In my last post we saw how a CMOS inverter is used as transimpedance amplifier. Now we will see how it can be used as transconductance amplifier. We will arrive to a similar circuit but we take a different path this time. Lets look what is a transconductance. It converts the input voltage to output current. The differential transconductance amplifier symbol is usually drawn as shown in figure 1(a). A simple MOS transistor can act as transconductance amplifier as shown in figure 1(b) and (c)
If the input is applied at the gate of the NMOS and the source is grounded, the output current will flow into the device. That means a negative of this current is flowing out of the device.
Lets look as figure 2(a). If the two transistors of the
cmos inverter are operating in the saturation region, then for a small incremental input voltage on the gate will creates incremental currents in the two transistors. The effective transconductance of the inverter is sum of the individual transconductances of the transistors.
To make sure that both transistors are operating in the saturation region, and for maximum swing we have to set the operating/DC voltages of the output and input to the same provided both transistors are balanced. (this can be easily obtained from VTCurve of the inverter).
This can be simply obtained by connecting the input and output of the inverter as shown the figure 3(b). But the problem is that when we apply incremental voltage at the input it is directly connected to output. That means the feedback connection should not be there for incremental signals. To realize this we can modify the feedback connection by adding a very large resistor so that in the incremental view the feedback is seen as disconnected, whereas in biasing it will be there. See figure 4.
The ideal transconductance amplifier should output the current irrespective the output voltage or load. This analog building block is very much used in the design of the active filters where it is combined with elements like OPAMPs, capacitors and resistors. One example of band pass filter is shown in figure 5
$$ \frac{V_{out}(s)}{V_{in}(s)} = \frac{s t_{2}}{1 + s t_{2} + s^{2} t_{1} t_{2}}$$
where $$t_{1} = \frac{C_{2}}{G_{m1}}$$ and $$t_{2} = \frac{C_{1}}{G_{m2}}$$
No comments:
Post a Comment