# Exams Help: Design a circuit that produces 1.25V using diodes and 1 current source.

#### babaliaris

Joined Nov 19, 2019
115
You have in your possession as many (identical) diodes as you want (with exponential I-V characteristics and one point $$(V_{D} = 0.7V, I_{D}=20mA)$$
with $$n=1$$ and $$V_{T} = 25mV$$. Remember that $$I_{D} = I_{ss} \cdot e^{\frac{V_{D}}{n \cdot V_{T}}}$$

Using one current source $$I_{in}=4mA$$ and by combining diodes in series and parallel, design a circuit that produces an output of $$V_{o} = 1.25V$$

Solution (as far as I went):

Using the I-V point I can calculate Iss: $I_{D} = I_{ss} \cdot e^{\frac{V_{D}}{n \cdot V_{T}}} <=> I_{ss} = I_{D} \cdot e^{\frac{-V_{D}}{n \cdot V_{T}}} = 20mA \cdot e^{\frac{-0.7V}{1 \cdot 25mV}} <=> I_{ss} = 1.38 \cdot 10^{-14}A$

I also attached a pdf file with the circuit that I tried to design but I couldn't in the end. I just can't think of how to combine diodes in parallel and series in order to produce the desired output.

Question 1)
Can you show me a solution and try to make understand how to approach similar problems? Do not show me the math, only the circuit and let me
do the math to see if it actually works.

Question 2)

How should I think when designing diode circuits?
For example, I know that if I put 2 diodes in parallel and one is reversed, they can't conduct current at the same time. Are any other
stuff like that, that I need to know and take into consideration while designing diode circuits?

Thank you!

PS: In the pdf, the Greek text is the exercise pronunciation which I translated for you at the beginning of this post.

#### Attachments

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#### Jony130

Joined Feb 17, 2009
5,332
First, try to find Vd voltage if Id current is equal to 4mA. So you will know a single diode voltage drop (Vd).

And it seems that the solution will look something like this:

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#### babaliaris

Joined Nov 19, 2019
115
Thank you so much!!! By doing what you told me for 1 diode and then for 2 in series I noticed that the voltage across each diode is the same plus the voltage across both of them is doubled while they are both at about 0.65 volts.

I also noticed that if you put more branches in parallel, (each branch has 2 diodes in series), the current in each branch is exactly the same so the current source is equally divided.

So with 4 branches you get 1mA in each branch which gives 1.24V across each branch.

The following PDF has my complete solution with all the mathematical proofs, including the proof that each branch has the same current.

I'm proud of this solution, it deserves an A+++++++++++

#### Attachments

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#### MrAl

Joined Jun 17, 2014
8,975
Thank you so much!!! By doing what you told me for 1 diode and then for 2 in series I noticed that the voltage across each diode is the same plus the voltage across both of them is doubled while they are both at about 0.65 volts.

I also noticed that if you put more branches in parallel, (each branch has 2 diodes in series), the current in each branch is exactly the same so the current source is equally divided.

So with 4 branches you get 1mA in each branch which gives 1.24V across each branch.

The following PDF has my complete solution with all the mathematical proofs, including the proof that each branch has the same current.

I'm proud of this solution, it deserves an A+++++++++++
This is interesting. I take it you MUST use 4ma exactly. That means in a string of 4 parallel branches there will be 1ma in each branch, exactly, and the voltage across each set of two diodes must be 1.25v exactly. However, that's not going to happen, at least not exactly. Your solution is probably good enough though so this discussion is just to make it a little more interesting.

For the first correction, we would have to place the original 4 strings of 2 diodes in series each in parallel just as planned, but then we'd have to place 26 strings of 3 diodes in series each in parallel., all in parallel with the first 8 diodes set. That gets us closer.
For the second correction, we'd have to take that and place another 33 strings of 4 diodes in each series string. That gets us very close but still not exact. To get exact there may be no real solution other than an infinite series.

The above two corrections require another 56 diodes (ha ha) for a total of 64 diodes.
If you are interested, try it, but keep in mind in order to simulate this exactly you have to create your own diode model that is simpler than the usual spice model. The new model will have to work exactly as the original diode formula for this exercise represents.

For extra credit calculate the next parallel string number of diode sets in parallel which would be made from series strings of 5 diodes each. That will get even closer to exactly 1.25 volts. The next string would be sets of 6 diodes each, etc., etc., until you obtain the desired accuracy of that 1.25 volt specification.

If you think this sounds crazy, then consider the mathematician who recently solved the old chess problem of 8 queens on a chess board none of which are allowed to attack any other according to the regular rules of chess. The solution on a regular chess board is not that difficult and even finding ALL solutions is not that hard. He took it one giant step farther however, and solved it for chess board with millions of squares, and actually any number of squares not just the regular normal 64 chess squares we all know about. He spent FIVE YEARS working out this solution

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#### crutschow

Joined Mar 14, 2008
29,768
As close as I got (within 0.22%) simulating with a few (less than ideal) diodes:

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#### Ian0

Joined Aug 7, 2020
5,076
Just put two diodes in series, then heat or cool it until it reads exactly 1.25V.
Don’t forget that T (absolute temperature) appears in the equation.

