In Shockley's Equation what is e? Euler's Number?

Nope, can't be Eulers...

 

Can`t be Eulers? Why not? Did you ever hear about an exponential relationship between Vd and Id?

 

Is your google broken?

e = 2.71828182845904523536028747135266249775724709369995....

https://en.wikipedia.org/wiki/E_(mathematical_constant)

 

K, I'll do it again. Euler's gave me a very strange result... Also tried electron charge, nada, nope...

 

OK, I've got 40nA((e^(.5V/2*25.27mV))-1) = 0.000791736 which is 0.792mA which is close enuff!

Got it, thx guys

 

OK, I've got 40nA((e^(.5V/2*25.27mV))-1) = 0.000791736 which is 0.792mA which is close enuff!
View attachment 287348
Got it, thx guys

Hi,

For some reason i am just seeing this now.
But this should be interesting for you.

Next try it without including the "-1" in your first equation, then compare results. This will show you why not including the "-1" is a good approximation for many problems (but not all).

 

Or perhaps an even better way to drive the point home is to ask at what Vd does the -1 become a negligible factor.

First, we have to decide what is negligible. In practice, our results are seldom good even to 1% -- it takes quite a bit of effort to achieve even that. So let's start there.

For that -1 to represent 1% of the result, we need e^x to be 100, making x = 4.61. If the ideality factor, n, is unity, that means that Vd is about 120 mV.

Since we typically write results to three sig figs, we could say that we want the -1 to represent no more than 0.1% of the result. Using that criterion, we reach that point when the diode voltage is more than about 180 mV.

If we use an ideality factor of 2, that doubles to 360 mV.

So, unless we are working well below the transition voltage, we can completely neglect the -1.

 

OK, I've got 40nA((e^(.5V/2*25.27mV))-1) = 0.000791736 which is 0.792mA which is close enuff!
View attachment 287348
Got it, thx guys

One thing to watch out for is order of operations. As written above, you actually have

e ^ (((0.5 V)/2) * 25.27 mV)

when what you meant was

e ^ ( (0.5 V)/(2 * 25.27 mV) )

Sloppiness here tends to bite you when entering equations into a program or a spreadsheet because we inadvertently assume that the tool will interpret them the way we meant as opposed to the way we actually wrote it.

 

So, unless we are working well below the transition voltage, we can completely neglect the -1.

Yes - and the big advantage is that we can make use of one of the the most important expression in transistor applications:
gm=Ic/Vt.

So we have a very simple relation between the DC quiescent collector current Ic and the transconductance gm which is the key parameter for calculating the voltage gain for each of the three basic amplifier alternatives (common base/emitter/collector).

* Mathematical background: The transconductance gm is the slope of the Ic=f(Vbe/Vt) characteristics. And because the differential quotient of an e-function is again an e-function, we arrive at this simple and versatile relation.
During this mathematical manipulation we ignore, of course, the "-1" (as mentioned before) - and the error is totally negligible because we are working with Vbe-values in the range of 0.7 Volts.
As shown earlier (WBahn) at Vbe=0.18 volts the error due to this simplification would be only 0.1%.
At a voltage of app. Vbe=0.7 volts it is, therefore, much, much smaller.

 

Or perhaps an even better way to drive the point home is to ask at what Vd does the -1 become a negligible factor.

First, we have to decide what is negligible. In practice, our results are seldom good even to 1% -- it takes quite a bit of effort to achieve even that. So let's start there.

For that -1 to represent 1% of the result, we need e^x to be 100, making x = 4.61. If the ideality factor, n, is unity, that means that Vd is about 120 mV.

Since we typically write results to three sig figs, we could say that we want the -1 to represent no more than 0.1% of the result. Using that criterion, we reach that point when the diode voltage is more than about 180 mV.

If we use an ideality factor of 2, that doubles to 360 mV.

So, unless we are working well below the transition voltage, we can completely neglect the -1.

