Introducing a realistic opamp model, UNDER CONSTRUCTION

This opamp model is a differential amplifier with realistic properties. It measures the difference in voltage between its two input ports. It amplifies that difference by the gain factor $\text A$ and puts the result on the output port.

$\sf{vo} = \text A (\sf v$+ $- \,\,\sf v$-$)$

This realistic opamp model has two power supply pins, $\small{\sf{Vs+}}$ and $\small{\sf{Vs-}}$. The power supplies limit how far the output voltage, $\sf{vo}$, can swing. $\sf{vo}$ always has to be between the two supply voltages. For more details see the Circuit Sandbox opamp model.

You can modify various properties by double-clicking on the opamp symbol in the Circuit Sandbox. This model allows you to adjust Gain, Rin (resistance between $\sf v$+ and $\sf v$-), Rout (resistance in series with the output).

## Prepare a simple circuit

Open this incomplete realistic opamp schematic in an new browser tab.

Double-click on the opamp. Give it a name like “Op1”.

The gain of this opamp has been set low so you can see how a differential amplifier works. What is the gain?

The initial gain setting is $\text A = 10$.

Opamps have gains in the many $1000$’s, but we start low to get a feel for how it works.

Add another opamp from the parts bin. Open up its properties with a double-click.
What is the gain of an opamp fresh from the parts bin?
What is the input resistance?
What is the output resistance?

A default opamp in Circuit Sandbox has a gain of $\text A = 100{,}000$.

The input resistance is $1\,\text M\Omega$. (The resistance between $\sf v$+ and $\sf v$-.)

The output resistance is $1\,\Omega$. (Resistance in series with the output pin.)

You can leave these values alone most of the time.

We don’t need this opamp any more. Please delete it.

Build a circuit,

• Power: This opamp model requires two power inputs. Representative power supplies are provided. Connect power to the opamp,
• Double click each voltage source and set the voltage of both to $12\,\text V$. A typical opamp supply is between $5$ an $15$ volts.
• Move the two ???— labels so they connect to the opamp. Double click to change their names to $\sf Vs+$ and $\sf Vs-$. This makes the connection to the power supply.
• Alternatively, delete the node labels and wire the supplies directly to the opamp.
• Add a ground symbol to the node between the two sources. (Select the ground symbol and rotate it by tapping R on the keyboard.)
• Inputs: Connect the two input voltage sources,
• $\sf v$+ to the opamp’s + input
• $\sf v$- to the opamp’s - input
• Add a ground symbol to the bottom of each input source.
• Move the voltage probe so it touches $\sf{vo}$.
finished schematic

Do the rest of this exercise with your own schematic.

If you want you can compare it to the one I drew — Willy’ schematic.

## Performance questions

With the inputs as given, figure out in your head what $\sf vo$ will be.

Hint: Fill in the opamp equation, $\sf{vo} = \text A \,[(\sf{v+}) - (\sf{v-})]$.

Perform a DC operating point analysis (click on DC).
What is $\sf vo$? Did you guess right?

$\sf{vo} = \text A \,[(\sf{v+}) - (\sf{v-})]$

$\sf vo = 10 \,(1.1 - 1)$

$\sf vo = 10 \,(0.1)$

$\sf vo = 1\,\text V$

It actually comes out $0.98\,\text V$. I don’t know why, yet. The gain appears to be 9.8 instead of 10.

The output voltage is $10$ times the difference between $\sf v$+ and $\sf v$-.

Change $\sf v$+ to $3\,\text V$. What is going to happen to $\sf vo$?

The expected answer is $(3 - 1)\times 10 = 20\,\text V$. But the result is actually close to $12\,\text V$. Why is that?

That’s because the output cannot exceed the power supply. The opamp is saturated.

Change $\sf v$+ to $-3\,\text V$. What will happen to the output?

### Voltage transfer plot

Change $\sf v$- to $0\,\text V$.
Change $\sf v$+ to a triangle wave with these parameters,

• Initial value: $-4$
• Plateau value: $+4$
• Frequency (Hz): $0.5$

Add a second voltage probe to $\sf v$+. Change it’s color to something different than the other probe.

Click on TRAN and run a transient simulation for 1 second. This is an approximation to a voltage transfer plot. Sweeping the input voltage from $-20$ to $+20$ shows you the output voltage starts in negative saturation at the negative power rail, then becomes linear with a slope of GAIN, and finally saturates at the positive rail.

Special probe setting: Select the $\sf v$+ probe you added. Change its color to “x-axis”. Perform the same TRAN again.

Notice the input probe now appears as the x-axis of the plot. This is a proper voltage transfer plot, $v_{out}$ vs. $v_{in}$.

Voltage transfer circuit model

### More to come with realistic opamp model

Under construction.