# dear god why cant i understand diodes

A

#### [email protected]

Jan 1, 1970
0
Okay. This should be simple. I have been designing circuits for years
but I have never had a satisfactory understanding of this:

I have a circuit. Its +5V connected to one end of a resistor, the other
end of the resistor is connected to the anode of a diode, the cathode
of the diode is connected to ground.

I also have a graph showing diode current versus diode forward voltage.

So my question is, if I pick some arbitrary value for the resistor, how
do I determine exactly how much current will be flowing through the
diode using the graph?

Yes I know you can predict it roughly by using the nominal diode
forward voltage (i.e. 0.6V) and then calculating the resultant voltage
drop across the resistor and therefore the current through the resistor
and thus the diode, but thats very imprecise and not what I am trying
to do here. I want to be able to predict the diode current for all
kinds of values of resistors, not just values which result in a diode
forward voltage in the vicinity of the knee.

Thanks!

Asa

S

#### [email protected]

Jan 1, 1970
0
Yes I know you can predict it roughly by using the nominal diode
forward voltage (i.e. 0.6V) and then calculating the resultant voltage

I've always thought it was 0.7V for typical diodes. I usually double
check with the spec sheet though since this is not always the case.
drop across the resistor and therefore the current through the resistor
and thus the diode, but thats very imprecise and not what I am trying
to do here. I want to be able to predict the diode current for all
kinds of values of resistors, not just values which result in a diode
forward voltage in the vicinity of the knee.

Can't remember but open any text book on semi-conductor electronics and
it will show you how to calculate p-n junction voltage. Lots of
calculus here. I remember my professor going through the math on the
white board resulting in the approxmation of 0.7V. At which point my
brain decided to forget the math and remember the 0.7V.

J

#### Jim Thompson

Jan 1, 1970
0
Okay. This should be simple. I have been designing circuits for years
but I have never had a satisfactory understanding of this:

I have a circuit. Its +5V connected to one end of a resistor, the other
end of the resistor is connected to the anode of a diode, the cathode
of the diode is connected to ground.

I also have a graph showing diode current versus diode forward voltage.

So my question is, if I pick some arbitrary value for the resistor, how
do I determine exactly how much current will be flowing through the
diode using the graph?

Yes I know you can predict it roughly by using the nominal diode
forward voltage (i.e. 0.6V) and then calculating the resultant voltage
drop across the resistor and therefore the current through the resistor
and thus the diode, but thats very imprecise and not what I am trying
to do here. I want to be able to predict the diode current for all
kinds of values of resistors, not just values which result in a diode
forward voltage in the vicinity of the knee.

Thanks!

Asa

A precise solution will require either a simulator, or that you be
handy with Newton's Method of Successive Approximations (which you can
easily set up on a hand-held calculator).

...Jim Thompson

J

Jan 1, 1970
0
Okay. This should be simple. I have been designing circuits for years
but I have never had a satisfactory understanding of this:

I have a circuit. Its +5V connected to one end of a resistor, the other
end of the resistor is connected to the anode of a diode, the cathode
of the diode is connected to ground.

I also have a graph showing diode current versus diode forward voltage.

So my question is, if I pick some arbitrary value for the resistor, how
do I determine exactly how much current will be flowing through the
diode using the graph?

You draw a load line. That is, on the same graph as the diode's behavior,
draw the current through the resistor vs. the voltage across the _diode_: If
should have I=0 at the open circuit voltage (5V), and 5V/your resistor value
at V=0.

Since the parts are in series, they carry the same current, and hence where
the two curves intersect you can read off the current through both of them and
the diode voltage (...leaving 5V-V_{diode} as the resistor voltage, of
course).

This can be easily extended to two non-linear components in series. For three
components you could, at least conceptually, draw a 3D graph and find the
intersection of all three curves, although by that point you might as well sit
down and get a computer to iteratively obtain the solution.

You could sit down and fit the graph to the various parameters in the diode's
equation, but personally before I did that I'd be trying to get a SPICE model
out of the vendor which will be somewhat more accurate anyway.

---Joel

S

#### Spehro Pefhany

Jan 1, 1970
0
Okay. This should be simple. I have been designing circuits for years
but I have never had a satisfactory understanding of this:

I have a circuit. Its +5V connected to one end of a resistor, the other
end of the resistor is connected to the anode of a diode, the cathode
of the diode is connected to ground.

I also have a graph showing diode current versus diode forward voltage.

So my question is, if I pick some arbitrary value for the resistor, how
do I determine exactly how much current will be flowing through the
diode using the graph?

