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15th March 2006, 06:12 PM  #1 
diyAudio Member
Join Date: Sep 2005

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? 
15th March 2006, 09:39 PM  #2 
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Join Date: Apr 2003
Location: London

Something like this?
xpro 
16th March 2006, 06:04 AM  #3 
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Join Date: Dec 2005
Location: Kuala Lumpur

Two bipoler transistors wired as a thyristor or a neon tube

17th March 2006, 04:38 AM  #4 
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Join Date: Sep 2005

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. 
17th March 2006, 09:57 AM  #5 
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Join Date: Jun 2004
Location: Central CA

Using opamps, how about this?
And make R1,2,3,4,5 < R6,7,8,9,10,11,12,13 Tom 
17th March 2006, 12:14 PM  #6 
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Join Date: Jun 2004
Location: Central CA

Would this work???
Is the resistance  R1* R3 / R2 ? Assuming R1=R5 and R2=R4 . Tom 
17th March 2006, 01:09 PM  #7  
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Quote:
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Nigel Goodwin 

17th March 2006, 08:13 PM  #8 
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Join Date: Sep 2005

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.) 
17th March 2006, 09:28 PM  #9  
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Quote:
__________________
Nigel Goodwin 

18th March 2006, 10:27 AM  #10 
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Join Date: Jun 2004
Location: Central CA

rtarbell,
The four opamp 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 opamp 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 opamp 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 opamp 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 opamp version of two second generation current conveyors forming a negative impedance converter or nic. See the attachment. Tom 
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