# Operational amplifiers

## Homework Statement

http://desmond.imageshack.us/Himg853/scaled.php?server=853&filename=opamp.jpg&res=medium [Broken]
relevant equations
v=ir (ohm's law)
kirchoff's current law

## The Attempt at a Solution

i'm concerned about the bottom left node in this picture. current into an ideal op amp is zero because resistance is infinite so there should be no voltage drop, so the node according to the input of the op amp should be +16 volts. but i do not know how to handle the +10 V floating around down there, it seems to indicate that the node is 10 volts and not 16. how can i deal with these two seemingly conflicting notions?

i also have a second question:

http://desmond.imageshack.us/Himg807/scaled.php?server=807&filename=opamp1.jpg&res=medium [Broken]

Is I = 0 amps in this image since it is connected to ground? i see no other way around it. we can't define the voltage of the node of the output as some unknown and solve it that way, because we know it is 0 because it is connected to ground. if it isn't 0 then i would be able to define the current on the left branch as I=(14/500)=((14-v1)/3000)) and solve for v1 as -70 volts, so it would be -70/5000.. but i still think it should be zero. i just don't think my teacher would put something so simple so i feel as if i am wrong. edit:[ upon further review it seems that i should define the two inputs as unknowns and go from there, i'm so used to having inputs that are connected to ground that i naturally assume the input is 0, however i still do not understand how the output node isn't 0]

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gneill
Mentor

## Homework Statement

http://desmond.imageshack.us/Himg853/scaled.php?server=853&filename=opamp.jpg&res=medium [Broken]
relevant equations
v=ir (ohm's law)
kirchoff's current law

## The Attempt at a Solution

i'm concerned about the bottom left node in this picture. current into an ideal op amp is zero because resistance is infinite so there should be no voltage drop, so the node according to the input of the op amp should be +16 volts. but i do not know how to handle the +10 V floating around down there, it seems to indicate that the node is 10 volts and not 16. how can i deal with these two seemingly conflicting notions?
The "floating 10V" is referenced to ground by convention. The 16V supply, on the other hand, you can see where both its terminals are connected, and neither is ground. So the lower left node is at +10V potential above ground. What does that make the upper left node?
i also have a second question:

http://desmond.imageshack.us/Himg807/scaled.php?server=807&filename=opamp1.jpg&res=medium [Broken]

Is I = 0 amps in this image since it is connected to ground? i see no other way around it. we can't define the voltage of the node of the output as some unknown and solve it that way, because we know it is 0 because it is connected to ground. if it isn't 0 then i would be able to define the current on the left branch as I=(14/500)=((14-v1)/3000)) and solve for v1 as -70 volts, so it would be -70/5000.. but i still think it should be zero. i just don't think my teacher would put something so simple so i feel as if i am wrong. edit:[ upon further review it seems that i should define the two inputs as unknowns and go from there, i'm so used to having inputs that are connected to ground that i naturally assume the input is 0, however i still do not understand how the output node isn't 0]
I will be zero if the output of the op-amp is zero. If the output of the op-amp is nonzero then the current will not be zero. The op-amp's output is not grounded -- it's connected via a load resistor (5KΩ) to ground.

EDIT: You should always label the inverting and non-inverting inputs on the op-amps.

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I'm curious. How do you solve the first problem.

I tried the virtual short method.
For the first opamp, the voltage is -6V at the 50 ohm resistor.
So the -ve terminal is also at -6v.
For the second opamp, the volage is 10v at the 600 ohm resistor.
So the ove terminal is also at 10v.
But what after that?

gneill
Mentor
I'm curious. How do you solve the first problem.

I tried the virtual short method.
For the first opamp, the voltage is -6V at the 50 ohm resistor.
So the -ve terminal is also at -6v.
For the second opamp, the volage is 10v at the 600 ohm resistor.
So the ove terminal is also at 10v.
But what after that?
We should probably wait for the OP to show some progress on the problems before we move further ahead.

I'll give you a small hint though: Use the other property of ideal op-amp inputs to infer a couple of other potentials nearby.

We should probably wait for the OP to show some progress on the problems before we move further ahead.

I'll give you a small hint though: Use the other property of ideal op-amp inputs to infer a couple of other potentials nearby.
I tried to find the current thru the 200 and 300 ohms resistors. But didn't get anywhere.
-6v at one end of 200 ohm res, but the other end is not connected directly to output of the opamp1. Opamp output current is flowing both in 400 ohm and 200ohm resistor.
Looking at the ckt, the 3resistors- 400, 100 and 500 form a voltage divider, no?

gneill
Mentor
I tried to find the current thru the 200 and 300 ohms resistors. But didn't get anywhere.
-6v at one end of 200 ohm res, but the other end is not connected directly to output of the opamp1. Opamp output current is flowing both in 400 ohm and 200ohm resistor.
Looking at the ckt, the 3resistors- 400, 100 and 500 form a voltage divider, no?
How much current can flow into or out of the input terminals of an ideal op-amp?

