# Spatial Matrix

Discussion in 'General Electronics Chat' started by T.Jackson, Nov 22, 2011.

Nov 22, 2011
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2. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
14
Then you do this one ...

:: OHM's Law Quiz ::

Requirements:

• 6 resistors ...
• 5 specific Vrefs of: 1, 2.2, 3, 3.3, and 4.5VDC
• IT = 20mA
• Calculate value of R1 - R6
Code ( (Unknown Language)):
1.
2. [FONT=Courier New]              4.5V      3.3V      3V        2.2V      1V
3.                |         |         |         |         |
4.   5V o--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--o GND
5.           R1        R2        R3        R4        R5        R6[/FONT]
6.
Show all workings, natural voltages measurable from ground.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Is there a trick?

R6=1/20mA=50Ω
R5=(2.2-1)/20mA=60Ω
R4=(3-2.2)/20mA=40Ω

and so on ....

R1=(5-4.5)/20mA=25Ω

I'm reluctant to open a zip file for fear of viruses.

4. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
14
That's wrong.

No need to fear of virus. I don't do that.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Many thanks for the interesting challenge.

If you mean my answer to your question is wrong then you might be kind enough to help me correct my errors.

6. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
14
Let me get a calculator and type it all out.

7. ### SgtWookie Expert

Jul 17, 2007
22,182
1,728
No, no calculators!

You must use a slide rule.

8. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
14
:: Problem Solution ::

Code ( (Unknown Language)):
1.
2.
3. [FONT=Courier New]Step 1.  Calculate RT
4.
5.             RT = V / IT
6.                = 5 / 0.02
7.                = 250Ω
8.
9. Step 2. Calculate R1  R5
10.
11.            Using Formula: R = (V - Vref) / V x RT
12.
13. Therefore ...
14.
15.              R1 = V  Vref1
16.                 = 5 - 4.5
17.                 = 0.5 / V
18.                 = 0.1 x RT
19.                 = 25Ω
20.
21.              R2 = Vref1  Vref2
22.                 = 4.5 - 3.3
23.                 = 1.2 / V
24.                 = 0.34 x RT
25.                 = 60Ω
26.
27.              R3 = Vref2  Vref3
28.                 = 4.5 - 3.0
29.                 = 0.3 / V
30.                 = 0.06 x RT
31.                 = 15Ω
32.
33.              R4 = Vref3  Vref4
34.                 = 3.0 - 2.2
35.                 = 0.8 / V
36.                 = 0.16 x RT
37.                 = 40Ω
38.
39.              R5 = Vref4  Vref5
40.                 = 2.2 - 1.0
41.                 = 1.2 / V
42.                 = 0.24 x RT
43.                 = 60Ω
44.
45.   Step 3. Calculate Trimming R
46.
47.              R6 = RT   (R1 + R2 + R3 + R4 + R5)
48.                 = 250  (25 + 60 + 15 + 40 + 60)
49.                 = 50Ω[/FONT]
50.

9. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
14
I think that's right. Been a while.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782

I notice I got the same answers as you did despite being "wrong".

11. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
14
Arh see when I just see numbers with no working I automatically assume it to be insanity and totally wrong.

You need to explain how you arrive at the answer.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I understand. I'll keep that in mind. Cheers!

13. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
14
How would you do it if the Vref was required across each resistor instead?

14. ### Wendy Moderator

Mar 24, 2008
20,765
2,536
Smell the smoke?

15. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Depends on how much current was needed at each voltage point.

i.e. set I=1mA and you have a lot higher resistor values, set I=20 A and you have very low resistor values (and a lot of heat).

16. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
He showed his work/formulas. The steps between could be done mentally instead of a calculator after making a few hundred voltage dividers for other circuits.

17. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
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Well he's extremely gifted if he can do all of that mentally. I sure as hell cannot.

18. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
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I actually remember it all as a 'protocol' (set of step-by-step rules) -- that I was formally taught.

1. Decide on IT
2. Find RT
3. R = (the voltage that you want / the voltage that you have) x RT

I have memorized many things such as resistor colour codes in the form of sequential colours.

When you study, you write down on a piece of paper many times and you never forget.

Last edited: Nov 23, 2011
19. ### steveb Senior Member

Jul 3, 2008
2,433
469
Actually, t_n_k is gifted at seeing easy solutions to complex problems. We've seen that here before. This problem might not be all that complex, but still a person might not see the simple approach right away. Often we take the hard way, out of habit.

In general, problems have easy solutions and difficult solutions, so it is important to not just dive straight into a problem and do it by the standard methods. Often a little thinking reveals a viewpoint that makes the whole problem seem trivial.

A computer program that was developed to solve very general resistor network problems, might not be able to find t_n_k's simplifying viewpoint on this problem, and it might write out a bunch of simultaneous equations and then solve them in a "brute-force" way. I'd be willing to bet that if students were given this problem on a test, some would dive straight in with simultaneous equations and waste 10 minutes that could be spent on the other problems. The calm guy (or girl) would take time to look carefully and then see the easy way.

20. ### T.Jackson Thread Starter New Member

Nov 22, 2011
328
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It is hard to teach an old dog new tricks boss.