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Given a resistive element that fluctuates between 0 and 10kOhms, a PNP transistor, a 20V source and a bunch of resistors, is it possible to construct a circuit with an output voltage as follows:

1) Linear output with respect to the resistance of the variable element, at least within the 2 to 5kOhm range.

2) Output voltage is 0V when the variable element is at 2kOhms

3) Output voltage is 5.12V when the variable element is at 5kOhms

i.e. we want to use these elements to construct a circuit that has some \$V_o(R) = \frac{5.12}{3000}*(R-2000) \$

I am very unskilled in circuit design. Can anyone give me any pointers?

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    This isn't homework, it's part of a research project. I designed a working circuit with an opamp but my professor said that they are hard to work with, and gave me some transistors to use instead. –  May 30 '13 at 17:25
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    I'm guessing the professor isn't an EE professor... – Dave Tweed May 30 '13 at 17:29
  • I've been fiddling on paper with schematics on and off for the better part of 3 days, I've put a good deal of effort into trying to find the solution. –  May 30 '13 at 17:30
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    Your professor doesn't know what he is doing, at least when it comes to electronics. – Olin Lathrop May 30 '13 at 17:52
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    Your verbal description and your formula are off by 3 orders of magnitude! – Olin Lathrop May 30 '13 at 17:54
  • Out of curiosity, what is the age of the person that told you to not use op-amps? – Anindo Ghosh May 30 '13 at 18:23
  • How are my verbal and formulaic descriptions different? –  May 30 '13 at 18:27
  • Furthermore how is this an unreasonable question for this website? –  May 30 '13 at 18:45
  • The formula should be \$V_o(R) = \frac{5.12}{3000}*(R-2000) \$. The question is unreasonable because you already had a working solution, and the reason for seeking a different one is specious. – Dave Tweed May 30 '13 at 18:55

2 Answers2

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Use the PNP transistor to create a current source that drives 1.70666 mA through the variable resistance.

Then, you just need to subtract off 3.41333 V of offset, which can be done by tying the bottom end of the resistance to a negative supply of that value. Or you could set up a second current source (current mirror) that drives the same current through a fixed 2K resistor and take the output voltage as the difference between them.

schematic

simulate this circuit – Schematic created using CircuitLab

Note that 5.45KΩ can be created as the parallel combination of 10KΩ and 12KΩ. See this question if you have a limited selection of resistor values.

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Dave Tweed
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This isn't a complete answer since your constraints are silly. However, consider that you can get a voltage proportional to a resistance by running a fixed current thru the resistance. In your case you want a 3 kΩ difference to cause a 5.12 V change. 5.12 V / 3 kΩ = 1.71 mA, which is the fixed current that would get you the desired sensitivity.

The current source thru the resistance under test by itself will result in a 0-based voltage. In other words, 0 V is a result of 0 Ω. 2 kΩ will result in 3.41 V, so you'd have to subtract off this amount to get the signal you want. This is where a opamp would be really useful. Look up something called a "level shifter".

Olin Lathrop
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