Braski
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cheenu2002 said:
you could identify the direction of the current variations induced by the variations of the small signal voltage applied to the input terminal as they flow in the output resistance of the first stage (in a Norton-like representation).
In your design, for example, the small signal current of QP2 is pushed in the output resistance, so if it increases the voltage rises up! The small signal current of QP1 flows across the mirror and is pulled from the output resistace. An increase of that current causes the output voltage to go down! So, an increase of Vinp causes the current flowing through QP1 to decrease, and the rising of the output voltage as a consequence. The same, if you think to connect +Vd/2 to Vinn, an increase of Vd causes the output voltage to decrease.
But you have another gain stage (common source) that causes another sign inversion. So i think that Inn becomes the non-inverting input.
You could also take into account the sign inversions of the signal. Let's consider the signal applied to Inp. From the gate to the drain of QP1 we have a sign inversion. From the gate to the drain of QN0 we have the second sign inversion. Finally, to reach the output, we have another sign inversion from the gate to the drain of QN2. It appears clear that 3 sign inversions causes Inp to be the inverting input. Inn sees only 2 sign inversions to reach the output that makes it to be the NON-inverting input.
If you consider ONLY the diff pair, Inp remains the non inverting input, and Inn remains the inverting one.
You can always reason this way.
Hope i've been clear, it's my first post.
