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# help on derive Iout in terms of the resistor R1 and (W/L).

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#### imperza

##### Junior Member level 3
Hi all,
I want to ask help here on how derive the current Iout in terms of the resistor R1 and (W/L) . The figure shows a supply independent biasing.

Hi,
The original idea is to use same size nmos pair and different size pmos pair. The right-hand pmos should be stronger than the other one.
For the 1st order calculation (assume perfect saturation characteristic for both nmos and pmos, and vdd is higher than the min value which is required to keep all mosfet in saturation):
assume left side pmos (P1) size W/L, right one (P2) n*(W/L);
I(P1)= beta*(W/L)* (Vod_p1)^2 ...(1)
I(P2)= beta*(n*W/L) * (Vod_p2)^2 ...(2)
The nmos mirror makes sure I(P1)=I(P2)=Iout,
so Vod_p1=sqrt * Vod_p2 ...(3)
Considering the voltage drop on the resistor, we have Vod_p1 - Vod_p2 = Iout * R ...(4)
Use (2) (3) (4), easy to get Iout= 4/(n*beta*(R^2)*W/L).

Hope it helps.
Rui

Hi all,
I want to ask help here on how derive the current Iout in terms of the resistor R1 and (W/L) . The figure shows a supply independent biasing.

View attachment 86550

imperza

### imperza

Points: 2
Let I1 = I2 = Iout = I

Vgs1 = I*R1 + Vgs2
Vtp + Vdsat1 = I*R1 + Vtp + Vdsat2
Vdsat1 = I*R1 + Vdsat2
Vdsat1 - Vdsat2 = I*R1 (1)
Note: Vdsat = 2*I / gm (2)
From (1) and (2) ----> 2*I/gm1 - 2*I/gm2 = I*R1
Choose gm2 = 2gm1 ---> R1 = 1/gm1
and W2/L2 = 4* W1/L1

imperza

### imperza

Points: 2
Hi again, tq for insight, could anyone here help me how to size the components so that the circuit produces current IOUT ~ 1 µA.

Appreciated.

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