hi mayank,
                 first time one person trying to think differently to give ans for this question. thank you. But i didn't understand how different carrier densities at the end because different potential means different electric field, and during equilibrium how voltage will drop?

Thanks,
rajasekhar.

Hi again raja,
                     Lemme try to elaborate a bit more...Say you have two ideal voltage sources whose -ve terminals are connected to GND & +ve terminals are hanging as floating(Not connected as of now, No connections, no leakages, no arching nothin...). Then you can assume these points to be ,say, A = 2V + terminal , B = 3V +ve terminal) and assert that these act as POINT CHARGES and have separate electric fields of their own...say k.q1/r² and k.q2/r² But force exerted on one due to another will always be same --> k.q1.q2/r²...Since the reference GND wrt which the Potentials of points A and B are different [recollect V = k.q/r], this itself means that the amount of charge present at these nodes is different...So I guess that explains the different charge densities at both the points..

Now, as soon as you connect them through a conductor, The electric field between the two points becomes the -ve gradient of the potential between two points. [It always is the -ve gradient of pot diff b/w 2 points but till the time they were unconnected, we could have argued that individual E fields exist and are differentiable]. Now the E field becomes (3V-2V)/ Length of the conducting wire...

But, what if the connecting wire is a superconducter ? ZERO resistance...Perfect e- gas....
Two processes will come into play -->

1>   Diffusion process  ---> as i explained earlier...Charge carriers e-s diffuse from higher density to lower density -- until dynamic equilibrium is attained...without any external force...

2>  E Field b/w two points ---> As soon as you connect the nodes, There will exist a E field between two points which will point from +ve node to -ve node (3V node to 2V node) because of different charge densities at both the points...[remember from prev post..why did E field generate in d 1st place ?? --> coz of diff charge densities.] This E field will cause e-s to move from 3V end to 2V end...reducing E field magnitude in the process..until finally an equilibrium is attained when both potentials become equal...

Both there processes will ensure that the superconducting wire...will essentially settle at a common voltage...(ultimately making both the points appear as SHORTED)...But what makes them a SHORT  ?? --> These two processes....

If the wire has a finite resistance...Then also initally there will be a voltage drop across the wire...But eventually, both the ends will settle to same voltage levels, because of the above mentioned processes...Differnece being that In this case, the diffusion process will be confronted by the finite resistance of the wire...

Hope that explains it all....Still welcoming suggestions....These are ultra-basic topics involving physics...any1 can be wrong anytime  :P

Anyways, Did this help raja or made it even more confusing ?

cheers,
Mayank....

Hi ,
      Great... mail me at singhmayank19@gmail.com...Anyways, raja...i was just curious what's cedt -- name of some organisation ?? You working in industry / academics ?

As per your reply, In superconducting case...It's correct..atleast according to me...In normal resistance wire...Only Initially Potential difference will exist...I will try to explain in the mail itself...
regards,
Mayank.

raja.cedt wrote on Oct 26^th, 2009, 9:56pm:


Regardless these are OTAs or OPAMPs both can be modeled by two voltage sources with a finite impedance (high for the OTA, low for the OPAMP and least at low freq). (see drawing)
Note: Although the OTA is usually modeled as a controlled current source and therefore it has an output shunt impedance this assumption is still valid.

The non-zero output impedance is a valid assuption since we are talking of actual amplifiers (as Jerry suggested before this circuit does not make sense for ideal OPAMPS, and even for ideal OTAS it does not make sense to consider both have different input referred offsets....otherwise they would not be ideal )

So if V1 is the first amp and V2 the second amp,  then V1= 10mv and V2= 20mv. Therefore Vo will be the voltage division of these two voltages. I suggested that if the open loop gain is not greatly affected by the different unbalances, then Z1 and Z2 will be similar and Vo should be close to 15mv. Otherwise Vo wil be set by the corresponding voltage division

Best
Tosei

loose-electron wrote on Oct 28^th, 2009, 3:52pm:
That assumes a 2-transistor inverter. If you need an inverter with a lot of drive strength, you might want to use three stages to reduce input capacitance; and I have seen standard cell libs offering such inverters...
BOE

Hi,
      thanx Jerry....I understood the functioning...
      @ Boe : Can you explain what do you mean by 3-stage Inverters ??

regards,
mayank.

Hi jerry,
          Ohh, it's that  8-) ...I usually call it  clock/data buffer-chain....Increases drive....I got it...I thought it was some fundoo Inverter design with 3 MOSes like some current-starved kinda thing...  ;)

thanx,
Mayank.