Yes, the 400, 100, and 500 Ohms resistors form voltage dividers.

How much current can flow into or out of the input terminals of an ideal op-amp?
Infinite! But how do I apply this?
Say, for example the bottom opamp can supply infinite current. So all the current flowing thru the 500 ohm and beyond is supplied by this opamp.
so the voltage at the other end of 300Ω is almost equal to 10v??

gneill
Mentor
Infinite!
No, that's exactly wrong The INPUTS of an ideal op-amp conduct NO current. They have infinite impedance.

No, that's exactly wrong The INPUTS of an ideal op-amp conduct NO current. They have infinite impedance.
Ok. Please ignore my last comment. I was thinking ideal opamp output.

I already thought of the infinite input impedance, but where do I apply?
No current flows into the bottom opamp input, so the 10v at 300Ω resistor is like a voltage source. Dead end!

gneill
Mentor
Ok. Please ignore my last comment. I was thinking ideal opamp output.

I already thought of the infinite input impedance, but where do I apply?
No current flows into the bottom opamp input, so the 10v at 300Ω resistor is like a voltage source. Dead end!
Is it? If no current flows through the 300Ω resistor, what's the potential difference across it?

Is it? If no current flows through the 300Ω resistor, what's the potential difference across it?
Zero. So 10v at the 300Ω and 100Ω node and -6v at the 200Ω and 100Ω.

So 16v across 100Ω. I=0.16A

So bottom opamp output is 80V, top opamp is 64v?

gneill
Mentor
Zero. So 10v at the 300Ω and 100Ω node and -6v at the 200Ω and 100Ω.

So 16v across 100Ω. I=0.16A

So bottom opamp output is 80V, top opamp is 64v?
Very close. Remember to add the potential difference across the 500 Ω resistor due to the current through it to the known potential at the confluence of the 300 and 100 Ω resistors.
Similarly, add the potential drop across the 400 Ω resistor to the known potential at its bottom end.

Very close. Remember to add the potential difference across the 500 Ω resistor due to the current through it to the known potential at the confluence of the 300 and 100 Ω resistors.
Similarly, add the potential drop across the 400 Ω resistor to the known potential at its bottom end.
I had the answer right the first time. But then, I thought the voltages are too high, something must be wrong. 90V out of an opamp, that's not normal!
Anyway, this is just a homework problem. Not a lab ckt.
Thanks!

gneill
Mentor
I had the answer right the first time. But then, I thought the voltages are too high, something must be wrong. 90V out of an opamp, that's not normal!
Anyway, this is just a homework problem. Not a lab ckt.
Thanks!
90V for an op-amp looks unlikely because we're used to dealing with the usual ubiquitous op-amp IC's running at relatively low supply voltages. But op-amps can be constructed with other technologies, too, for higher voltages and currents. The operational amplifier is more important for its concept than its particular implementation.

In my excitement, I forgot about the second problem.
Looks more difficult than the first.

The 6k and 2k look like gain resistors. The 1k can be ignored since no current flows in to the opamp.
I tried to ignore the load and analyze. But got nowhere.
From the 14v source all the current flows thru the 5k and 2.5k resistor.

gneill
Mentor
In my excitement, I forgot about the second problem.
Looks more difficult than the first.

The 6k and 2k look like gain resistors. The 1k can be ignored since no current flows in to the opamp.
I tried to ignore the load and analyze. But got nowhere.
From the 14v source all the current flows thru the 5k and 2.5k resistor.
That's a 500 Ω resistor, not 5k.

Redraw the circuit so that it appears more "typical" looking. Use KCL at the two nodes that will share the input terminal potential (call them both V1 perhaps). Proceed to solve for the output voltage. #### Attachments

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For the inverting input side

V1/6 = (Vo-V1)/2
This will reduce to V0=1.33V1

For the non-inverting side

(14-V1)/0.5 = - (V1-V0)/2.5

or is it
(V1-14)/0.5 = (Vo-V1)/2.5

gneill
Mentor
(14-V1)/0.5 = - (V1-V0)/2.5

or is it
(V1-14)/0.5 = (Vo-V1)/2.5
You should be able to sort out the math by being consistent with current directions and potentials when applying KCL. At a given node the currents should sum to zero.