can admiitance be negative

i want to knw that for a simple dc ckt , can value of admittance be negative.
if so how..
i solved a numerical and found its y parameters for a two port nwk and value comes out to be negative which is same as in book...

Hi,

An active circuit can present a negative admittante or impedance.
For instance, a NIC (negative inmitance converter) presents a negative admittance.
Regards

Z

As zorro sad, active components can have negative impedance/admittance due to (inherent) positive feedback. It is used to create oscillators and circuits with hysteresis. All two-terminal devices with negative resistance slope (tunnel diodes, discharge lamps, etc) have negative admittance in some part of their operating curve.

It is important what WimRFP said about two-terminal devices (even passive) having negative slope (negative incremental admittance or resistance).

I overlook that Pooja told about a two-port network, not two-terminal.
In this case, with the conventions normally used for two-ports, even a serier resistor has negative Y12 and Y21 parameters.

Regards

Z

Zorro: To be honest, I also thought that Pooja referred to Y11 and Y22 (as Y21 and Y121 are negative in may many cases, look to your resistor example).

thanks WimRFP & zorro
"even a serier resistor has negative Y12 and Y21 parameters"..
it means resistance is negative and hence admittance...
i understand the explanation given for 2 terminal devices but not for two port nwk..
what is the meaning of negative resistance here and why it came and how it is possible ?

Hi pooja_khubbar

Y12 abd Y21 are not admittances but transfer admittances or transadmittances.
Given a two-terminal device, an admittance is the ratio between the current circulating across it and the voltage between its terminals.
Instead, a transfer admittance is the ratio between a current and a voltage, but in different parts of the circuit.
For a two-port, the parameter Y21 is defined as I2/V1 when V2=0 (i.e. port 2 short circuited)
In this two-port (this is the one we were speaking about):

o-------------/\/\/--------o<--- I2
V1..port1........................port2
o------------------------------o

with a short circuit in port2, I2=-V1/R, (look at the direction of the arrow, that by convention enters the network).
So, Y21=I2/V2=-R. With R positive, Y21 is negative.
The same (inverting the ports) is for Y12.

Not than Y11=Y22=1/R, i.e. positive.
Y11 and Y22 are true admittances, rather than transadmittances:
Y11=I1/V1 when V2=0
Y11=Y22=1/R

Regards

Z