnitude smaller than the one the problem specifies?

Which is more reasonable -- that the original problem has a current of about 5 μA, or that your modified problem has a current of about 4300 A?



Why not? The current will be about 5 μA and the voltage across the diode will be a fraction of a millivolt.



No, it's not. Not in diodes that we come across in normal every day life. But since Is is proportional to area, it is certainly conceivable to build one, if not as a single device, then by putting a sufficient number of them in parallel.

Looking at the data sheet for a 1N4001, we see that at elevated temperatures we can get reverse saturation currents large enough that the number of parallel diodes needed to get to 1 mA isn't that extreme at all. I imagine it would be even easier with high power diodes.



Doesn't a reverse current of 5 μA seem a LOT more reasonable than claiming that, somehow, 1000+ V is going to magically appear across a diode connected to a 5 V source?

Remember, Is is NOT the reverse current that WILL flow the MOMENT Vd goes negative.

Draw the load line, if nothing else.

Hi,

I dont think you understood me right, and i have to admit i did not incorporate enough detail most likely.

First, you should have at least noted that ONE of the results i gave was 5ua.

Second, did you take the time to calculate the FORWARD current through that diode when Vd is 0.7v?
Try that, let me know what you get. If i made an error in calculation i apologize in advance and i'll surely correct it.

Third, i NEVER quoted 1000v or something. Not once; NEVER

 

Hi,

I dont think you understood me right, and i have to admit i did not incorporate enough detail most likely.

First, you should have at least noted that ONE of the results i gave was 5ua.

It was just a result that you said you "might be inclined" to take. And I did take note of it -- I asked why that wouldn't be a more reasonable answer that a voltage of 1005 V. In other words, why just be inclined to take it?

Second, did you take the time to calculate the FORWARD current through that diode when Vd is 0.7v?
Try that, let me know what you get. If i made an error in calculation i apologize in advance and i'll surely correct it.

First, what's the point of calculating the forward current through that diode when Vd is 0.7 V? That's just an arbitrary voltage and in this problem whether the diode is forward or reverse biased it will not be operating anywhere near that high a voltage, so what does it matter what current would be required to get the forward voltage up that high.

Did you calculate the forward current when Vd is 0.2 mV, or 1 V, or 10 V, or 100 mV? Or any other voltage? Why would you? They are all arbitrary voltages with absolutely no bearing on the problem, even IF the diode in this problem had been forward biased.

Did you stop to consider that nowhere did the problem even claim that this is a silicon diode (or any other particular type of diode)? So what's magical about 0.7 V?

Third, i NEVER quoted 1000v or something. Not once; NEVER

You're right, what you said was this:

Yeah it is reverse biased with a reasonably high reverse voltage that's how the exponential becomes near zero and leaves us with -1 inside the brackets.

You just said that it will have a reasonably high reverse voltage across it. The 1005 V figure came from the numerical solution that was posted by TheEngineer. I guess I just assumed you were accepting that as the correct answer and were trying to explain why it was true.

Even so, I still take exception to it. Even if you consider 259 μV to be "a reasonably high reverse voltage", it's nowhere near high enough to take the exponential to near zero and leave us with a -1 inside the brackets.

I'm also still interested in why the rating of the diode has to be over five million amps or why a one megohm resistor in series won't work.

 

It was just a result that you said you "might be inclined" to take. And I did take note of it -- I asked why that wouldn't be a more reasonable answer that a voltage of 1005 V. In other words, why just be inclined to take it?



First, what's the point of calculating the forward current through that diode when Vd is 0.7 V? That's just an arbitrary voltage and in this problem whether the diode is forward or reverse biased it will not be operating anywhere near that high a voltage, so what does it matter what current would be required to get the forward voltage up that high.

Did you calculate the forward current when Vd is 0.2 mV, or 1 V, or 10 V, or 100 mV? Or any other voltage? Why would you? They are all arbitrary voltages with absolutely no bearing on the problem, even IF the diode in this problem had been forward biased.

Did you stop to consider that nowhere did the problem even claim that this is a silicon diode (or any other particular type of diode)? So what's magical about 0.7 V?



You're right, what you said was this:



You just said that it will have a reasonably high reverse voltage across it. The 1005 V figure came from the numerical solution that was posted by TheEngineer. I guess I just assumed you were accepting that as the correct answer and were trying to explain why it was true.

Even so, I still take exception to it. Even if you consider 259 μV to be "a reasonably high reverse voltage", it's nowhere near high enough to take the exponential to near zero and leave us with a -1 inside the brackets.

I'm also still interested in why the rating of the diode has to be over five million amps or why a one megohm resistor in series won't work.

Hi again,

Ok thanks but that's a lot to talk about all at once so lets take it one or two at a time.

First, reasonably high voltage is 5v. Why? Because it creates a large negative exponent and that makes the exponential part zero. Try it.

Second, 0.7v is a typical forward voltage for silicon diodes so it is informative to look at the forward current with 0.7v. You can use less if you like but it should be at least 0.5 i would think. That tells us that the diode must be made to handle a huge current, which is so high it seems unreasonable. In fact, the Is=0.001 is the unreasonable part as typically this is 1 nano unit or even less like femto units. Change Is to 1 nano unit and see how much this problem changes with a 1 Megohm resistor.

Third, "M" is often taken to be milli in simulators, even though it is capitalized.

I guess you came up with 5ua too?
The reason i said that was because if we have such a high current diode then it will look like a short in reverse for a 1Megohm resistor, and as we all know the diode can NOT conduct more current than is available, so it defaults to 5ua approximately.

