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# current mirror gain

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#### zxcv2201

##### Member level 1
I know that the current mirror's gain is determined by the ratio of W/L.
I want to know if the ac gain in the current mirror is also like this.

I want to find the gain in the following circuit.
If a small signal goes in with the IREF signal, is there a small signal gain at the Iout stage?

Depends on M2/M1 ratio.
Also, Vb impacts the Current ration until M2 gets into saturation.

Let's complicate the problem...

For large signal, mosfet works in saturatation region:
• Ids1 = 0.5*Kn*W1/L1*(Vgs-Vth)^2
• Ids2 = 0.5*Kn*W2/L2*(Vgs-Vth)^2
Hence,
• Ids1/Ids2 = (W1/L1) / (W2/L2) = Ratio1/Ratio2

Now turn to small signal analysis, gm of a mosfet is defined by "gm=ids/vgs = delta_Ids/delta_Vgs"
• gm1 = 2*Ids1/(Vgs-Vth)
• gm2 = 2*Ids2/(Vgs-Vth)

The small current "iin" injected into "X" is fisrt converted to "vin" by gm1, then converted to "iout" by gm2, hence
• iout/in = gm2/gm1 = Ids2/Ids1 = Ratio2/Ratio1

All discussion above assume ”gds“ is much smaller than "gm“ , the interesting frequency of the signal "iin" is much smaller than "gm/(2*pi*cp)"
# cp is mainly contributed by the gate capacitance of M1 and M2
# gds is defined by "gds = ids/vds"
# the function of M3 is used to decrease the effective gds of M1 observed at "P", then the current source behaves more like a ideal one

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