How do I do calculation in CircuitLab? How do I provide +Vcc and -Vcc to the OpAmp?

Question

Here is my simple circuit. I would like to test whether the output is correct as per manual calculation. According to manual calculation, Gain=3.1, Vo=10.54V.

My question are:

• How do I do that calculation?
• Is my way to do +Vcc and -Vc connection correct? Which both should be connected to +12V and -12V respectively.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

According to manual calculation, Gain = 3.1, Vo = 10.54 V.

Nope. You're feeding 4 V in so a gain of 3.1 would give an output of 12.4 V. Something wrong somewhere.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Redrawn schematic to show the circuit blocks.

How do I do that calculation?

Break down the circuit into its functional blocks.

1. The input signal is 4 V and fixed. That's easy.
2. Next you have a potential divider. (This is the part missing in your overall gain calculation.) The output of that is given by the ratiometric division $$ V_O = \frac {R1}{R1 + R2} Vg $$.
3. This is followed by, what should now be obvious, a non-inverting amplifier stage. The output of this is given by $$ V_O = ( 1 + \frac {R_f}{R_s} ) V_I $$
4. Next is the load. This doesn't affect the output provided we don't overload it. 27 kΩ will be fine.
5. The power supply becomes important because an op-amp output can't go above the positive supply voltage and can't go below the negative supply voltage. In most cases it can't get within a couple of volts of either supply rail so we'll come back to this later.

The output of your circuit will be the product of all the stages that affect the signal; the divider and the non-inverting amplifier. $$ V_O = V_g \frac {R1}{R1 + R2}( 1 + \frac {R_f}{R_s} ) $$

Is my way to do +Vcc and -Vc connection correct? Which both should be connected to +12V and -12V respectively.

Now we come back to the power supply.

• Your input signal of 4 V is driving the output to 10.5 V which is only 1.5 V from positive supply. You need to check the op-amp datasheet to see if it is capable of getting that close to positive supply. Some are. Some aren't.
• The power supply limits the maximum input signal. If V_g > 4 V then you need to raise the 12 V to a higher voltage.
• As V_g is decreased towards 0 V the op-amp may reach the lower output limit. If this is a problem you need to choose an op-amp that can output to negative rail or use a negative supply rail.

^{simulate this circuit}

Figure 2. Split rail powered version.

Figure 3. Internals of the ancient 741 opamp. Source: Wikipedia.

Most opamps will have an output arrangement similar to the push-pull arrangement of the old 741. Others will have FET transistors rather than BJTs. In either case if the top transistor (red oval) is turned on the output will be pulled to positive rail. If the bottom transistor (green oval) is turned on the output will be pulled to negative rail. How close they get depends on the exact output configuration and the driving circuitry.

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Answering the questions in the title:

How do I do calculation in CircuitLab?

For this example you just run the DC solver. This is simplest if you attach a NODE to your input, V_g, and your output, R_o. Simulate | DC Solver | Click each of the nodes to add them in | Run.

How do I provide +Vcc and -Vcc to the op-amp?

See my Figure 2.

Answer 2

Well, we have the following circuit (and we assume an ideal model of an OPAMP):

^{simulate this circuit – Schematic created using CircuitLab}

Using KCL, we can write:

$$ \begin{cases} \text{I}_4=\text{I}_-+\text{I}_2\\ \\ \text{I}_x=\text{I}_++\text{I}_3\\ \\ \text{I}_\text{o}=\text{I}_4+\text{I}_5 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_x=\frac{\text{V}_x-\text{V}_+}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_--0}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_+-0}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{o}-\text{V}_-}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_\text{o}-0}{\text{R}_5}\\ \\ \end{cases}\tag2 $$

Notice: in the ideal OPAMP circuit we assume that \$\text{I}_+=\text{I}_-=0\$ and \$\text{V}_+=\text{V}_-\$.

Now, the gain is defined by:

$$\text{G}:=\frac{\text{V}_\text{o}}{\text{V}_x}\tag3$$

We can find an expression for the output voltage \$\text{V}_\text{o}\$, by solving the systems of equations:

$$\text{V}_\text{o}=\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag4$$

So, we get:

$$\text{G}=\frac{1}{\text{V}_x}\cdot\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}=\frac{\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag5$$

In your case we get:

$$\text{G}=\frac{68000\cdot\left(30000+63000\right)}{30000\cdot\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

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Solving it, in general, gives (notice that \$\text{V}_+=\text{V}_-=\text{V}_\text{p}\$:

In your case (using your values):

I checked my solution using LTspice and I got it right.