Yes I know you can predict it roughly by using the nominal diode
forward voltage (i.e. 0.6V) and then calculating the resultant voltage
drop across the resistor and therefore the current through the resistor
and thus the diode, but thats very imprecise and not what I am trying
to do here. I want to be able to predict the diode current for all
kinds of values of resistors, not just values which result in a diode
forward voltage in the vicinity of the knee.

One way is graphically- draw the "load line" across the V-I curve for
the diode. This method was used all the time in the evil old days.

Another way is to write out your nonlinear equations and solve them,
numerically or otherwise. Or shove it into Spice with a suitable diode
model and let the program do the heavy lifting (numerical solution).

You can manually iteratively solve using your initial guess of 0.6V or
whatever. For example, if you are using Vf = k*T/q*ln(I/Is) for an
ideal diode, find the approximate current from 5V-0.6V, then find the
forward voltage, correct for a better approximation of the current
from 5V-Vf, rinse and repeat. The initial guess does not have to be
very accurate if the supply voltage is high relative to the diode
forward voltage. You might also not be using an ideal diode, so the
results might not match reality (a diode-connected transistor is
closer to an ideal diode than a diode typically is)>

Thanks!

Asa

Best regards,
Spehro Pefhany

J

Jan 1, 1970
0
Spehro Pefhany said:
One way is graphically- draw the "load line" across the V-I curve for
the diode. This method was used all the time in the evil old days.

Ironically enough, I was never made to use a load line in circuits classes,
but I was made to use one in an electric machines class where you were trying
to match a motor and generator or somesuch and they wanted to include various
non-linear effects!

G

#### Genome

Jan 1, 1970
0
Jim Thompson said:
A precise solution will require either a simulator, or that you be
handy with Newton's Method of Successive Approximations (which you can
easily set up on a hand-held calculator).

...Jim Thompson
--

****!

DNA

G

#### Genome

Jan 1, 1970
0
Genome said:
****!

DNA

Or did I miss something.... worry worry worry.

DNA

J

Jan 1, 1970
0
J

#### Jim Thompson

Jan 1, 1970
0
Or did I miss something.... worry worry worry.

DNA

Of course, Deary ;-) A diode is NON-LINEAR.

...Jim Thompson

R

#### Ron Russell

Jan 1, 1970
0
In order to calculate a diodes current at any given point, you must use the
third approximation method.

The factor you are missing is "Bulk Resistance" the N/P wafers have a bulk
resistance which at low currents is insignificant, but as the current
increases the voltage drop Vbr becomes a factor.
This is why the knee voltage is sloped.

So a diodes voltage drop is Vf = .7 + Vbr

Most rectifier diodes drop ~1 volt at the max. forward current, so the total
bulk resistance would be .3/If

Including bulk resistance in your calculations will give you very accurate
values.

Hope this helps,

Ron

B

#### Bob Monsen

Jan 1, 1970
0
Okay. This should be simple. I have been designing circuits for years
but I have never had a satisfactory understanding of this:

I have a circuit. Its +5V connected to one end of a resistor, the other
end of the resistor is connected to the anode of a diode, the cathode
of the diode is connected to ground.

I also have a graph showing diode current versus diode forward voltage.

So my question is, if I pick some arbitrary value for the resistor, how
do I determine exactly how much current will be flowing through the
diode using the graph?

Yes I know you can predict it roughly by using the nominal diode
forward voltage (i.e. 0.6V) and then calculating the resultant voltage
drop across the resistor and therefore the current through the resistor
and thus the diode, but thats very imprecise and not what I am trying
to do here. I want to be able to predict the diode current for all
kinds of values of resistors, not just values which result in a diode
forward voltage in the vicinity of the knee.

Use the lambertW function:

Id = lambertW(l, Is*R/Vt/m*exp(5)*exp(Is*R))/R - Is

(m is the diode 'fudge factor'..., between 1 and 2. Precise enough for you?

Simple, right? How could you have missed that? (xmupad just figured it out
for me...)

---
Regards,
Bob Monsen

Zero is the number of objects that satisfy a condition that is never
satisfied. But as never means "in no case", I do not see that any progress
- Poincare

D

#### Dan984

Jan 1, 1970
0
Bob said:
Use the lambertW function:

Id = lambertW(l, Is*R/Vt/m*exp(5)*exp(Is*R))/R - Is

(m is the diode 'fudge factor'..., between 1 and 2. Precise enough for you?

Simple, right? How could you have missed that? (xmupad just figured it out
for me...)

