UCC28070: PFC design using shunt base current sensing

Hello Haresh, 

Thank you for your interest in using the UCC28070 with a shunt-resistor in the MOSFET source leg. 

In both the CT design and the shunt-R design, the CSA and CSB input are looking for a Vs signal that is an equivalent representation of the MOSFET drain current Id.
Re-arranging your equation above, Rs/Nct = Vs/Ip, where Ip is the highest peak value of Id.    (To maximize signal with respect to noise, Vs is typically set to ~3.6Vpk.)

Whether CT design and the shunt-R design, Vs/Ip should be the same. 
In a CT design, the turns-ratio makes a large current look small to apply to a large resistance to get Vs.
In a shunt design, you need an amplifier to make the same Vs from a large current through a small resistance.

With a shunt, Ip*Rs*Av = Vs, so Vs/Ip = Rs*Av.  Av is the gain of the shunt amplifier. 
Now we need to distinguish between Rs used with CT and Rs used as a shunt.  I propose "Rs_ct" and "Rs_sh".

This means that Rs_sh*Av = Rs_ct/Nct.  

==> Now, everywhere in equations where you see Rs/Nct (meaning Rs_ct/Nct), you can substitute Rs_sh*Av.  
==> And, everywhere in equations where you see Nct/Rs_ct, you can substitute 1/(Rs_sh*Av).  
With these substitutions, you can calculate Rsyn, Rimo, and Rzc.  

How do you choose Av?  First you choose Rs_sh value based on the power dissipation that you can tolerate. 
The UCC28070 was designed to use CTs so that a high current-sense signal level can be obtained with negligible power loss.  
Using shunt sensing will incur power loss.  The same Vs signal level would result in excessive power loss unless amplification is used. 
But there is a trade-off to be made:  too small Rs_sh value needs high amplifier gain (Av).  High Av requires wide bandwidth and will also amplify noise, and could be costly.
Using low Av can be less expensive but needs higher Rs_sh with more loss.  

Choose an acceptable loss for Rs_sh (remember that there are two of them), then determine Av needed for the same Vs.  
If this Av is too high (i.e. suitable op-amp costs too much), then consider to trade-off a little less gain for a little more power loss or a little lower Vs peak signal, or both. 

Regards,
Ulrich

Hello Haresh, 

Yes, I think a two-stage gain can work.  But if they are the same op-amp, I think both stages will need the same bandwidth, so I suggest making their gains equal: 3.162 and 3.162, for example.
Also, consider that each op-amp will have its own input voltage and current offsets so the effect of all offsets may be larger than with a single op-amp.  

You can certainly give it a try, and see if the results are acceptable. 

Regards,
Ulrich