Circut containing diode- calculate the Id and Vd

Jony130

Joined Feb 17, 2009
5,089
There is something wrong with your answer. If ID = IS =1mA than the voltage dropa across resistor is V = I*R = 1mA *1MΩ = 1000V which is impossible in a circuit supply from a 5V source.
 

wayneh

Joined Sep 9, 2010
16,125
Some years ago I was trying to model a rectifier, and created an Excel spreadsheet that contains the diode formula and some other stuff. It's pretty rough and was never meant for sharing but I think you'll have no trouble using it if you're familiar with Excel and maybe you'll find it useful. If not, just toss it.
 

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wayneh

Joined Sep 9, 2010
16,125
So, the only right way is to solve the attached system of equations? Isn't there any approximation to make it easier? Is that diode in reverse or forward state?
I don't think there's an analytical solution. You have to solve by iteration and convergence.

About the only simplification I see is that, with 5V, you know the diode is conducting and current-limited by the resistor, not the diode. So most of the time you could guess the diode voltage drop will be ~0.6V for a silicon diode (at such low current, you might guess 0.7V for amp-level current). So that's a starting point for estimating the current and then converging on a more precise estimate.
 

WBahn

Joined Mar 31, 2012
24,970
So, the only right way is to solve the attached system of equations? Isn't there any approximation to make it easier? Is that diode in reverse or forward state?
The diode in this case is reverse biased. Let's take things in two chunks.

Let's first switch the diode around so that it would be forward biased.

The value of Is given, namely 1 mA, is completely unrealistic. Typical values of the scale current are in the picoamp to nanoamp range. This will result in an unrealistically low diode voltage drop (namely, the same voltage drop at 1 mA that a typical diode has at 1 pA).

So our first approximation is that the diode drop is 0 V and we would get a current of 5 μA (i.e., 5 V / 1 MΩ).

What value are you using for the ideality factor. Modern diodes tend to be close to 1, but historically a value of 2 was used for diodes. So let's use that. Vt is about 26 mV at room temperature, so let's go with that.

Solving the diode equation for 5 uA then gives a forward voltage of 259 μV.

That's probably close enough. But if not, we can then use this as a new starting point.

Vd = 259 μV which means the voltage across the resistor is 4.99974 V making the current 4.99974 μA, making the diode voltage 259 μA. So we have agreement to three sig figs.

That's the forward biased case.

In reverse bias, the we do essentially the same thing. I'll give you a chance to see if you can make some progress on that one.
 
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WBahn

Joined Mar 31, 2012
24,970

MrAl

Joined Jun 17, 2014
6,788
Hi,

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.
 

WBahn

Joined Mar 31, 2012
24,970
Because of the fact that the diode is reverse biased.
Look at the diagram. Vd is defined as the forward voltage across the diode. So if it is reverse-biased, Vd will be negative. If you then go and artificially change the sign in the equation, you are saying that magnitude of the current goes UP as the voltage across the diode becomes more negative.

That the over equal has a negative sign added is required because the diagram defines Id to be the reverse current through the diode.
 

WBahn

Joined Mar 31, 2012
24,970
Hi,

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.
So are you really claiming that if you put a 5 V source across a 1 MΩ resistor in series with this diode such that it is reverse biased that Vd will be +1005 V across the diode?
 

WBahn

Joined Mar 31, 2012
24,970
Because of the fact that the diode is reverse biased.
I went back and looked at your Wolfram link and see that it shows two solutions -- I was only seeing a single solution (the 1005 V) when I looked before.

It's showing the second solution at the right magnitude, but the wrong sign if you are claiming that x is the reverse voltage across the diode (to justify the negative sign in the exponent).

But I'm not understanding why it is claiming two solutions in the first place. The curve is monotonic, so there should only be, at most, one solution.

Ah.. it took a bit of pondering, but you also are not setting up the equation to satisfy KCL quite right. You also need that minus sign out front.
 

WBahn

Joined Mar 31, 2012
24,970
Oh crap, I didn't see that! :eek: I wouldn't have done well on this test.
Actually, in this case you might have. The diode is being operated so far down the curve that it is pretty linear.

With n=2 and at room temperature, at zero volts the diode looks like a 52 Ω resistor.

In series with 1 MΩ, that's lost in the weeds. But, for completeness sake, that would be a current of

5 V / (1 MΩ + 52 Ω) = 4.99974 μA and a voltage of 259.99 μV, which is pretty much exactly what the iterative approach got.

At these levels, the diode is symmetric about the origin, so forward versus reverse bias makes no difference. So depending on whether or not you took the polarity of the signals consistent with the drawing (and just mentally turned the diode around), you would have gotten the correct answer or been off by a minus sign.

EDIT: Fix typos.
 
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MrAl

Joined Jun 17, 2014
6,788
So are you really claiming that if you put a 5 V source across a 1 MΩ resistor in series with this diode such that it is reverse biased that Vd will be +1005 V across the diode?
Hi,

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.
Or else there is something else wrong. I dont think it is typical at all to see Is=0.001 amps.

In all fairness for a diode rated that high i might be inclined to take it as a short circuit when used with a resistor of high value like 1Megohm. That would make the reverse current Id equal to 5/1000000 amps which is 5ua.

Now we see what happens when textbooks do not give any numerical results to compare.
 
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WBahn

Joined Mar 31, 2012
24,970
Hi,

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.
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?

I dont think it works with a 1 megohm resistor in series.
Why not? The current will be about 5 μA and the voltage across the diode will be a fraction of a millivolt.

Or else there is something else wrong. I dont think it is typical at all to see Is=0.001 amps.
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.

In all fairness for a diode rated that high i might be inclined to take it as a short circuit when used with a resistor of high value like 1Megohm. That would make the reverse current Id equal to 5/1000000 amps which is 5ua.

Now we see what happens when textbooks do not give any numerical results to compare.
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.
 
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