I wanna check my answer with you guys .. and also need help .
View attachment New.pdf

 

6a) Your graph is close to correct, but the voltage where it switches from low to high is incorrect.
6b i) correct!
6b ii) Think about an ideal op-amp how much current is flowing into the input pin? Consider the input pin of an ideal op-amp has infinite impedance.
6b iii) Close, but incorrect, remember the voltage on the inverting input is the same as the voltage on the non-inverting input. Also remember to divide by 20k not 20.
6b iv) If you figure out the 6b ii) and 6b iii) then using KCL the answer should come to you.
6b v) Close again, but remember the voltage on the inverting input is the same as the voltage on the non-inverting input.
6b vi) Repeat all of the 6b questions with the new parameters and you will have your answer.

 

6a) Your graph is close to correct, but the voltage where it switches from low to high is incorrect.
6b i) correct!
6b ii) Think about an ideal op-amp how much current is flowing into the input pin? Consider the input pin of an ideal op-amp has infinite impedance.
6b iii) Close, but incorrect, remember the voltage on the inverting input is the same as the voltage on the non-inverting input. Also remember to divide by 20k not 20.
6b iv) If you figure out the 6b ii) and 6b iii) then using KCL the answer should come to you.
6b v) Close again, but remember the voltage on the inverting input is the same as the voltage on the non-inverting input.
6b vi) Repeat all of the 6b questions with the new parameters and you will have your answer.

Ok ..

I dont get what are you trying to say on 6b i, ii, iii..

Can you please explain more !!!

Because this is my weakest area in electronics.

I dont get anything at all in Op Amp

 

6a) Your graph is close to correct, but the voltage where it switches from low to high is incorrect.
6b i) correct!
6b ii) Think about an ideal op-amp how much current is flowing into the input pin? Consider the input pin of an ideal op-amp has infinite impedance.
6b iii) Close, but incorrect, remember the voltage on the inverting input is the same as the voltage on the non-inverting input. Also remember to divide by 20k not 20.
6b iv) If you figure out the 6b ii) and 6b iii) then using KCL the answer should come to you.
6b v) Close again, but remember the voltage on the inverting input is the same as the voltage on the non-inverting input.
6b vi) Repeat all of the 6b questions with the new parameters and you will have your answer.

Okay for 6 ii)

It will be 0V because in op Amp no current in input right?

for 6 iii)
No current at input which means V+ = V- so V- will be 2V
Voltage drop across R1 will be Vin - V+ so it will be 6V.

so 6/30Kohm will give me 0.002A Which will flow into R1

for 6 iv) it will be same as R1 current because no current is going into input right?

for 6 V) i did different way but

i got 6V right??

 

Ok ..

I dont get what are you trying to say on 6b i, ii, iii..

Can you please explain more !!!

Because this is my weakest area in electronics.

I dont get anything at all in Op Amp

Then this is what you need to be focusing on -- strengthing your understanding of opamp and why we can analyze opamp-based circuits the way we typically do.

An opamp is a device in which Vout = A(Vp - Vn) where Vp is the voltage at the non-inverting input and Vn is the voltage at the inverting input. The coefficient A is the voltage gain is an typically 100,000 or more (sometimes 10 million or more). So let's call it A=1million for sake of discussion.

An opamp (generally not even an ideal opamp) can't put out a voltage that lies outside of its supplies. So Voutmax = +15V and Voutmin = -15V (in the case of the circuits in your problems).

Q1) What is the difference between Vp and Vn just as the output reaches +15V?

Q2) What is the difference between Vp and Vn just as the output reaches -15V?

Q3) If the voltage at the inverting input is 3V, what is the voltage at Vp just as the output reaches -15V?

Q4) If the voltage at the inverting input is 3V, what is the voltage at Vp just as the output reaches +15V?

Q5) What does the graph in Problem 6a look like, taking all of the above into account?

 

Then this is what you need to be focusing on -- strengthing your understanding of opamp and why we can analyze opamp-based circuits the way we typically do.

An opamp is a device in which Vout = A(Vp - Vn) where Vp is the voltage at the non-inverting input and Vn is the voltage at the inverting input. The coefficient A is the voltage gain is an typically 100,000 or more (sometimes 10 million or more). So let's call it A=1million for sake of discussion.

An opamp (generally not even an ideal opamp) can't put out a voltage that lies outside of its supplies. So Voutmax = +15V and Voutmin = -15V (in the case of the circuits in your problems).

Q1) What is the difference between Vp and Vn just as the output reaches +15V?

Q2) What is the difference between Vp and Vn just as the output reaches -15V?

Q3) If the voltage at the inverting input is 3V, what is the voltage at Vp just as the output reaches -15V?

Q4) If the voltage at the inverting input is 3V, what is the voltage at Vp just as the output reaches +15V?

Q5) What does the graph in Problem 6a look like, taking all of the above into account?

Ans. 1 --> 0 V
Ans.2 --> 0 V

AM I RIGHT?

 

If we use infinite gain, but I specifically stipulated a gain of one million.

What does the difference have to be such that, with a gain of one million, the output will be +15V.

Don't just guess at an answer and ask if your guess is right -- show some work.

 

If we use infinite gain, but I specifically stipulated a gain of one million.

What does the difference have to be such that, with a gain of one million, the output will be +15V.

Don't just guess at an answer and ask if your guess is right -- show some work.

difference has to be million / 15 right?

 

From your answers given in your first post. It looks like you have not paid attention to the resistors prefix values. 30Kohm is not equal to 30ohm, but 30000ohm. Always make it a habit of using Scientific notation then typing in those numbers. Most calculators have an EXP button. That let you type in 30K as 30E+03

 

difference has to be million / 15 right?

Is this a guess or do you have a reason for saying this?

When WBahn comes back on-line he will tell you to check your units - then you will see why "million / 15" is not right. I am just saving WBahn the trouble!

 

Is this a guess or do you have a reason for saying this?

When WBahn comes back on-line he will tell you to check your units - then you will see why "million / 15" is not right. I am just saving WBahn the trouble!

Vout = A(Vp - Vn)

So a = 1 mil. Vout = 15 so
15/ 1 million to get difference between Vp and Vn

 

Is this a guess or do you have a reason for saying this?

When WBahn comes back on-line he will tell you to check your units - then you will see why "million / 15" is not right. I am just saving WBahn the trouble!

Twould appeat that my reputation precedeth me!

But to make it official:

You are looking for something that has to have units of volts, right?

Your answer, had you tracked your units, would have been

Vout = 1 million / 15V, or 66,700 V^-1

I don't know what that is, but 1/V certainly is not a unit of voltage. If the units are wrong, you KNOW the answer is wrong.

 

Vout = A(Vp - Vn)

So a = 1 mil. Vout = 15 so
15/ 1 million to get difference between Vp and Vn

Vout = A(Vp-Vn)

You want to solve for the difference, namely (Vp-Vn). Divide both sides by A:

Vout/A = (Vp-Vn)

If this is indicative of your math skills, you need to take a step back and really spend some time improving them. Find a basic Algebra text (lots of online material available) and work your way through it working problems until you understand each concept you come to.