The power of exponentials

Thread Starter

WBahn

Joined Mar 31, 2012
29,979
A recent post in Homework Help had a solution that claimed that the voltage across a diode was 10.5 V. How reasonable/unreasonable is this?

So it got me to pondering the following problem:

If I have a diode that obeys the diode equation ideally (i.e., no equivalent series resistance and no heating effects) that conducts 1 mA of current with 0.7 V across it at room temperature, what is the voltage across it when every electron in the known universe is flowing through it every second?

What is your gut feel, without making any attempt to estimate it at all?

What is your best mental estimate?

What is you best estimate after running some numbers?

Did the results surprise you?

Feel free to post all of your reasoning and work. But try to come up with your answers before reading anyone else's responses. Mine are in the following spoiler.

My gut feel -- What prompted me to think of the problem in response to the proposed solution is that 10 V struck me as being enough to achieve this. In fact, I wouldn't have been surprised to find that it is a few volts less than this but would not have argued too much if someone claimed that it was a bit more than this. But I would have absolutely balked if someone claimed it was 100 V because I think I have a decent appreciation for the rate of growth associated with exponential relationships.

My best mental estimate -- I know that there are approximately 10^80 protons/neutrons/electrons in the known universe. I also know that at room temperature the current in a diode increases by an order of magnitude for every 60 mV increase in forward voltage. I also know that there are about 10^19 electrons in a coulomb. So assuming that 10^80 is the number of electrons in the universe that would be 10^61 A or 10^64 mA. So 64 orders of magnitude at 60 mV each. That would in turn be about 3.8 V more than the 0.7 V at 1 mA, or about 4.5 V.

Running the numbers -- The best estimate I can find for the number of hydrogen atoms in the observable universe is 10^80, so I can just use that as the number of electrons by assuming that the universe as a whole is neutrally charged. Probably not too far off. With an electronic charge of -1.601e-19 C/e-, the current we are hypothesizing is 1.602e+61 A.

As is almost always done at anything except extremely small currents, I will neglect the "-1" in the diode equation. Given a known voltage, Vo, at a known current, Io, we can find the voltage at other currents with

\(
V_d \; = \; V_0 \; + \; V_t \ln\( \frac{I}{I_0} \)
\)

Vt is the thermal voltage at room temperature, which is 25.85 mV.

Plugging all this in we get.

\(
V_d \; = \; 0.7 \, V \; + \; \(25.85 \, mV \) \ln\( \frac{1.602 \times 10^{61} \, A}{1 \, mA} \)
v_d \; = \; 4.52 \, V
\)

This is a result worth remembering -- it would take less than 5 V to push every electron in the universe through a diode. In fact, 5 V would push all of the electrons in about 100 million universes!

So when people are getting answers that are much more than a volt or so (unless they are taking into account effective series resistance), you can trot out this little factoid.

Did the result surprise me -- I don't think so. I think I would have not been surprised by anything between 5 V and 15 V, so this is slightly below the low side of that range, but not enough to shock me. If anything, it was that last little calculation -- that an extra half a volt to get to an even 5 V, increases it from one universe to a hundred million universes -- that made the biggest impact.
 
Last edited:

joeyd999

Joined Jun 6, 2011
5,237
A recent post in Homework Help had a solution that claimed that the voltage across a diode was 10.5 V. How reasonable/unreasonable is this?

So it got me to pondering the following problem:

If I have a diode that obeys the diode equation ideally (i.e., no equivalent series resistance and no heating effects) that conducts 1 mA of current with 0.7 V across it at room temperature, what is the voltage across it when every electron in the known universe is flowing through it every second?

What is your gut feel, without making any attempt to estimate it at all?

What is your best mental estimate?

What is you best estimate after running some numbers?

Did the results surprise you?

Feel free to post all of your reasoning and work. But try to come up with your answers before reading anyone else's responses. Mine are in the following spoiler.

My gut feel -- What prompted me to think of the problem in response to the proposed solution is that 10 V struck me as being enough to achieve this. In fact, I wouldn't have been surprised to find that it is a few volts less than this but would not have argued too much if someone claimed that it was a bit more than this. But I would have absolutely balked if someone claimed it was 100 V because I think I have a decent appreciation for the rate of growth associated with exponential relationships.

My best mental estimate -- I know that there are approximately 10^80 protons/neutrons/electrons in the known universe. I also know that at room temperature the current in a diode increases by an order of magnitude for every 60 mV increase in forward voltage. I also know that there are about 10^19 electrons in a coulomb. So assuming that 10^80 is the number of electrons in the universe that would be 10^61 A or 10^64 mA. So 64 orders of magnitude at 60 mV each. That would in turn be about 3.8 V more than the 0.7 V at 1 mA, or about 4.5 V.

