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Old 15th March 2006, 05:12 PM   #1
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Default Two port negative resistance element

Okay, this isn't exactly audio, but it could be used for audio purposes.

==> I have seen several circuits utilizing positive feedback that have negative input impedance in a one port model (meaning that one end of the negative resistor model is ALWAYS tied to ground).

I'm looking for a negative resistance in a two port model where both ends of the negative resistor model are left floating. Does anyone know of a schematic or a paper online that addresses such a model, or how to build one?
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Old 15th March 2006, 08:39 PM   #2
x-pro is offline x-pro  United Kingdom
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Something like this?

x-pro
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File Type: gif l_diode1a.gif (8.1 KB, 181 views)
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Old 16th March 2006, 05:04 AM   #3
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Two bipoler transistors wired as a thyristor or a neon tube
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Old 17th March 2006, 03:38 AM   #4
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I should have included a schematic, my fault for not doing so. What I'm looking for is a negative resistance element that is floating and can supply power. Take a look at the included schematic attached. If I did my math correctly, the input resistance (as seen by the source Vin) is:

Rin = - R1 * R3 / R2

Thus, the entire circuit can be modeled as a negative resistor (with value equal to -R1 * R3 / R2) tied to ground from the Vin viewpoint. I'm looking for a circuit that acts as a negative resistor with neither end grounded.
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Old 17th March 2006, 08:57 AM   #5
Tom2 is offline Tom2  United States
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Using op-amps, how about this?
And make R1,2,3,4,5 < R6,7,8,9,10,11,12,13

Tom
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File Type: png negativeres2.png (8.0 KB, 123 views)
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Old 17th March 2006, 11:14 AM   #6
Tom2 is offline Tom2  United States
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Would this work???
Is the resistance - R1* R3 / R2 ?
Assuming R1=R5 and R2=R4 .

Tom
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File Type: png negativeres3.png (5.6 KB, 111 views)
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Old 17th March 2006, 12:09 PM   #7
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Quote:
Originally posted by rtarbell
I'm looking for a circuit that acts as a negative resistor with neither end grounded.
Perhaps you should give us more details about what you're trying to do? - it makes no sense to me so far!.
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Old 17th March 2006, 07:13 PM   #8
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Yes! Mr Tom2, I think you have it! I did the hand calculations, and I think your second circuit works! I didn't trty the first one you proposed, it got a bit long for me on paper.

I had a professor in school that taught a class called Opamp Design. It was a class about lots of neat circuits that can be built with opamps (simple amplifiers, generail impedancec converters, negative impedance converters, etc.). The negative resistance circuit really interested me because he showed us the circuit I posted before, and he derived the input resistance for us. I wanted to take it a step further because the model is only for a one port circuit, but Tom2 has found a way to make it differential!

(You can use negative resistance circuits in applications such as oscillators. The circuit can cancel out the positive resisance in a crystal oscillator itself so that the only elements that are left are the L and the C, and thus the charge will slosh back and forth indefinately.)
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Old 17th March 2006, 08:28 PM   #9
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Quote:
Originally posted by rtarbell
(You can use negative resistance circuits in applications such as oscillators. The circuit can cancel out the positive resisance in a crystal oscillator itself so that the only elements that are left are the L and the C, and thus the charge will slosh back and forth indefinately.)
Isn't that just an amplifier with positive feedback? - which was what your opamp circuit looked like as well?.
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Old 18th March 2006, 09:27 AM   #10
Tom2 is offline Tom2  United States
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rtarbell,

The four op-amp circuit is simple to understand.
Let R1,2,3,4,5 be 100 ohm.
Let R6,7,8,9,10,11,12,13 be 100k ohm.
Let a voltage difference of dv appear between in1 and in2.
Then by the op-amp action of U1 and U2, dv occurs across R1.
This creates a current through R1 which is the same current
through R4 and R5.
Since R1=R4=R5 then dv also occurs across R4 and R5.
The op-amp circuit of U3,R6,R7,R10,R11 and R2 form a voltage to
current converter, i.e. the voltage across R4 occurs across R2.
Thus a current flows through R2 equal to dv/R2 and
since R1=R2 then this current is also equal to Dv/R1.
The same reasoning also applies to the lower op-amp U4 circuit
as a voltage to current converter.
Thus the currents at the in1 and in2 nodes will be equal and
opposite in sign and equal to +/- dv/R1.
Then for a positive dv, current flows out of in1 and into in2.
Thus the input Z is -R1.

The circuit is an op-amp version of two second generation current
conveyors forming a negative impedance converter or nic.
See the attachment.

Tom
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File Type: png negativeres1.png (3.7 KB, 77 views)
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