Operational Transconductance Amplifier: OTA Basics - 1

Operational Transconductance Amplifier - OTA is a very important building block in analog design.

As we see from the previous posts the transconductance amplifier converts an input voltage to output current. The word "Operational" refers to the fact that it amplifies the difference voltage. The transconductance $g_{m} \propto I_{Bias} $

Using the OTA as a central element we can build application circuits like filters, voltage amplifiers, integrators etc. Here we will see a few filter/integrator circuits using OTA. 

We will find the transfer functions of these circuits intuitively without doing full analysis. The basic idea is 

1. The input voltage $V_{in}$ at input node causes an output current $i_{o}$. 
2. Output current $i_{o}$ flows through the resistance $r_{oeq}$seen at output node causing output voltage $V_{o} = i_{o} r_{oeq} $       

Lets see the recipe for finding the transfer function

1. Short the output to the ground and find the conductance seen at input node. Say $G_{m}(s) ^{\dagger}$.
2. Short the input to the ground and find the conductance seen at the output node. Say $G_{o}(s)$
3. Transfer function is $$\frac{V_{o}(s)}{V_{in}(s)} = \frac{G_{m}(s)}{G_{o}(s)}$$

$^{\dagger}$ When the input is connected to inverting terminal the $g_{m}$ is replaced by $-g_{m}$ 

Figure 1 shows the integrators. 

For 1(a). The transconductance seen from input to output is $g_{m}$ and the conductance seen into the output node is $sC$. 

$$\therefore \begin{equation} \frac{V_{o}(s)}{V_{in}(s)} = \frac{g_{m}}{sC} \end{equation}$$ 

For 1(b). The transconductance seen from input to output is $g_{m} + sC_{2}$ and the conductance seen into the output node is $s(C_{1} + C_{2})$. 

$$\therefore \frac{V_{o}(s)}{V_{in}(s)} = \frac{g_{m} + sC_{2}}{s(C_{1} + C_{2})} $$ 

Whenever there are more than one paths for the signal we can expect zeros in the transfer function. In 1(b) there are two paths from the input to output via transconductance and capacitor $C_{2}$  

In both circuits in figure 1 have poles at origin. i.e., the 20dB rolloff starts soon after the 0Hz with $\infty$ DC gain. In 1(b) though there are two capacitors in the circuit they are in parallel w.r.t the output node and hence we have a first order transfer function rather than second order. 

Now we will see how to shift the pole away from the origin to increase the useful bandwidth. Refer to the figure 2. A resistor R is added to the output node in parallel with the capacitance. Since R is connected to output node and GND there is no change in the positions of the zeros in the transfer function. Only the position of the poles will change as the conductance seen at the output changed.  

Figure 2 shows the lossy - integrators.

For 2(a). The transconductance seen from input to output is $g_{m}$ and the conductance seen into the output node is $sC + \frac{1}{R}$. 

$$\therefore \begin{equation} \frac{V_{o}(s)}{V_{in}(s)} = \frac{g_{m}}{sC + \frac{1}{R}} \end{equation}$$ 

For 2(b). The transconductance seen from input to output is $g_{m} + sC_{2}$ and the conductance seen into the output node is $s(C_{1} + C_{2}) + \frac{1}{R}$. 

$$\therefore \frac{V_{o}(s)}{V_{in}(s)} = \frac{g_{m} + sC_{2}}{s(C_{1} + C_{2}) + \frac{1}{R}} $$ 

So far we have seen the first order transfer functions, now refer to the figure 3. The two capacitors are independent and see different resistances, giving the hint there will be two poles in the transfer function. 

The transconductance seen from input to output is $g_{m} + \frac{sC_{2}}{sC_{2}R_{2} + 1}$ and the conductance seen into the output node is $sC_{1} + \frac{1}{R_{1}} + \frac{sC_{2}}{sC_{2}R_{2} + 1}$. 

$$\therefore \frac{V_{o}(s)}{V_{in}(s)} = \frac{g_{m} + \frac{sC_{2}}{sC_{2}R_{2} + 1}}{sC_{1}+ \frac{1}{R_{1}} + \frac{sC_{2}}{sC_{2}R_{2} + 1}} $$ 

Lets verify the sanity of the above transfer function. If we put $R_{2} = 0 \Omega $ the transfer function reduces to the first order as of the circuit 2(b). 

Lets find the zeros and poles of the transfer function. Lets rewrite the transfer function 

$$\frac{V_{o}(s)}{V_{in}(s)} = \frac{g_{m} + s[C_{2}(g_{m}R_{2} + 1)]}{C_{1}C_{2}R_{2}s^{2} + s(C_{2} + C_{1}) + \frac{1}{R_{1}}} $$ 

We will apply quadratic root approximation to find the poles. When the roots of the quadratic equation $ax^{2} + bx + c = 0$ are far from each other say $|r_{1}| > 10| r_{2}| $ then $r_{1} = \frac{-b}{a}$ and $r_{2} = \frac{-c}{b}$. To make sure the above conditions valid assume $R_{1} >> R_{2} $ so that the non-dominant pole is very far away from the dominant pole. 

By using the above approximations 

• DC gain = $g_{m} R_{1}$
• zero $z_{1} = \frac{-1}{C_{2} (\frac{1}{g_{m}} || R_{2})}$
• dominant pole $p_{0} = \frac{-1}{(C_{1} + C_{2})R_{1}}$
• non-dominant pole $p_{1} = \frac{-1}{(\frac{C_{1}C_{2}}{C_{1}+C_{2}})R_{2}}$ 

• 0dB frequency $\omega_{u} = \frac{g_{m}}{C_{1} + C_{2}}$

For the system to be stable the relative positions of the poles and zeros are very important. In a second order system like above they should be $$ |p_{0}| \ll |p_{1}| , |z_{1}|$$