Hello, At this manual time 2:20 they say that if Zout of the filyer will be at the same impedance as Zin then well have oscilations.
What is the logic behind it?
Is there a way to simulate it in ltspice?
Thanks.

 

Negative dynamic input resistance is a cause.

https://www.analog.com/en/resources/analog-dialogue/raqs/raq-issue-211.html

 

Negative dynamic input resistance is a cause.
https://www.we-online.com/catalog/media/o109021v410 AppNotes_ANP008_NegativeInputResistanceOfSwitchingRegulators_EN.pdf
https://www.analog.com/en/resources/analog-dialogue/raqs/raq-issue-211.html

Really nice, didn't know that. I always wondered how to tune a certain filter to not oscillate under dynamic loads, but there it is. It seems that it's possible to set a negative resistance in LTSpice. It remains for me to analyse such behaviors and verify the equations.
Thank you for that.

 

Really nice, didn't know that. I always wondered how to tune a certain filter to not oscillate under dynamic loads, but there it is. It seems that it's possible to set a negative resistance in LTSpice. It remains for me to analyse such behaviors and verify the equations.
Thank you for that.

I've banged my head on this problem more than once.
https://forum.allaboutcircuits.com/threads/dealing-with-negative-resistance.107553/

 

When SPICEing around the inductors might have zero ohms and the caps might have zero ESR. They will ring.
If you are looking at input filters in SPICE add the R to the L and Cs.
I sometimes I add damping resistors to the filter because the inductors have very low resistance.

 

Hello ,Given the article below at the photo every component has negative input impedance. Rin=-Uin^2/Pin
where this formula is valid?
As I see it every device obeys p=U*I so it has negative input resistance.
Where did I go wrong in the logic?
Thanks.


https://www.analog.com/en/resources/analog-dialogue/raqs/raq-issue-211.html

 

Is there a way to simulate it in ltspice?

Below is a way to simulate a simple negative impedance (current reduces with increasing voltage) in LTspice using the voltage-to-current source.
The G1 transconductance gain value (here 1) sets the value of the negative impedance.

 

Hello crutschow,few question:
1.impedance is a complex numberI think the meaning is negative resistance.
by ohms law R=V/I ,from the plot below both I and V are positive ,how is the resistance is negative?
2.why the Rin expression below is unique toa motor.
Thanks.

https://www.analog.com/en/resources/analog-dialogue/raqs/raq-issue-211.html

 

impedance is a complex number I think the meaning is negative resistance.

hi yef,

Rethink your definition of impedance.
Consider it as Vector parameter,

E
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html

Added a video.
https://www.google.com/search?clien...ate=ive&vld=cid:2bec8542,vid:l0f6c3AkTvA,st:0

 

1.impedance is a complex numberI think the meaning is negative resistance.

Yes, the circuit generates a negative impedance with no reactive component so it is a negative resistance.

how is the resistance is negative?

The current reduces as the voltage increases.
It's what the input to a switching-regulator does.

What is your definition of a negative resistance?

 

Hello Crutschow,my logic is R=V/I for example R=-5V/1A=-5Ohm
not reduction in value but the actual value, where am I wrong in the theory?

 

Hello Crutschow,my logic is R=V/I for example R=-5V/1A=-5Ohm
not reduction in value but the actual value, where am I wrong in the theory?

Your formula is just a reversal of the current through a normal resistance, which requires an amplifier.
Your logic is flawed.

Google AI says -- Negative resistance is an electrical characteristic where an increase in voltage across a device causes a decrease in current, creating a negative slope on a V-I graph.

That's what the simulated circuit does.