Running the numbers -- The best estimate I can find for the number of hydrogen atoms in the observable universe is 10^80, so I can just use that as the number of electrons by assuming that the universe as a whole is neutrally charged. Probably not too far off. With an electronic charge of -1.601e-19 C/e-, the current we are hypothesizing is 1.602e+61 A.

As is almost always done at anything except extremely small currents, I will neglect the "-1" in the diode equation. Given a known voltage, Vo, at a known current, Io, we can find the voltage at other currents with

\(
V_d \; = \; V_0 \; + \; V_t \ln\( \frac{I}{I_0} \)
\)

Vt is the thermal voltage at room temperature, which is 25.85 mV.

Plugging all this in we get.

\(
V_d \; = \; 0.7 \, V \; + \; \(25.85 \, mV \) \ln\( \frac{1.602 \times 10^{61} \, A}{1 \, mA} \)
v_d \; = \; 4.52 \, V
\)

This is a result worth remembering -- it would take less than 5 V to push every electron in the universe through a diode. In fact, 5 V would push all of the electrons in about 100 million universes!

So when people are getting answers that are much more than a volt or so (unless they are taking into account effective series resistance), you can trot out this little factoid.

Did the result surprise me -- I don't think so. I think I would have not been surprised by anything between 5 V and 15 V, so this is slightly below the low side of that range, but not enough to shock me. If anything, it was that last little calculation -- that an extra half a volt to get to an even 5 V, increases it from one universe to a hundred million universes -- that made the biggest impact.
Good work, @WBahn.

Can you now possibly tell me how many angels can dance on the head of a pin?
 

wayneh

Joined Sep 9, 2010
17,496
So what's the answer in our real world? I mean, what's the highest ∆V we can ever see across real, commercial diodes operating within specifications? In that other thread (a homework problem), there's a 1.5V drop across the diode. Even that seemed suspicious and made me wonder if there is a real diode that would show 1.5V at 350mA. It would have to be dissipating quite a bit of heat.
 

Thread Starter

WBahn

Joined Mar 31, 2012
29,979
In the real world, you can construct diodes to have just about any voltage at a particular target current -- you just have to balance the doping, the junction dimensions, and the current density. If you want to provide really high reverse voltage breakdown protection, you generally have to live with high forward voltages. It's far from unheard of to have diodes that have dozens of volts across them when forward biased before you even get into the ohmic voltage drops associated with the parasitics.

You might find this interesting:

http://www.semtech.com/images/datasheet/s1kwxkax.pdf

I suspect these are actually diode arrays, given the scaling of the parameters.
 

The Electrician

Joined Oct 9, 2007
2,971
So what's the answer in our real world? I mean, what's the highest ∆V we can ever see across real, commercial diodes operating within specifications? In that other thread (a homework problem), there's a 1.5V drop across the diode. Even that seemed suspicious and made me wonder if there is a real diode that would show 1.5V at 350mA. It would have to be dissipating quite a bit of heat.
Are you asking about the forward drop across a diode consisting of a single junction? If so, the Semtech rectifier family WBahn referenced doesn't really answer your question.

I emailed Semtech asking if members of that family are actually many junctions and they answered in about 1 hour with this response: "The S1KW series of HV rectifiers are, as you suggested, a series/parallel stack of discrete diodes, encapsulated."
 

MrAl

Joined Jun 17, 2014
11,396
A recent post in Homework Help had a solution that claimed that the voltage across a diode was 10.5 V. How reasonable/unreasonable is this?

So it got me to pondering the following problem:

If I have a diode that obeys the diode equation ideally (i.e., no equivalent series resistance and no heating effects) that conducts 1 mA of current with 0.7 V across it at room temperature, what is the voltage across it when every electron in the known universe is flowing through it every second?

What is your gut feel, without making any attempt to estimate it at all?

What is your best mental estimate?

What is you best estimate after running some numbers?

Did the results surprise you?

Feel free to post all of your reasoning and work. But try to come up with your answers before reading anyone else's responses. Mine are in the following spoiler.

My gut feel -- What prompted me to think of the problem in response to the proposed solution is that 10 V struck me as being enough to achieve this. In fact, I wouldn't have been surprised to find that it is a few volts less than this but would not have argued too much if someone claimed that it was a bit more than this. But I would have absolutely balked if someone claimed it was 100 V because I think I have a decent appreciation for the rate of growth associated with exponential relationships.

