Negative resistance with switched capacitor

[pseudockb]

capacitor negative resistance

I know this is very fundamental but I could not grasp it correctly. The circuit in the left shows a positive resistor while the one at the right shows a negative resistor. Here's my analysis on the positive resistor.

Assuming from S1 is on from 0 to T/2 but off from T/2 to T,
S2 is off from 0 to T/2 but on from T/2 to T.
V1 is the voltage at the left port while V2 is the voltage at the right port.

The current i1 from V1 to V2 is then given by
C[V(T/2)-V(0)]/T = C[(V1-V2)-0]/T
As such, we get the equivelent resistance of T/C

Similarly, when I performed the same analysis on the right hand side circuit, I got:
i1=C[V(T/2)-V(0)]/T = C[V1-(-V2)]/T

I do not understand how does this circuit give a negative resistance value? Is my analysis correct? Thanks

 

[easytarget]

Yes, your analysis is correct. For the negative resistance, if we say the left plate has the positive polarity, then:
q(0) = C(V1-0)=CV1
q(T) = C(0-V2)=-CV2
I = dq/dt = C(-V2-V1)/T

Edit: so R = -T/C

Edit#2: MY ANSWER IS WRONG. Ignore my crappy answer, sorry.

 

[fantaci]

pseudockb:

your analysis is right, but your concept is wrong. I think this is not a negtive resistor, instead it is a negtive transresistor which is from v1 to v2.

Accutually, the exact model is not just a simple resistor connected btween V1 and V2. You can refer to Phillip E. Allen`s book [cmos analog circuit design, second edition]page535.

Anyway, in most application, the node V2 will be bonded to the minus input of CMOS AMP(the positive input is grounded), so you can view V2 as gound. In this case, you can see that the current from V1 to V2 is -C/T*V1, so the equivalent transresistor from v1 to v2 is -T/C.
That where the negtive transresistor comes from. It is called the transresistor, I think, because it does not model all the current from v1, but just the current flow from v1 to v2 or vice verse.

Hope it helps.