# simple op amp

Discussion in 'Homework Help' started by jstrike21, Feb 3, 2010.

1. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
The assignment is attached.
Problem 6 is the one i am struggling with. Im just not sure how to solve this problem. Any suggestions?

I set up an equation for the current at the node going into the negative terminal.
i1=(Vi-V1)/(1/sc)+(Vi-V1)/(10k)
I then set V1 to zero because V2(voltage at positive terminal) - V1=0 and V2 has to equal 0 so V1 is also 0.
THen i got the equation:
i1= (sc+1/10k)Vi

Is this right so far?

Also I need help on number 5.... It has an input that goes into the + and an input the goes into the - im not sure what to do.

• ###### EE 230 HW 4 Spring 2010.pdf
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Last edited: Feb 3, 2010
2. ### Fraser_Integration Member

Nov 28, 2009
142
5
Can't help with problem 6 I'm afraid bit complicated for me, but 5 is just a non-inverting amplifier.

The inverting input will have a voltage which is a voltage divider function of the two resistors. Set that equal to Vi, and you will have Vo/Vi = 1 + R1/R2, i'll leave you to figure out which resistor value goes where.

3. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
thats what i did at first
Vo/Vi=1+500k/10k
Vo=51Vi

but this seems wrong because i didnt use V2

4. ### thyristor Active Member

Dec 27, 2009
94
0
This is just a simple inverting amplifier with a capacitor across the input resistor. You're on the right lines but you have made an algebraic slip up.

If X is the parallel combination of the 10K resistor and the capacitor, then:

i = Vin/X and also i = -Vout/500K since the input terminal is at ground potential for the reasons you outlined.

so Vout/Vin = - 500K/X

I'll leave you to calculate X

5. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Yep. You need to reexamine your solution for number 5. Your answer should contain the term V2 in it.

If you can post your work in solving number 5 we could be of grater assistance.

hgmjr

6. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
I dont see whats wrong with the algebra....
Vo/Vi= -(sc+(1/10k))*(500k)
right?

7. ### thyristor Active Member

Dec 27, 2009
94
0
Actually, that's not correct because the resistor chain doesn't go to 0v but, instead, to V2 as shown, so the calculation is a little more complex.

Calculate the input voltage at the -ve terminal by simple voltage divider maths (call this V3) and then Vout must be V1 - V3

8. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
My book has no examples with two inputs like this problem. I dont know how I am supposed to start this problem.

9. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
I calculated the current at the node and that was i1=(sc+(1/10k))V1) so wouldn't I just multiply this by 500k to get Vout?

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Like thyristor has indicated, you can start by calculating the voltage at the negative input to the opamp. This equation will contain Vout and V2. Can you at least give that a try and then post you result.

hgmjr

11. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The current is not what you need to calculate. You need to calculate the voltage at the negative input.

hgmjr

12. ### thyristor Active Member

Dec 27, 2009
94
0
Apologies, yes that is correct. I was flicking between qs. 5 & 6 and confused myself.

13. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
If i make V3 voltage at -ve input then say:
V3=V2(500k/510k)
but this doesnt use Vout

14. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
Ok what really confused me about this problem is the input of 2u(t) i'm not quite sure how to use this for the input.

15. ### thyristor Active Member

Dec 27, 2009
94
0
That's incorrect.

What is the voltage across the 10K and 500K resistor in total. That should start you on the right lines. Come back and tell us.

16. ### hgmjr Moderator

Jan 28, 2005
9,030
214
That is because your expression for V3 (which you have equated to V-) is in error.

hgmjr

17. ### thyristor Active Member

Dec 27, 2009
94
0
Just substitute 2u(t) for Vin.

2u(t) merely means that Vin is a time dependent function, ie: its frequency may change and we know that the amplifier in question has a frequency dependent transfer function.

Sep 24, 2009
104
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19. ### hgmjr Moderator

Jan 28, 2005
9,030
214
2u(t) is the expression for a "step" function. It says that for t < 0, the signal amplitude is 0. For t >= 0, the function has an amplitude of 2.

hgmjr

20. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
so is it proper use to put my answer as Vo=-100.1u(t) or do i just leave the u(t) off?