Charles daEngineer

Feedback Configuration Derivations

Published on Sun Oct 09 2022

Introduction

Using Nullors for feedback network design is so easy! To understand it, all we have to do is analyze the current divider circuit and the voltage divider circuit. Then, using nullators and norators I can invert the circuit to achieve gain from the source to the load.

Attenuation

Voltage Divider circuit where two series resistors are in series with a Voltage Source

The above Voltage circuit has the governing equations:

$\begin{array}{r} (1) & i_{o u t} = \frac{v_{i n}}{R_{1} + R_{2}} \end{array}$

and:

$\begin{array}{r} (2) & v_{o u t} = \frac{R_{2} v_{i n}}{R_{1} + R_{2}} \end{array}$

Notice that the equations always divide the input voltage, so this circuit is called a voltage divider.

Current Divider Circuit where two parallel resistors are in parallel with a Current Source

The above Current Circuit has the governing equations:

$\begin{array}{r} (3) & v_{o u t} = \frac{R_{1} R_{2} i_{i n}}{R_{1} + R_{2}} \end{array}$

and:

$\begin{array}{r} (4) & i_{o u t} = \frac{R_{1} i_{i n}}{R_{1} + R_{2}} \end{array}$

Notice that the equations always divide the input current, so this circuit is called a current divider.

These two circuits attenuate, so the output is always a fraction of the input. So to achieve a gain I would like my measured input to be at the output of these circuits and I would like the output to be at the input of these circuits. Effectively I would like to invert these circuits.

Amplification based on Inversion

Voltage to Voltage Amplifier

Starting with the voltage divider I would like to treat the source in that circuit as the output; so I attach a free variable in parallel with it and replace the source with a resistor. As a first case I would like my input to be a voltage, that is; I would like to guarantee that the voltage across $R_{2}$ to be equal to my input voltage.

Voltage amplifier circuit using a nullor

The image above shows a nullator (a zero symbol) between the source voltage and $R_{2}$ , which guarantees that the voltage across the nullator is zero and guarantees the current through is also zero. By doing this I am forcing the voltage across $R_{2}$ to be $V_{s}$ .

The image also shows a norator (an infinity symbol) in parallel with the load impedance. While the current is determined by the load impedance; the voltage across the norator is determined by equation $2$ : $v_{s} = \frac{R_{2} v_{o}}{R_{1} + R_{2}}$ Solving for $V_{o}$ we get the following equation: $\begin{array}{r} (5) & v_{o} = (\frac{R_{1}}{R_{2}} + 1) v_{s} \end{array}$

Current to Voltage Amplifier

To force the current in $R_{2}$ to be $i_{s}$ we can place a nullator in series with $R_{2}$ and place $i_{s}$ in parallel with the nullator. By doing this I am also forcing the voltage across $i_{s}$ to be zero.

Trans-impedance circuit using a nullor

Now I can directly use equation $1$ by setting $i_{s} = i_{o u t}$ and $v_{o} = v_{i n}$ :

$\begin{array}{r} (6) & i_{s} = \frac{v_{o}}{R_{1} + R_{2}} \end{array}$

Solving now for $v_{o}$ we get:

$\begin{array}{r} (7) & v_{o} = i_{s} (R_{1} + R_{2}) \end{array}$

Current to Current Amplifier

For measuring current at the output I take the inverse of the equations provided by the current divider circuit. Now we are measuring current at the output so the norator must be in series with the load. Similarly for measuring current at the input a nullator must be in series with $R_{2}$ and place the source current in parallel with the nullator.

Current amplifier circuit using a nullor

Notice that I am forcing $i_{R_{2}} = i_{s}$ so we can directly use equation $4$ :

$\begin{array}{r} (8) & i_{s} = \frac{R_{1} i_{o}}{R_{1} + R_{2}} \end{array}$

Solving for $i_{o}$ we get the following equation:

$\begin{array}{r} (9) & i_{o} = (\frac{R_{2}}{R_{1}} + 1) i_{s} \end{array}$

Voltage to Current Amplifier

The output doesn't change from the Current Amplifier. But, at the input I am using a voltage so I would like to put a zero between the source voltage and the resistor $R_{2}$ .

Trans-admittance circuit using a nullor

With the nullator I am forcing $v_{R_{2}} = v_{s}$ so I can use equation $3$ :

$\begin{array}{r} (10) & v_{s} = \frac{R_{1} R_{2} i_{o}}{R_{1} + R_{2}} \end{array}$

Solving for $i_{o}$ we get our solution:

$\begin{array}{r} (11) & i_{o} = (\frac{1}{R_{1}} + \frac{1}{R_{2}}) v_{s} \end{array}$

Conclusion

Using a current divider, and a voltage divider we can synthesize all possible gain transfers using nullators and norators to invert the circuit.