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  1. #1
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    Voltage drop across cascode amplifier

    I can say that I understand simple current mirrors fairly well, but what i do not understant is the minimum output voltage requierd for the circuit to work properly...for the transistors to be in saturation.

    NMOS cascode current mirror

    Vout=V(D4). Vds_sat is the saturation voltage of the transistor M2 and M4. Vt is the treshold voltage.

    Why is the minimum output voltage, on the drain of the M4, equal to the Vout_min=Vt+2*Vds_sat, why not only 2*Vds_sat?

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    What will be the voltage drop for self cascode current mirror. How is it equal to 2vdsat.


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  2. #2
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    Dominik Przyborowski's Avatar
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    Re: Voltage drop across cascode amplifier

    Cascode transistor and it biasing can be considered as so-called "gate follower".
    In the first figure, source terminal of Q4 copying the source potential of Q3, which is nothing more like Vgs of Q1=Vdsat1+Vth1. So, this cascode current source working properly as long as Vout-VS(Q4) > Vdsat4. However VS(Q4) is already equal to Vdsat1+Vth1. When we decrease the Vout below Vth+2Vdsat, cascode transistor Q4 is in linear region (as long Vout>Vgs(Q1) potential at source of Q4 is constant) and the circuit acts as simple current mirror.



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  3. #3
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    Re: Voltage drop across cascode amplifier

    Thanx for your reply. What about the second circuit. The drain of upper transistor is connected to gate of the lower transistor. So will vout = vgs1 (lower transistor). Both transistor will be cutoff if i consider this case



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