---
Regards,
Bob Monsen

Zero is the number of objects that satisfy a condition that is never
satisfied. But as never means "in no case", I do not see that any progress
- Poincare

P

#### Paul Burke

Jan 1, 1970
0
Yes I know you can predict it roughly ...thats very imprecise and not what I am trying
to do here. I want to be able to predict the diode current for all
kinds of values of resistors, not just values which result in a diode
forward voltage in the vicinity of the knee.

You'll need to know a lot about the diode, including parameters that
vary with temperature. And the temperature will vary with ambient and
the precise cooling (package to air, through the legs to the PCB). And
the parameters vary from one unit to another. And the parameters you
need probably won't be in the data sheet, except as graphical curves for
a typical unit, for example the fairly ordinary 1N4001:
http://www.fairchildsemi.com/ds/1N/1N4001.pdf

In other words, there's not a lot of point. Even if you can predict it
correctly for the diode you got out of the drawer, the next one will be
different. If your design relies on the diode for something crucial,
change the design.

Paul Burke

J

#### Jim Thompson

Jan 1, 1970
0
In order to calculate a diodes current at any given point, you must use the
third approximation method.

The factor you are missing is "Bulk Resistance" the N/P wafers have a bulk
resistance which at low currents is insignificant, but as the current
increases the voltage drop Vbr becomes a factor.
This is why the knee voltage is sloped.

So a diodes voltage drop is Vf = .7 + Vbr

Most rectifier diodes drop ~1 volt at the max. forward current, so the total
bulk resistance would be .3/If

Including bulk resistance in your calculations will give you very accurate
values.

Hope this helps,

Ron

Dear God, Indeed :-(

...Jim Thompson

Q

#### qrk

Jan 1, 1970
0
Use the lambertW function:

Id = lambertW(l, Is*R/Vt/m*exp(5)*exp(Is*R))/R - Is

(m is the diode 'fudge factor'..., between 1 and 2. Precise enough for you?

Simple, right? How could you have missed that? (xmupad just figured it out
for me...)

---
Regards,
Bob Monsen

Zero is the number of objects that satisfy a condition that is never
satisfied. But as never means "in no case", I do not see that any progress
- Poincare

The closed form solution for the lambert function is?

J

#### Jim Thompson

Jan 1, 1970
0
The closed form solution for the lambert function is?

Now, now, Mark! Details, details ;-)

Apparently Kevin, et al, believe looking it up in a table makes it OK
;-)

...Jim Thompson

B

#### Bob Monsen

Jan 1, 1970
0
The closed form solution for the lambert function is?

a bit too large to fit into the margins of this email...

---
Regards,
Bob Monsen

There once was a man from Hornepayne,
Who tried to transform the whole plane,
It bent a meridian
So it wasn't Euclidean,
And frustration drove him insane.
- Anonymous

B

#### Bob Monsen

Jan 1, 1970
0
Now, now, Mark! Details, details ;-)

Apparently Kevin, et al, believe looking it up in a table makes it OK
;-)

What is the closed form for the ln, cube root, or sin functions again? Your
calculator has approximations for them encoded in its little brain. Doing
the same for the lambertW function is not beyond the pale, except that
it is not nearly as useful, so nobody except the mathcad guys thought
of it.

I was reading about Archimedes last night. He managed to bound the
solutions to various roots using geometric methods, on his way to bounding
the value of PI. He was truly an astonishing fellow, who used the 'method
of exhaustion' to prove the various relations between PI and the area of a
circle, volume of a sphere, etc. He proved that the volume of a cylinder
of height and diameter D is 3/2 the volume of a sphere of diameter D.
Also, the surface area of this cylinder is 3/2 times the surface area of
this sphere...

---
Regards,
Bob Monsen

Our minds are finite, and yet even in those circumstances of finitude, we
are surrounded by possibilities that are infinite, and the purpose of human
life is to grasp as much as we can out of that infinitude.

J

#### Jim Thompson

Jan 1, 1970
0
What is the closed form for the ln, cube root, or sin functions again? Your
calculator has approximations for them encoded in its little brain. Doing
the same for the lambertW function is not beyond the pale, except that
it is not nearly as useful, so nobody except the mathcad guys thought
of it.
[snip]

If it ain't on my cheapy TI calculator it doesn't exist ;-)

...Jim Thompson

Replies
3
Views
1K
Replies
1
Views
1K
B
Replies
14
Views
2K
J
F
Replies
5
Views
1K
Frank Olson
F
B
Replies
8
Views
1K
Matt Ion
M