MrAl:
That is a modification to an existing DC/DC converter board. They come with a feedback divider already on the PCB and I believe R2 is a part of that divider. The idea is, I presume, to nudge the output of that divider.
ifonlyeverything:
Just skimming over your calculations, there is one thing that should ring alarm bells in your head. You have 2 unknowns, but 6 equations!
As for the solution, you have those currents:
I
1 = (V
PSU - 2.5 [V]) · G
1I
2 = 0.001 [A]
I
3 = (V
DAC - 2.5 [V]) · G
3In case you did not encounter the notation yet, G = 1/R (
electrical conductance). This just saves us from having 1/R everywhere in equations — there is no conceptual or numerical difference.
From KCL:
I
1 - I
2 + I
3 = 0
… equals:
(V
PSU - 2.5 [V]) · G
1 - 0.001 [A] + (V
DAC - 2.5 [V]) · G
3 = 0
You have two constraints:
I. V
DAC = 3.3 [V], V
PSU = 4 [V]
II. V
DAC = 0 [V], V
PSU = 6 [V]
That gives us a system of two equations:
(4 [V] - 2.5 [V]) · G
1 - 0.001 [A] + (3.3 [V] - 2.5 [V]) · G
3 = 0
(6 [V] - 2.5 [V]) · G
1 - 0.001 [A] + (0 [V] - 2.5 [V]) · G
3 = 0
From that you calculate G
1 and G
3, which are inverses of resistances you want.
Since this is just two-dimensional space, the system can be solved visually. See the attachment. The solution is around (G
1 = 0.5 mS, G
3 = 0.3 mS). That corresponds to resistances of about 2000 Ω and 3333 Ω.
Since you have used a linear solver yourself, let’s use linear algebra approach as the second one. The entire problem is written as: A·x = b:
⎡ 4 - 2.5 3.3 - 2.5 ⎤ ⎡ G₁ ⎤ ⎡ 0.001 ⎤
⎣ 6 - 2.5 0 - 2.5 ⎦ · ⎣ G₃ ⎦ = ⎣ 0.001 ⎦
⎡ 1.5 0.8 ⎤ ⎡ G₁ ⎤ ⎡ 0.001 ⎤
⎣ 3.5 -2.5 ⎦ · ⎣ G₃ ⎦ = ⎣ 0.001 ⎦
The solution is, x = A
-1·b, equals (
see Wolfram Alpha):
⎡ 0.000503817 ⎤
⎣ 0.000305344 ⎦
Which, this is not a coincidence, is a point — just like with the graphical method. The corresponding resistances are: 1985 Ω and 3275 Ω.
Finally, the high school approach, step by step:
⎧ (4 [V] - 2.5 [V]) · G
1 - 0.001 [A] + (3.3 [V] - 2.5 [V]) · G
3 = 0
⎩ (6 [V] - 2.5 [V]) · G
1 - 0.001 [A] + (0 [V] - 2.5 [V]) · G
3 = 0
⎧ (1.5 [V]) · G
1 - 0.001 [A] + (0.8 [V]) · G
3 = 0
⎩ (3.5 [V]) · G
1 - 0.001 [A] + (- 2.5 [V]) · G
3 = 0
⎧ (1.5 [V]) · G
1 - 0.001 [A] + (0.8 [V]) · G
3 = 0
⎩ (3.5 [V]) · G
1 - 0.001 [A] + (- 2.5 [V]) · G
3 = 0
Substracting:
(1.5 [V] - 3.5 [V]) · G
1 - (0.001 [A] - 0.001 [A]) + (0.8 [V] - (-2.5 [V])) · G
3 = 0
-2 [V] · G
1 + 3.3 [V] · G
3 = 0
Calculating G
1:
G
1 = (-3.3 [V] · G
3) / -2 [V] = 1.65 · G
3Substituting G
1:
1.5 [V] · G
1 - 0.001 [A] + 0.8 [V] · G
3 = 0
1.5 [V] · (1.65 · G
3) - 0.001 [A] + 0.8 [V] · G
3 = 0
2.475 [V] · G
3 - 0.001 [A] + 0.8 [V] · G
3 = 0
(2.475 [V] + 0.8 [V]) · G
3 - 0.001 [A] = 0
G
3 = 0.001 [A] / 3.275 [V]
Plugging G
3:
1.5 [V] · G
1 - 0.001 [A] + 0.8 [V] · G
3 = 0
1.5 [V] · G
1 - 0.001 [A] + (0.8 [V] · 0.001 [A] / 3.275 [V]) = 0
1.5 [V] · G
1 - 0.001 [A] + (0.0008 [A] / 3.275) = 0
1.5 [V] · G
1 + (-0.002475 [A] / 3.275) = 0
G
1 = 0.002475 [A] / 4.9125 [V]
… which gives us the resistances of R
1 = 4.9125 / 0.002475 ≈ 1985 Ω, and R
3 = 3.275 / 0.001 = 3275 Ω.