#### MrAl

Joined Jun 17, 2014
8,975
As close as I got (within 0.22%) simulating with a few (less than ideal) diodes:

View attachment 259314
That's an interesting variation and requires less diodes for the approximate solution.

#### MrAl

Joined Jun 17, 2014
8,975
Just put two diodes in series, then heat or cool it until it reads exactly 1.25V.
Don’t forget that T (absolute temperature) appears in the equation.
Yeah sure, rather than string 4 to 8 diodes together, just use two and build a sophisticated controlled temperature oven to heat them up until you get the right voltage. Sure, easy

#### Ian0

Joined Aug 7, 2020
5,076
Yeah sure, rather than string 4 to 8 diodes together, just use two and build a sophisticated controlled temperature oven to heat them up until you get the right voltage. Sure, easy
Easier than trying to buy a diode specified at exactly 0.7V @ 20mA @ 25°C.
If we are using hypothetical diodes, I can use a hypothetical temperature controller.

#### MrAl

Joined Jun 17, 2014
8,975
Easier than trying to buy a diode specified at exactly 0.7V @ 20mA @ 25°C.
If we are using hypothetical diodes, I can use a hypothetical temperature controller.
Ok sure, no problem.

In that case, i propose a single diode with exactly 1.250000000... volt drop at all currents and all temperatures and all humidity levels and all external radiation levels and has existed since an infinitely long time before the big bang and will exist for all eternity

Hey ya know in real life a voltage reference diode would fit the job pretty well.

#### Ian0

Joined Aug 7, 2020
5,076
Hey ya know in real life a voltage reference diode would fit the job pretty well.
that’s exactly what I thought when I saw the question.

#### MrAl

Joined Jun 17, 2014
8,975
that’s exactly what I thought when I saw the question.
Yeah interesting that many of them have the default value of 1.25 volts too without any divider resistors. Would be simple. I guess the lesson for the student here is to understand the diode and how it works with current and volt;age.

#### Ian0

Joined Aug 7, 2020
5,076
Yeah interesting that many of them have the default value of 1.25 volts too without any divider resistors. Would be simple. I guess the lesson for the student here is to understand the diode and how it works with current and volt;age.
That's a matter of physics. If you make a bandgap reference circuit, that's the voltage you end up with. I can't remember the exact maths but it's something to do with temperature changes cancelling out. Someone on this forum will know!
All other references are a 1.22V bandgap with an amplifier.

#### MrAl

Joined Jun 17, 2014
8,975
That's a matter of physics. If you make a bandgap reference circuit, that's the voltage you end up with. I can't remember the exact maths but it's something to do with temperature changes cancelling out. Someone on this forum will know!
All other references are a 1.22V bandgap with an amplifier.
Someone on this forum will know. Yes, that includes me
I actually designed a voltage reference using a different method too.
The old National Semiconductor parts catalog had an article about this too back in the early 1980's. I may still have the physical book somewhere.
I used to be very interested in this devices but moved on to other things a quite a while back. Now i just buy one if needed

#### MrAl

Joined Jun 17, 2014
8,975
This discussion is about finding a solution for 2.5 volts not 1.25 volts.
If anyone still wants to think about this it turns out more interesting than i thought at first.
For example, here is an equation that solves a rectangular matrix of diodes Ns by Np where Ns are the number in series and Np is the number of strings of Ns diodes in parallel:
2.5=0.025*log((2.8985507246376813e11)/Np)*Ns
Keep in mind if you want to try different strings with a lot of diodes you probably have to recalculate that constant in the log() function to a greater accuracy. Everything else is exact.

It seems like a 4x4 matrix of diodes is the most economical, getting to an accuracy of 0.025 percent.
Going to 5 series diodes in each string requires 597 diode strings in parallel, and gets down to 0.0036 percent, and going to 6 series diodes in each string requires 16747 parallel strings and gets down to 0.000035 percent.
Going to 7 series diodes in each string requires almost 200000 diode strings and gets down to 6.7e-6 percent, and last going to 8 series in each string requires over a million parallel diode strings and brings the error down to 2.4e-6 percent, not too much improvement vs the number of added diodes.

This all makes me wonder if there is a most elegant solution that gets extremely close, like say to an error of maybe 1e-16 or so.
Also, this was only for a rectangular matrix, we could also explore a triangular matrix, circular matrix, etc., see what comes out of it. The rectangular matrix was easy the triangular will be a bit harder with another variable.

What i noticed about the solutions is that the closer Np comes to an integer in the calculation, the better the approximation turns out to be. That would mean that if Np could come out to a perfect integer, it would be an exact solution. This probably wont work for a rectangular matrix, but perhaps some other arrangement might do it.

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#### MrAl

Joined Jun 17, 2014
8,975
As close as I got (within 0.22%) simulating with a few (less than ideal) diodes:

View attachment 259314
Hello again,

Unfortunately when we use the original diode model from this thread we get an error of greater than 3 percent with this arrangement of diodes. The voltage comes out to 1.292167 volts with those seven significant digits.
Try creating a model that matches the model of this thread see if you can verify that.