In the wide swing cascode current mirror shown below, the transistor Q4 is in the linear region of operation according to the voltages at gate and drain. Is it not necessary for transistor to operate in saturation region? Is it ok to operate in linear region as I am not finding any issue?
Electronic – Is the transistor Q4 in the wide swing cascode current mirror operating in linear region
analogcircuit analysiscurrent-mirrortransistors
Best Answer
The pre condition for being in the saturation region is the following:
$$V_{GS}-V_{TH} = V_{OV} < V_{DS}$$
By applying only \$V_{OV}\$ to the gate of \$Q_2\$ you would be in the threshold between cut-off and linear regions, therefore an additional \$V_t\$ is applied. Based on that you can say that \$Q_2\$ is at least conducting.
In order to make sure that \$Q_2\$ is in the saturated region, its \$V_{DS}\$ must be greater than its \$V_{OV}\$, and this is accomplished with the \$V_{BIAS}\$. In this case \$V_{BIAS}\$ has to account for the \$V_{OV}\$ of \$Q_4\$ and the minimum required \$V_{DS}>V_{OV}\$ of \$Q_2\$, hence:
$$V_{BIAS} = V_{OV,Q_4} + V_{DS,Q_2}$$
$$V_{BIAS} = V_{OV,Q_4} + V_{OV,Q_2} + V_{t}$$
Assuming that all transistors have the same characteristics:
$$V_{BIAS} = 2\cdot V_{OV} + V_{t}$$
Now it is clear that \$Q_2\$ is in the saturated region. What about \$Q_4\$?
\$V_{BIAS}\$ ensures that \$Q_4\$ is either in the linear region or saturated one by applying an effective \$V_{GS,Q_4}\ = V_{OV}\$. In order to drive \$Q_4\$ into saturation, you just have to apply a voltage slightly larger than its \$V_{OV}\$, hence:
$$V_{D,Q_4} > V_{OV} + V_{t}$$
Based on that you can say that \$Q_4\$ is in saturation.