Op-amp as a comparator (analog-to-digital conversion)

[ichabodgrant]

Homework Statement

Okay, This time it's an exam question. I am doing some past papers and there are questions about comparators which are not introduced in this course (this course is preliminary course for engineering students who are interested in electrical, electronic or computer engineering). So I guess the instructor wants us to know the logic of a comparator just within the 3 hours examination (in which not all 3 hours are for that particular question...)
Now I find it difficult... maybe I am not familiar with the concepts yet... But anyway I have tried to solve it...Here are the details and my attempts:

First, we are given an op-amp circuit for building a radio (the situation is that you are in a lost island which is rich of op-amps and resistors but no logic gates... so are you going to create logic gates AND and OR using just the above materials):

To find V_x, I use the node method and finally get the value of V_x in terms of V_A and V_B:

V_x = - (R_f / R_a) (V_A + V_B)

Then, assuming R_a = 2R_f and V_cc = 5 V, we are now relating the unknowns voltage with real values:

What I get are 0, -2.5, -2.5 and 5 from top to bottom.

Third, now we are go to transform voltages to logic values:
Let V_y = - V_x:

What I write are R₁ = R_b, R₂ = 3R_b, R₃ = 2R_b, R₄ = 2R_b, IN₁ connects to V_cc, IN₂ connects to GND.

Then, it comes to the comparator question:

What I am sure about is that V_s+ = +V_cc, V_s- = GND...
But how to know which V- and V+ should be respectively?

AND gate in logic means y = ab. How to see this from the op-amp above?

I know that 5 V is "true" which represents logic value "1"; 0 V is "false" which represents logic value "0".
So this thing should either output 5 V (V_cc) or 0V (GND)...

So is V₊ = V_y and V_- = V_th?

Now, other than AND gate, we also can create an OR gate using the above materials...

Since I totally don't understand the logic of 4d, I also don't understand this. But I guess it mentions V_th this way, which shows that V_th is used as a reference for comparing? But why is just changing V_th can be resulted in two different logic gates functions?

Here comes the final part: to design a 3-input AND gate using no more than 4 op-amps...
Once again I don't understand the logic, so I don't know what to do.

Homework Equations

For Q4d, V_out = K (V₊ - V_-) where K is a huge gain about 10⁵.

The Attempt at a Solution

The attempts are written above.

 

[KataKoniK]

Hello, it seems like you are struggling with understanding the logic behind comparators and how they can be used to create logic gates. I will try my best to explain it to you.

Firstly, a comparator is a type of op-amp circuit that compares two input voltages and outputs a digital signal based on the comparison. In your case, the inputs are VA and VB, and the output is Vx. The output Vx will be either a high voltage (5V) or a low voltage (0V) depending on the comparison of VA and VB.

Now, let's look at the circuit you provided. The first thing to note is that the op-amp is configured as a non-inverting amplifier, meaning that the output voltage will be the same as the input voltage (with some amplification). So, the output Vx will be equal to the input voltage at the positive terminal of the op-amp, which is (VA + VB). The negative terminal of the op-amp is connected to ground, so Vx will be equal to - (Rf / Ra) (VA + VB).

Next, you mentioned that you are given the values of Ra = 2Rf and Vcc = 5V. Using these values, we can simplify the equation for Vx to Vx = -2.5 (VA + VB).

Now, to understand how this circuit can be used as a comparator, we need to look at the output Vx in terms of logic values. As you correctly stated, 5V represents a logic value of 1 and 0V represents a logic value of 0. So, if Vx is greater than 0, the output will be 5V, indicating a logic value of 1. If Vx is less than 0, the output will be 0V, indicating a logic value of 0. This is the basic principle of a comparator.

Moving on to the AND gate, we can see that it is essentially a combination of two comparators. In this case, V+ and V- will be the outputs of the two comparators, and the output of the AND gate will be the output of the second comparator. So, for an AND gate, the output will be 5V only when both inputs (VA and VB) are 5V. Otherwise, the output will be 0V.

Similarly, for an OR gate, the output will be