Hi,
lets answer your questions bottom up
Why in the equation the |Vtp| is shown?
well because:
the lowest value of Vicm should not be lower than the voltage at the drain of Q1 (-VSS + VGS3 = -VSS + Vtn + VOV3) by more than Vtp
Vcim ≧ -Vss + Vtn + Vov3 - |Vtp|.
I can't understand what they mean should not be lower than the voltage at the drain of Q1 by more than Vtp. I know that the NMOS current mirror needs some voltage to be in saturation, say 150mV. I know that Vgs = Vov + Vth, say Vov = 50mV (for weak inversion) and Vth around 500mV.
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sorry, clicked the sent button accidently...
In a way you answered your question yourself
but I will tried to find different words...
what they mean with
should not be lower than the voltage at the drain of Q1 by more than Vtp? This is simply the condition for the PMOS to stay in saturation.
To be in saturation the channel needs to be pinched off on one side. The channel below the gate only exists if the voltage difference is bigger than Vth (considering the right polarity...). So to remain in saturation the voltage difference between gate and one side (which is then referred to as drain) is not allowed be bigger than Vth.
As you stated: to determine the absolut allowed minimum value for Vcim, you start at Vss. Considering the voltage needed for the current mirrow (also to be in saturation) you come up with a maximum value at the drain of your input/diffpair transistor. You substract Vtp which gives you the minimal allowed value of Vcim. If Vcim would further decrease both sides of the transistor would be "on" = " full conducting channel below the gate" so no pinch off on one side and hence no saturation.
I hope it becomes clearer now
P.S. What I am a little confused about is the first minus in your equation: Vcim ≧ -Vss + Vtn + Vov3 - |Vtp|.
Vcim ≧ Vss + (Vtn + Vov3) (=Vgs for the current mirrow) - |Vtp|