Hi,

Yes good idea really.
I mentioned this for the interest but also for the resulting simplification. Getting rid of that "-1" term makes a lot of diode related problems much simpler to solve because then there is a simple solution for both Id and Vd as well using log and anti log functions.

And to drive the point home, we have the difference between the approximation and the exact functions:
Is*e^(Vd/NVT)-Is*(e^(Vd/NVT)-1)

which of course equals, Is, and Is is usually very small.

Also interesting is the ratio f1/f2 and we end up with:
1-e^(-Vd/NVT)

which is similar to a capacitor charging, and so we can estimate 1 percent for Vd=5*NVT.
That with VT=0.026 means we have about 1 percent for Vd=0.13*N.

I also get about 180mv for 0.1 percent (N=1):
Vd=0.1796*N



I hope i did all this right

 

Turned out the problem was the calculator was giving a bad result for . Once I broke down the equation and solving for e^(.5V/(2*25.27mV)) first I was able to get the correct result from the calculator. If my memory serves me correctly...

 

@WBahn the calculator was expected to MDAS order of operations on (.5V/2*25.27mV) which should have been to multiply 2*25.27mV before dividing it into 0.5. It's been a while but I remember also having a problem with the -1 part of the equation on the calculator. Whatever it was I did manage to resolve it by breaking down the equation to use the calculator to resolve it.

 

@WBahn the calculator was expected to MDAS order of operations on (.5V/2*25.27mV) which should have been to multiply 2*25.27mV before dividing it into 0.5. It's been a while but I remember also having a problem with the -1 part of the equation on the calculator. Whatever it was I did manage to resolve it by breaking down the equation to use the calculator to resolve it.

Order of operations, as done by any calculator I know of that imposes order of operations, follows the rule that multiplication and division are done before addition and subtraction, but within those groups, they are done left to right. Thus multiplication and division have equal precedence (as do addition and subtraction).

Exponentiation is higher precedence that any of those operations, but how exponentiation is performed varies from one calculator to another.

If you do 3^2-1, then by order of operations that is (3^2) - 1, which is 8, and not 3^(2-1), which is 3.

My guess is that you were multiplying the result of the exponentiation by the 40 nA and then subtracting off the 1 (which would either be 1 A or 1 nA, depending on how you treated units).

Breaking down the equation is always a good fallback when things aren't going well.

 

Yes, I suspected it had to do with using an equation for the exponential factor instead of a simple number. Who knows... But I got there in the end at least.

 

Yes, I suspected it had to do with using an equation for the exponential factor instead of a simple number. Who knows... But I got there in the end at least.

Due to the rules most calculators follow you could have done this:
x=0.5/2/0.02527

and that would be equivalent to:
x=0.5/(2*0.02527)

because it would first divide 0.5 by 2, then divide that result by 0.02527.
So in other words:
a/(b*c)=a/b/c

Some people don't like doing it that way but i do sometimes when i don't feel like using the parentheses. It's faster to just use two forward slashes. It may not be as clear to read though for some people that don't expect to see that.

For the calculators i have used in the past, and that was many different makes and models, i find that the more expensive ones always follow the rules of algebra while some of the cheapies do not and that includes the order of operations. Sometimes they just work left to right with no preference for any operators. Some of them get that part right, but not the exponentiation. The more expensive ones (maybe $10 USD and up) usually do it all right. It's always good to check though by doing a few examples you know should work right.

 

The one I use the most is the TI-36X Pro which has an e^x function. But using the equation as the power was giving me garbage. Breaking the full equation down and solving the power equation to use a simple number for the power of e worked. It just didn't seem to like using the equation for the power factor. Which in turn led me to thinking that maybe the e they were using was something other than Euler's number. Ah well, I have it worked out now...

 

The one I use the most is the TI-36X Pro which has an e^x function. But using the equation as the power was giving me garbage. Breaking the full equation down and solving the power equation to use a simple number for the power of e worked. It just didn't seem to like using the equation for the power factor. Which in turn led me to thinking that maybe the e they were using was something other than Euler's number. Ah well, I have it worked out now...

Hi,

That's great to hear.