How did you get 5ua?

 

Hi again,

Ok thanks but that's a lot to talk about all at once so lets take it one or two at a time.

First, reasonably high voltage is 5v. Why? Because it creates a large negative exponent and that makes the exponential part zero. Try it.

So? If it has 5 V across it, then the resistor has NOTHING across it and NO current flows, and so the diode has 0 V across it. Where is it written in stone that a diode has to have a reasonably high voltage across it when reverse biased?

Second, 0.7v is a typical forward voltage for silicon diodes so it is informative to look at the forward current with 0.7v. You can use less if you like but it should be at least 0.5 i would think.

Why? Are you saying that it is impossible to operate a diode with, say, 100 mV of forward bias?

Why should it be at least 0.5 V? Again, the problem does NOT state that it is a silicon diode. What it it's a germanium diode? Should it still be at least 0.5 V? What if it is nothing more than a fictitious diode dreamed up by whoever made the problem, perhaps specifically to make it more difficult to get a close answer by using the conventional assumptions or blindly throwing it into a simulator without modifying the device model?

That tells us that the diode must be made to handle a huge current, which is so high it seems unreasonable.

Without necessarily agreeing, so what if it is (either of those)? It is the diode that was specified for the problem.

In fact, the Is=0.001 is the unreasonable part as typically this is 1 nano unit or even less like femto units.

Not arguing that -- in fact I addressed it explicitly in my first post in this thread.

Change Is to 1 nano unit and see how much this problem changes with a 1 Megohm resistor.

It doesn't matter how much changing the problem changes the problem.

Changing the problem does not solve the problem that was given.

The problem that was given has a very easy to get solution that is perfectly reasonable for the device parameters given.

Third, "M" is often taken to be milli in simulators, even though it is capitalized.

Yes. So? This is pretty clearly not a schematic capture from a simulator.

I guess you came up with 5ua too?
The reason i said that was because if we have such a high current diode then it will look like a short in reverse for a 1Megohm resistor, and as we all know the diode can NOT conduct more current than is available, so it defaults to 5ua approximately.

How did you get 5ua?

I didn't exactly hide how I got it, see Post #10.

I only did the forward bias case in that post since I wanted to give the TS an opportunity to solve the actual problem on their own. But I addressed it a few posts later in Post #18.

 

Hi,

First, i dont care if it is a schematic capture or not. 1M COULD be 0.001, that doesnt mean that is HAS to be that.

Second, when we apply 0.5 or 0.7v we see what kind of current rating we are dealing with. If we are dealing with a high current diode then then we can apply the approximation of a short for a resistor of 1 Megohm. If you dont see how this works i cant help it
Notice how it actually WORKS though. Now try a lower current rated diode and see that this approximation does not work anymore.

Third, we can certainly write the entire equation. Solve the original for Vd, then drop in the resistor voltage Vr=Id*R, then equate to Vcc, then solve that equation for Id numerically. That gives a numerical solution. If the ideality factor is 1.2 we get around 4.99984ua.

NOW, why is this 4.99984ua so darn close to 5ua, the approximation after considering the current rating of the diode.

It's ok if you want to solve the "Third" note above, that's fine with me, i did it too. If that's the only way you want to do it that's up to you.

Side note, the "Third" note above is actually how we start out if we wanted to curve fit a more real life diode that has contact resistance also. We can solve for that resistance as well as Is and N given the data sheet curve.

 

Perhaps it would help me see where you are coming from if you could explain the following:

Seeing as this diode must have a rating of something like 5.5 Meg amps i took the resistor R to be 0.001 ohms not 1 megohm.
I dont think it works with a 1 megohm resistor in series.

Why don't you think it works (with the given diode parameters) with a 1 megohm resistor in series?

What was your reasoning behind this thought?

If someone had that diode and hooked up that circuit, what did you think would happen? Or, more to the point, what would constitute it "not working"?

 

Perhaps it would help me see where you are coming from if you could explain the following:



Why don't you think it works (with the given diode parameters) with a 1 megohm resistor in series?

What was your reasoning behind this thought?

If someone had that diode and hooked up that circuit, what did you think would happen? Or, more to the point, what would constitute it "not working"?

Hi,

Oh ok i see what you mean. I should have provided more info there. Maybe in too much of a hurry.

I said that because if we follow the 'directions' of that first post verbatim, then we end up with that 1000v across the resistor which as you noted just cant be right at all. So i said it cant work. The only way we can make it work then i think is if we note that since the current *would* be too high for the circuit and that a diode is a completely passive device, then we could consider it a short circuit. The only problem then is that the voltage we use to determine this goes to zero and so it still doesnt really work.

However, if we abandon that first post 'directions' then we find solutions.

 

Hi,

Oh ok i see what you mean. I should have provided more info there. Maybe in too much of a hurry.

I said that because if we follow the 'directions' of that first post verbatim, then we end up with that 1000v across the resistor which as you noted just cant be right at all. So i said it cant work. The only way we can make it work then i think is if we note that since the current *would* be too high for the circuit and that a diode is a completely passive device, then we could consider it a short circuit. The only problem then is that the voltage we use to determine this goes to zero and so it still doesnt really work.

However, if we abandon that first post 'directions' then we find solutions.

The first post 'directions' were the TS's attempt to solve the problem.

 

The first post 'directions' were the TS's attempt to solve the problem.

Hi,

Oh geeze

It looked like the proper font for a description of how to solve it.
However, i do see the first part of that being interesting and worth while to think about, how the exponential part goes to zero.