# Transfer function and bode plot of an op-amp circuit

Given is a op-amp circuit which has to be solved using the nullor model to find the transfer function as shown below:

Using the nullor model of an ideal op-amp I tried to find the transfer function and plot frequency and phase response graphs. I change variables and indexes to a more common used once.

The resistor R1 has no effect as there is no current and no voltage drop.

As I don't know if this can be even built I have know idea if this is correct or if this is more a theoretical approach in order to use the nullor model. The original circuit using an op-amp with this forth input at the bottom of the op-amp symbol which only shows that the nullor model has to be used. I couldn't find any thing similar to that symbol and it seem like that only my professor uses it which doesn't really help.

• Two questions: (1) Where are the supply voltage(s)? (2) Opamp with voltage output or OTA with current output (note the current source symbol in the 2nd figure)?
– LvW
Commented Sep 5, 2015 at 14:56
• @LvW: Vs is the supply voltage and can be of any kind but not further clarified and it is a Op-amp with voltage output. The current source symbol is used as the open loop voltage gain is infinite. The nullor model uses the nullator and norator for ideal op-amps. I just drew the op-amp triangle around it so it is clear where I used the nullor.
– Phil
Commented Sep 5, 2015 at 15:45
• The voltage source is just not drawn with its circle symbol.
– Phil
Commented Sep 5, 2015 at 17:22
• Ideal opamps have no supply voltages, and they can source infinite amounts of current, and have infinite voltages. They also have no noise, no offset and and no bias current. Commented Mar 8, 2016 at 18:01
• I know it's been a lifetime, but your answer is correct. The nullor is simply a more generalized abstraction than the op-amp. Commented Mar 31, 2023 at 12:52

The 4th pin could be one of two things:
A reference pin, similar to those found in instrumentation amps to offset the voltage.

The negative rail of the op amp (which I don't think it is, but could be)

Either solve it for both assumptions, or ask the designer of the question what they intended.

Also, your assumption of 0V for the voltage across the indicator doesn't look right. I would think that the opamp is going to drive some kind of current through the 4th pin.

• the drawing is misleading, but the bottom pin can very well the V- pin. Here's an implementation Maxim used for a current source using the V- pin: radiolocman.com/shem/schematics.html?di=462791 It would have been better that an actual nullor were drawn, as 99% of the textbooks ignore the supply pins when explaining op-amps. Commented Mar 31, 2023 at 12:55

A nullor has 4 pins as it also allows to reflect a series feedback connection at the output. The right-most terminals simply have the constraint that the currents flowing through them are equal. With op-amp symbols, people need to add an extra NMOS stage at the output to provide the series sensing path which is already included in the nullor abstraction. IMHO, it's more generalized abstraction of an amplifier than the op-amp.

You could also sense current via the supply pins of the op-amps, but that's a more advanced application of the op-amp and 99% of the typical electronics book do not explain that, they simply ignore the supply pins.

The mistake in your development was that you wrote 0V at the top of the inductor (but it seems that your result is correct). It should've been Vs, as shown below:

This is a simple consequence of the nullator at the input, which imposes the constraint that the voltage difference at the input terminals must be 0 (just like the op-amp).

In short, the transconductance or transadmittance transfer of this configuration is:

$$\frac{I_o}{V_s} = \frac{1}{-j\omega L}$$

And the voltage transfer function is simply the transfer above multiplied by load resistor:

$$\frac{V_o}{V_s} = \frac{I_o}{V_s}R_L = \frac{R_L}{-j\omega L}$$

I ignore whether a configuration like this has a practical application. It looks like a voltage and frequency-dependent current source with an RL load.

If you exchange the L with a resistor, and you replace the nullor with a transistor, you have the typical textbook common-emitter/source amplifier implementation.