# op amp function of a circuit #1

Discussion in 'Homework Help' started by notoriusjt2, Dec 13, 2010.

1. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1

now this one states "Assume the op amps are ideal" just like I am used to. But I did notice something funky with this circuit. The -Vin is connected to the + terminal on the bottom op amp. Does this matter? or is the op amp still going to function correctly?

2. ### steinar96 Active Member

Apr 18, 2009
239
4
It will still function correctly.

3. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
Greetings norotiusjt2,

What answer did you come up with? Be sure to show us your work so that we can determine any errors that you make.

hgmjr

4. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1

allright so lets arbitrarily assign Vin = 10V just for examination purposes. Therefore...

node1 = +5V
node2 = -5V

because these are ideal op amps V(inverting) = V(non-inverting)... so...

node3 = +5V
node4 = -5V

now here is where I am getting confused

at node3 is +5V. From there we have two resistors in parallel, a 1KΩ and a 4KΩ.

Since this is an ideal op amp there is no current flow into the +/- terminals of the op amp. which way is current flowing within those 3 resistors?

5. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
If you continue down the thought process you have already started with your statement that node3 = +5V and node4 = -5V (which is a correct assumption) then you will notice that the only drive source for node3 comes from the voltage at node6. Similarly, the only drive source for node4 comes from the voltage at node5. Since the drive in both cases is through a 4K resistor then you should be able to draw the necessary conclusion about the voltages that must exist at node5 and node6 to balance everything out.

6. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
but if we have 1 voltage and 1 resistor wouldnt we drop the full voltage across that resistor?

7. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
I'm am not clear on what you are referring to above.

hgmjr

8. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
say we have 10V at node6... wouldnt we drop the full voltage across that 4K resistor?

9. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
The question you may want to ask yourself is what voltage must be applied to the top of the 4K resistor between node3 and node5 to raise the voltage at node3 to +5V.

hgmjr

10. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
25V at node 5?

across the 4K we would drop to 20V then across the 1K we would drop to 5V

11. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
675
I think you realize that the voltage 3-4 is equal to the voltage 1-2. What is the current through the 1k, in terms of Vin? Since op amp input current is zero, what is the current through the 4k resistors, again in terms of Vin? If you can answer these questions, you can solve for Vout in terms of Vin.
If the algebra confuses you, let Vin=1V and go through the same exercise.

12. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
Greetings notoriusjt2,

Not exactly. Consider the current due to the voltage across the 1K resistor. No current is flowing into or out of node3. That means that the current must come from the voltage at node5 through the 4K resistor.

25V would not be enough to balance the current that must flow in the 1k resistor.

hgmjr

13. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
I realize that no current will flow in either direction at node 3 and node 4

originally I thought that the current would flow as in this diagram

from the positive to the negative polarity..

but when you said, "25V would not be enough to balance the current that must flow in the 1k resistor" I became confused

what do you mean when you say balance?

14. ### Ron H AAC Fanatic!

Apr 14, 2005
7,049
675
Did you see my post #11?

15. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
Let's take your value of 25V and see what we get.

With the voltage at node3 equal to +5V (your earlier correct conclusion) and the voltage at node5 at +25V from your above conclusion, then the voltage across the 4K resistor is 25 minus 5 which comes to 20V. Divide 20 by 4 to get a current of 5 milliamps. The current being pulled out of the terminal is 10 milliamps. 5 millamps is too low to balance out the 10 milliamps.

hgmjr

16. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
current through the 4kΩ = 5V/4000Ω = 1.25mA
current through the 1kΩ = 5V/1000Ω = 5mA

17. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
At least now I think we see where you are confused. The 1K resistor has 10V across it. That makes the current 10V/1000Ω = 10ma.

hgmjr

18. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
allright

40V/4000Ω=10mA

that would give us the correct balance

and 40 is Vin * 8

which is one of the answers except the answer listed is Vout=-8*Vin

is it negative because the polarities are reversed?

19. ### hgmjr Retired Moderator

Jan 28, 2005
9,030
218
Think about. If you have 40V across each of the two 4K resistors and you have 10V across the 1K, the total is 40V+10V+40V which is 90. Since your input is 10V then you will notice that you have an answer that corresponds to this gain.

hgmjr

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20. ### notoriusjt2 Thread Starter Member

Feb 4, 2010
209
1
I understand now

I was thinking that I had to find some sort of balance point for the current

In reality this is just a simple series circuit that I made way harder then it should have been.

so thats right... 40V+40V+10V = 90V across Vout

Vout = 9 * Vin