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Explain small signal analysis for MOSFET active load circuit.
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fig (a) : MOSFET differential with active load id = $\frac{gmvd}{2}$

fig (b) : small signal equivalent ckt.

fig ( c )

the circuit mainly consists of four transistors where $m_1$ $m_2$ are n – channel, devices and form the differential pair between IQ. The load circuits consists of transistor. $m_3$ and $m_4$, both p – channel devices, connected in current mirror configuration, a one side output is taken from the common drain of $m_2$ and $m_4$ when a common mode voltage of $v_1$ = $v_2$ = Vcm is applied, the current IQ splits evenly between $m_1$ and $m_2$ and I $d_1$ = $d_2$ = IQ/2. There are no gate currents.

$\therefore$ i$d_3$ = i$d_1$ and i$d_4$ = i$d_2$

If a small differential mode input voltage vd = $v_1$ - $v_2$ is applied

I$D_1$ = $\frac{IQ}{2}$ + id

I$d_2$ = $\frac{IQ}{2}$ - id where id = signal current

For small values of vd,

Id = $\frac{gmvd}{2}$

Since $m_1$ and $m_3$ are in series, we see that i$d_3$ = i$d_1$ = $\frac{IQ}{2}$ + id

Finally the current mirror consisting of M3 and M4 produces,

I $D_4$ = I $D_3$ = $\frac{IQ}{2}$ + id

Fig. (b) shows the small equivalent circuit at the drain mode of M2 and M4. If the output is connected to the gate of another MOSFET, which is equivalent to an infinite impedance at low frequency, the o/p terminal is effectively an open circuit. The circuit can be rearranged by

Vo = 2 $(\frac{gmvd}{2})$ ($ro_2$ ll $ro_4$)

The signal grounds at a common point.

Ad = $\frac{vo}{vd}$ = $\frac{gm (ro2 ll ro4 vd)}{vd}$

Ad = gm (ro2 ll ro4)

$\therefore$ Ad = gm $\frac{ro2.ro4}{ro2+ro4}$ = gm/$\frac{1}{ro2} + \frac{1}{ro4}$

= $\frac{gm}{go2+g04}$

Gm = $\sqrt[2]{KnID}$

= $\sqrt[2]{kniq}$

$go_2$ $\lambda_2$ IDQ2

= $\lambda_2$ $\frac{IQ}{2}$

$go_4$ = $\lambda_4$ IDQ4

= $\lambda_4$ $\frac{IQ}{2}$

then,

AD = $\sqrt[2]\frac{2 kniq}{IQ (\lambda_2 + \lambda_4)}$

= $\sqrt[2]\frac{2kn}{IQ}$ $\frac{1}{\lambda_2 + \lambda_4}$