Is negative resistance possible?

In a passive device, negative absolute resistance cannot exist. However, negative differential resistance, where an increase in voltage leads to a decrease in current or vice versa, is observed in a number of rather common systems, such as neon signage and fluorescent lighting, as well as some more esoteric ones like tunnel diodes. Below is a figure showing an I-V curve for a generic electrical discharge; notice the region between points D and G where the voltage decreases as the current increases. This is the region in which both fluorescent lighting and neon signage normally operate.

A current-voltage curve for a generic electrical discharge. (image source)

Negative absolute resistance can exist over limited ranges by using active elements. There's an op amp circuit commonly called a negative impedance converter that simulates a negative resistance, capacitance, or inductance by using an op amp and feedback:

schematic

simulate this circuit – Schematic created using CircuitLab

The two circuits above are equivalent, provided the op amp does not saturate.


My apologies to everyone, the original solution was wrong, I had the direction of the currents through R2 and R3 reversed. Solution now edited.


If we Measure all voltages relative to the common junction of the 2 ohm resistor, R2, the 8A current source and the 3 ohm resistor then:

  1. Summing the currents at the node at the top right gives the 37V voltage source supplying a current of 7A. (15-8)
  2. At the - end of the 37V source the 2 ohm resistor has 6A flowing through it therefore the current through R2 is 1A to make up the 7A.
  3. The top end of the 2 ohm resistor is at -12V (2 ohms X 6A). and hence R2 = 12 ohms (12V / 1A).
  4. The node at the top right is at 45V as 15A flows through the 3 ohm resistor.
  5. The other end of R3 (the + end of the voltage source) is at 37-12 = +25V (the voltage across R2 and the 2 ohm resistor {-12V} + the 37V source)
  6. The voltage across R3, Vs = 20V (45-25).
  7. -7A is flowing through R3 and hence R3 = -20/7 ohms, or approximately -2.86 ohms.

The more I look at it, the more I think the "-" in front of the 4 and 20 in the answers is just a dash (hyphen) not a minus sign.


To get the current and voltage values that were shown in the book.
You indeed need a negative resistance in the circuit. The R3 needs to have negative resistance \$R_3 = - \frac{20V}{7A} = -2.857\Omega\$

Because for the positive resistance we get this result:

enter image description here

As you can see the result is not even a close to the book solution.

But if we use a "real" negative resistance (negative impedance converter) instead.

The simulation result will match the book solution:

enter image description here