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    • What, When, Where, and Why?
    • Op Amps
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  • What and why?
  • Copper Rules of Op-Amps
  • It is important to note that the following rules only apply for ideal op-amps. That is, op-amps that meet the following requiem:
  • Golden Rules of Op-Amps
  • Practical Uses

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  1. Circuits

Op Amps

PreviousWhat, When, Where, and Why?

Last updated 5 years ago

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What and why?

Operational amplifiers (op-amps) are circuit components that help us measure and use voltage differences without potential resistance interacting with the rest of the circuit:

Copper Rules of Op-Amps

It is important to note that the following rules only apply for ideal op-amps. That is, op-amps that meet the following requiem:

  1. Infinite input impedance

  2. Infinite open-loop gain

  3. Zero output impedance

  4. Infinite bandwidth

  5. Zero noise

In the scope of this class, you will only deal with ideal op-amps. Doing so allows us to make use of the golden rules, which makes analyzing circuits much easier.

Golden Rules of Op-Amps

The golden rules are as follows:

  1. This holds regardless of whether or not the op-amp is wired in negative feedback.

  2. This rule only holds when the op-amp is wired in negative feedback.

Practical Uses

When we are taking in an analog voltage reading, or a range of voltage values, but we want to convert to a binary output, there is the comparator:

Op-amps are extremely useful in signal processing circuits, due to their ability to eliminate loading issues in our circuits. When being used for pure isolation, they are called buffers, depicted as below:

When we need a gain that is strictly greater than 1, we have the noninverting op amp:

Here from the golden rules, we know that:

This is great when we know that our input just needs to be stronger, a great example would be on a speaker. Another important note, in the name of thsi op amp, the gain of this configuration does not invert our output. This could be useful when working with a limited number of op amps (like in lab), or under other constraints.

When we want either a fractional gain, or a negative gain, we have the inverting op amp:

This is useful when we may want to divide our input, or simply need to negate it.

A common design goal might be to add two voltages together, and to do this we will need a passive summer:

Here we can use superposition to find:

And we're almost there, but this circuit is only averaging out our values, next we need to multiply the output by a factor of 2 (this factor comes from the number of resistors or voltage sources we are trying to sum):

Going back to our equation for a noninverting op amp, and using what we found earlier for the passive summer:

This is called a summer, or a summing circuit.

A more detailed description of op-amps on a functional level can be found .

I+=I−=0I_+ = I_- = 0I+​=I−​=0.

V+=V−V_+ = V_-V+​=V−​ .

Vout={VDD: if Vin>VrefVSS: if Vin<VrefV{out} = \left\{ \begin{array}{cc} V_{DD} & \text{: if } V_{in} > V_{ref} \\ V_{SS} & \text{: if } V_{in} < V_{ref} \end{array} \right.Vout={VDD​VSS​​: if Vin​>Vref​: if Vin​<Vref​​

You may notice that the behavior is undefined when Vin=VrefV_{in} = V_{ref}Vin​=Vref​, that is because we assume that our op amp has infinite gain. In most cases, you will not have to consider this, and in real life, noise will never keep your inputs at this critical point for long. But when an input is steadily approaching that point, consider what would happen the exact instant if it were to surpass the critical point? Remember back to both your homework and discussion for the timing circuit: as the positive feedback "comparator" approached a Vin−VrefV_{in} - V_{ref}Vin​−Vref​ of 0, our output (VoutV_{out}Vout​) would switch.

Where Vin=VoutV_{in} = V_{out}Vin​=Vout​.

Vin=VoutR2R1+R2  ⟹  Vout=Vin(1+R1R2)V_{in} = V_{out}\frac{R_2}{R_1+R_2} \quad \implies \quad V_{out} = V_{in}\left(1+\frac{R_1}{R_2}\right)Vin​=Vout​R1​+R2​R2​​⟹Vout​=Vin​(1+R2​R1​​)

The golden rules know no foul and we will use nodal analysis (current is all flowing in) at the node V−V^-V−:

Vin−0R1+Vout−0R2=0  ⟹  Vout=−VinR2R1\frac{V_{in} - 0}{R_1} + \frac{V_{out} - 0}{R_2} = 0 \quad \implies \quad V_{out} = -V_{in}\frac{R_2}{R_1}R1​Vin​−0​+R2​Vout​−0​=0⟹Vout​=−Vin​R1​R2​​

Vout=Vin1R1R1+R2+Vin2R2R1+R2V_{out} = V_{in1} \frac{R_1}{R_1+R_2} + V_{in2}\frac{R_2}{R_1+R_2}Vout​=Vin1​R1​+R2​R1​​+Vin2​R1​+R2​R2​​

Now, if we set R1=R2=RR_1 = R_2 = RR1​=R2​=R, we get

Vout=Vin1+Vin22V_{out} = \frac{V_{in1} + V_{in2}}{2}Vout​=2Vin1​+Vin2​​

V+=Vin1+Vin2R1+R2Vout=2V+  ⟹  Vout=Vin1+Vin2V^+ = \frac{V_{in1}+V_{in2}}{R_1+R_2} \quad V_{out} = 2V^+ \quad \implies \quad V_{out} = V_{in1} + V_{in2}V+=R1​+R2​Vin1​+Vin2​​Vout​=2V+⟹Vout​=Vin1​+Vin2​

here
Op-Amp Diagram
Buffer Diagram