My best mental estimate -- I know that there are approximately 10^80 protons/neutrons/electrons in the known universe. I also know that at room temperature the current in a diode increases by an order of magnitude for every 60 mV increase in forward voltage. I also know that there are about 10^19 electrons in a coulomb. So assuming that 10^80 is the number of electrons in the universe that would be 10^61 A or 10^64 mA. So 64 orders of magnitude at 60 mV each. That would in turn be about 3.8 V more than the 0.7 V at 1 mA, or about 4.5 V.

Running the numbers -- The best estimate I can find for the number of hydrogen atoms in the observable universe is 10^80, so I can just use that as the number of electrons by assuming that the universe as a whole is neutrally charged. Probably not too far off. With an electronic charge of -1.601e-19 C/e-, the current we are hypothesizing is 1.602e+61 A.

As is almost always done at anything except extremely small currents, I will neglect the "-1" in the diode equation. Given a known voltage, Vo, at a known current, Io, we can find the voltage at other currents with

\(
V_d \; = \; V_0 \; + \; V_t \ln\( \frac{I}{I_0} \)
\)

Vt is the thermal voltage at room temperature, which is 25.85 mV.

Plugging all this in we get.

\(
V_d \; = \; 0.7 \, V \; + \; \(25.85 \, mV \) \ln\( \frac{1.602 \times 10^{61} \, A}{1 \, mA} \)
v_d \; = \; 4.52 \, V
\)

This is a result worth remembering -- it would take less than 5 V to push every electron in the universe through a diode. In fact, 5 V would push all of the electrons in about 100 million universes!

So when people are getting answers that are much more than a volt or so (unless they are taking into account effective series resistance), you can trot out this little factoid.

Did the result surprise me -- I don't think so. I think I would have not been surprised by anything between 5 V and 15 V, so this is slightly below the low side of that range, but not enough to shock me. If anything, it was that last little calculation -- that an extra half a volt to get to an even 5 V, increases it from one universe to a hundred million universes -- that made the biggest impact.

Hi,

I had a harder time figuring out how you got the "spoiler" into your post so nicely :)

Your calculation looks nice, i did not try it myself yet, but my first response was different. My first response was that it could not be done based on the conservation of charge. If we ignore that though we do get an interesting result as you had shown. We also get a second Sun :)

LATER:
I got 7.4v with a typical diode, assuming 1e64ma of current.
If i put 1e64 ideal diodes in parallel i get 0.7v :)
Next we have to do a 20x20 diode grid with one series resistor :)

Also, try the same thing with a high power white LED.

1N4001 diode zero series resistance, 1e64 ma current:
Calculate: 5.01v
Spice: 4.95v
Of course adding the small series resistance brings the voltage up by a huge amount, more than 1e60 or so.
 
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Thread Starter

WBahn

Joined Mar 31, 2012
29,979
Your sim result of 4.95 V is in nice rough agreement with my hand calculations -- makes me feel more comfortable. I suspect the difference of 430 mV might be due to a combination of things -- an ideality factor slightly greater than one is probably the most likely. There's also an offset because the voltage at 1 mA is probably different than 700 mV (probably a bit less). The sim might also be running at a slightly different temperature than 300 K -- probably 298.15 K (25 C). Then, of course, there are lots of potential second and higher order effects in the model.

Curious how you arrived at the 7.4 V estimate.
 

MrAl

Joined Jun 17, 2014
11,396
Your sim result of 4.95 V is in nice rough agreement with my hand calculations -- makes me feel more comfortable. I suspect the difference of 430 mV might be due to a combination of things -- an ideality factor slightly greater than one is probably the most likely. There's also an offset because the voltage at 1 mA is probably different than 700 mV (probably a bit less). The sim might also be running at a slightly different temperature than 300 K -- probably 298.15 K (25 C). Then, of course, there are lots of potential second and higher order effects in the model.

Curious how you arrived at the 7.4 V estimate.

Hi there,

Well, if we halve the diode active die area we double the current density. So a diode with 1/2 the unit area would act like it had 2 times the current of the diode with the full unit area. So i think the numbers we are looking at can be all over the place, unless we assume something about the size of the die.

i threw some typical numbers in there to get the diode curve, then used 1e64/1000 as the current.
The numbers i used where fictitious because i did not get them from a spice model, but were something like i remembered. Of course i forced Rs=0 as you noted was part of the experiment.

I think i used 2e-10 for Is, and 1.7454 for N to get those figures (set Is=2e-10 and V=0.7 and I=0.001 and solved for N, with T approx 300K).
However, using the spice values for a 1N4148 (6e-9, 1.913) i get:
Raw calculation: 7.93v
Spice DC sim: 7.86v

Do they make a diode smaller than the 1N4148? I would guess that they do in an IC for the ESD diodes, but not sure if they make one in a stand alone package.

