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Is there a circuit either passive or not that can transform the value of a resistance with a fixed ratio ?

For example if the input resistance to the circuit is 5000 ohm and the ratio is 50:1 , the output resistance is 100 ohm.

and if the input resistance is 100k ohm , the output is 2k ohm and etc.

Elliot Alderson
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  • Do you need the 'resistor' to be floating, or can one side of it be grounded? If the latter, there are op amp circuits that do what you want, I believe. – Hearth Oct 15 '18 at 13:45
  • One side can be grounded .. would give me a sample circuit for that ? @Felthry – Ahmad Sayed Abdulrahman Oct 15 '18 at 14:00
  • https://en.wikipedia.org/wiki/Negative_impedance_converter This does what you want, but the ratio is required to be negative. I'm not sure how to make a positive one but it should be possible. Frankly, I can't figure out why you would want this rather than just getting a different resistor. – Hearth Oct 15 '18 at 14:18
  • @Felthry because it's not a resistor , it's a sensor which I want to transform the resistance between its terminals – Ahmad Sayed Abdulrahman Oct 15 '18 at 14:36

2 Answers2

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It is certainly possible to put together both resistance dividers and resistance multipliers.

A common approach is to use a bridge balance technique.

The one below was designed just to expose this concept. It can be easily reconfigured for multiplication or flipped "upside down" for operation close to the upper supply rail.

The virtual resistor is between the RPOS and RNEG terminals. R1 is the reference resistor whose value is divided.

The bottom end of the resistor doesn't need to be grounded; it just needs to be no higher than about 4V in this particular circuit. R7 is used to raise RNEG about 2V above ground to demonstrate this.

The resistor will not work well above at most a couple hundred Hz, due to the low speed of LM324. I'm only using LM324 since its input common mode voltage includes GND, and it works well enough for a demo.

For reasonable AC performance, OA1 needs to be bootstrapped: its (+) terminal should be at the midpoint of its supply rails. A symmetric bootstrap supply would do this job. You'd then use a fast, rai-to-rail I/O op-amp for OA1, running from 5V. The virtual resistor terminal voltage then extends from 2.5V to VCC-2.5V.

The MOS pass cell would need to be bipolar in applications that work with bipolar currents.

schematic

simulate this circuit – Schematic created using CircuitLab

The transfer curve for R1 swept from 0 to 100k ohm is:

The transfer curve of the resistance divider

The divider is 50, but due to poor gain of the op-amp, the divider is set to 49 to obtain accurate ratio at the output. A physical circuit would no doubt need a different correction.

R3 loads the diode D1 to ensure the gate is connected to the output of the op-amp at all times. The drop across D1 brings OA1's output into the valid output range. R5 is a bypass load to prevent latchup. This value could be higher in practice - this depends on the op-amp used, and on whether the circuit is simulated or breadboarded. CircuitLab has some DC operating point convergence problems that a lower R5 seems to help with.

The currents passed through this synthetic resistor are limited by dissipation on M1 and R4. Best dynamic performance - with an adequate op-amp - would be obtained at a current where M1 has best transconductance.

The bridge approach has a limitation: the minimum output resistance is R4.

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Yes, you can transform a resistance or impedance by using a transformer. However, it only works for AC.

A transistor is a portmanteau, which stands for transfer resistor. Transistors might get used for a so called buffer amplifier, which works for DC. Also it would be possible to use OP-amplifiers for such an inquiry (seeBuffer amplifier ).

Johannes
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