current-mode error amplifier gain
There isn't much to talk about. The UC3842 only uses peak current mode, so the inner current loop is built with just a comparator and the gain of the loop is rather low.
Basically, you should see the inner loop like a black box with the input being the error amplifier output, Vea and the output of the black box being simply the power supply output current. So, the output of the error amplifier controls the output current, like I showed in the previous post: Vea increases, it allows the output current to increase. The output current multiplied by the load resistance gives the output voltage. This is then fed to the voltage loop, the "outer" loop, which is really the error amplifier.
So now what is the gain of the current (inner) loop? First, let's assume that the inductor current ripple is very small, such that Ipk≈Iout.
Now as we have seen, the comparator will limit the peak inductor current to a certain value, simply by comparing the Vea to the voltage across the sense resistor, Rs. In reality, it compares Vea/3 to the voltage across Rs.
Generally, the sensed current is not the load current, but the primary current, thus, you sense a current that is N times lower, where N is the turns ratio of the transformer. Therefore, you can write:
Ipkprimary=Ipk/N
The voltage across Rs is Ipkprimary*Rs
So the comparator ensures Ipkprimary*Rs is always equal to Vea/3
Then, Vea/3=Ipkprimary*Rs=Ipk/N*Rs≈Iout/N*Rs
So this is the gain of the inner loop Gil=Iout/Vea (output/ input, and it has the dimension of a conductance)
Gil=N/(3*Rs) That is not a high gain, especially if N is low, such in the case of DC/DC converters.
With that, the output voltage is Vout=Rload*Iout≈Vea*N/(3*Rs)
As you can see, you do not have much control over it, since the turns ratio is usually fixed based on the input/ output voltages and maximum allowable duty-cycle, while Rs is chosen such that the voltage across it is about 0.7V at full load. That pretty much fixes your inner loop gain.
Thie is an idealized situation. We assumed that Ipk≈Iout, but in reality, the waveform will look much like my picture, wo there is some error there, since the peak is higher than the actual Iout. Also. I mentioned that ususally you sense the current on the primary side. That means that you will also see the magnetizing current of the transformer (another ramp) added on top of the actual reflected current. That will make the peak "look" higher than it actually is.
All this boils down to one thing: the current you are controlling is NOT the output current (that you wish you controlled), but a rather inaccurate image of it. The lower the output current, the greater the error.
Having said all that, current mode control is great and peak current mode control works just fine, being used in zillions of products.
(I hope I did not make any mistakes in my little dissertation here; you should double check that).