This is still an interesting way to look at the diodes and the unusual voltages we might see though, unless it is an LED which of course is a Light Emitting 'Diode'. I noticed that once in a while someone gets a little sloppy and just calls an LED a diode too, because i guess they figure it is a diode of a sort. I dont like to do that and i dont think i've ever done that and i dont think it is a good idea to do that.
 
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MrAl

Joined Jun 17, 2014
11,396
None, they just don't dance.
Hi,

This offbeat question can be easily answered :)

If N is the number of angels that each require area A to dance a minimum dance number and the pin in question has area B, then the number of angels that can dance on the head of a pin is:
N=B/A

Thus if each angel required 1 unit area for the dance and the pin had 100 such area units then the number would be:
N=100/1=100 angels.

So to answer this question exactly we have to know the type of angel dancing and what type of dance they are doing and the type of pin, and also last but not least have to be writing a fairy tale book :)

In this thread however we are considering a more serious idea about how far off the estimate of the voltage across a diode can be in real life, given some assumptions about the parameters that would have to be the least that would have to be considered. This came about because some estimates just seem too silly to be real or should even be in a quiz question, like the angel dance question.

This did get me thinking though about the area of the original diode vs another test diode. We should probably set the current density to be equal to the highest value the original diode probably had. If the original diode had max 350ma and a test diode has 100ma max, then the current density should be reduced by a factor of 3.5 for the new calculation with the new diode in order to match the current density. When i do this for my fictitious diode i get about 7.3 volts. The other would come down a little too. The 1N4001 would have to be a little higher though like by a factor of a little less than 3.
 
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joeyd999

Joined Jun 6, 2011
5,237
If N is the number of angels that each require area A to dance a minimum dance number and the pin in question has area B, then the number of angels that can dance on the head of a pin is:
N=B/A
You assume that angels obey some sort of exclusion principle. You assume incorrectly.

Angels that occupy the same space are considered "degenerate". And there are lots of them.
 

Thread Starter

WBahn

Joined Mar 31, 2012
29,979
I think i used 2e-10 for Is, and 1.7454 for N to get those figures (set Is=2e-10 and V=0.7 and I=0.001 and solved for N, with T approx 300K).
However, using the spice values for a 1N4148 (6e-9, 1.913) i get:
Raw calculation: 7.93v
Spice DC sim: 7.86v
I had used an ideality factor of 1 because I had read sometime back that modern diodes are close to this (transistor junctions pretty much always had been). When I was in school. the rule of thumb was that it was 2 for a diode.

I took a look at the data sheet for the 1N4148 (which may not qualify as a "modern diode") and got an ideality factor of right about 1.9.

https://www.fairchildsemi.com/datasheets/1N/1N914.pdf

Using that ideality factor, my figures for the problem posed would be 7.96 V instead of 4.52 V.

I just did some searching and didn't find much. I'm thinking that what I read years ago might have been applicable to some radically different kids of diodes.
 

MrAl

Joined Jun 17, 2014
11,396
You assume that angels obey some sort of exclusion principle. You assume incorrectly.

Angels that occupy the same space are considered "degenerate". And there are lots of them.
Hi again,

Ha ha, well that 'equation' N=B/A holds in any case, i just had shown one particular case where all the params were finite. If we take the area of an angel dancing to be infinitesimally small then we end up with:
N=limit(B/A) as A approaches zero

and then of course the result is:
N=infinity

so we end up with an infinite number of angels that can dance on the head of a pin, but since B is a constant we have:
N=B*limit(1/A) as A approaches zero

and so we see that the area of the object they are dancing on has no bearing on the outcome unless it too is infinitesimally small, and in that case we would not be able to solve this unless we could describe the two areas as functions and one had a faster rate of decline near zero.

So i guess the fairy tale ends here :)
 

MrAl

Joined Jun 17, 2014
11,396
I had used an ideality factor of 1 because I had read sometime back that modern diodes are close to this (transistor junctions pretty much always had been). When I was in school. the rule of thumb was that it was 2 for a diode.

I took a look at the data sheet for the 1N4148 (which may not qualify as a "modern diode") and got an ideality factor of right about 1.9.

https://www.fairchildsemi.com/datasheets/1N/1N914.pdf

Using that ideality factor, my figures for the problem posed would be 7.96 V instead of 4.52 V.

I just did some searching and didn't find much. I'm thinking that what I read years ago might have been applicable to some radically different kids of diodes.
Hi again,

Yeah i think that is what i remembered too, so i was going for something that had N near 2, and ended up with N around 1.7 or so. I only used one point on the diode curve too because all we had to go on was that one point at v=0.7v, but if we had a second point we could get it more accurate i think.
 
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