# (Ideal) Amplifier networks/nullors - determining the gain

I am having trouble understanding how the different amplifier networks actually do "their magic". In these networks a nullor is a unit that has always 0 voltage over the input and no current through it. The output has infinite power to match the requirement set. The "easiest" version -a voltage to voltage amplifier- I can easily understand:

Because the current and voltage over the input of the nullor is 0 the voltage at the output must follow the equation $$\V_s = V_{load} \cdot \frac{R_2}{(R_1 + R_2)}\$$

However I do not see how this works in other types of amplifiers; say the following standard "voltage-to-current" amplifier setup:
What is the gain here? And what is the equality equation to solve?

• This makes no sense without a clear definition of what "nullor" is. Commented Nov 5, 2013 at 15:48
• @Andy aka Nullor is a well know theoretical circuit component. Commented Nov 5, 2013 at 16:04
• @AndrejaKo I recognize the op-amp and transistor implementation but I've never heard of it before!! Commented Nov 5, 2013 at 16:07
• Long live the nullor! Commented Jan 14, 2022 at 14:12

Hmm, never heard of a Nullor before. Go figure. Sometimes it is easier to think of these things in terms of control systems. The amplifier tries to set the two input voltages equal to each other. One input voltage is the current command. The other one is the voltage across R1. So what it tries to do is set the voltage across R1 to the input voltage. The voltage on R1 comes from the current flowing through R1. The amplifier will increase the supply to the load until the voltage across R1 is equal to the input. In this case, Iload = Vin / R1. This circuit is called a transconductace amplifier because the gain is actually Iload / Vin = 1 / R1, which has the units of conductance.

• Sounds like an op amp to me Commented Jan 14, 2022 at 15:56
• @ScottSeidman Not really, the Op-amp doesn't lend itself for a series feedback connection at the output without using external components. The nullor allows that by imposing that the current through both output terminals are the same. They're used as a way to abstract an amplifier implementation. Read Ernst Nordholt Thesis for instance, it has its valuable uses. Commented Mar 31, 2023 at 9:40

The definition of a nullor is that it has zero voltage drop at its input, zero current flowing in or out its input and that any voltage/current could be delivered at its output. Knowing this, we can say that the voltage over R1 equals the source voltage, as zero current flows through the internal source resistor. This allows us to find the current through R1 using Ohm's Law. As a nullor should be examined as a two port (with a chain matrix consisting of only zero's), we know that the current flowing into one terminal of the port must equal the current flowing out of the other terminal on the same port, i.e. the current flowing through R1: Vsource/R1 = -Iload. Notice the minus sign due to the definition stated above.

A Nullor is a combination of a "Nullator" and a "Norator". Both are "pathological networks" (already introduced in the 1960th) which can be used to describe ideal active devices.

• A Nullator is characterized by V=I=0 (example: Ideal opamp diff. input)

• A Norator is described with V=x and I=y (both are arbitrary values, which must satisfy Kirchhoffs laws.)

• Hence, the Nullor - as a combination of both, as shown in the first diagram - represents an ideal opamp (with negative feedback and a load

• The second diagram represents a transconductance amplifier (ideal transistor resp. OTA) with feedback; the feedback voltage is developped across R1.

See here: https://de.wikipedia.org/wiki/Nullor

This concept is taught at TU Delft. I have met some fine designers who use this in their professional life to simplify invention of new circuits. One of them has a website where he wrote a book on design, and uses the nullor as a concept to continue with synthesizing new feedback topologies. You can find it at https://www.analog-electronics.eu/.

In short, you should consider that your input voltage will appear also at the top of R1. Finally, the current out of the nullor is the same as -Vin/R1. However, the output voltage will be load dependent. Therefore, Vout = Vin*R2/R1. This is to show that a V-I converter will always be load-dependent.

A very popular implementation of this circuit on the internet is the common-emitter amplifier with resistive degeneration.

simulate this circuit – Schematic created using CircuitLab

The reason why this is usually shown as a voltage amplifier is because, in this case, we have sort of "fixed" the load, assuming that whatever we drive has a higher input impedance than that of RL in the figure.

For your reference, this feedback configuration is also known as the "series-series" one.