oops i see many others posted while i was typing sorry if this is redundant... old jim
xconwing said:
how can you tell if it's an inverting or non-inverting op amp just by looking the the schematic
"Operational Amplifier" unfortunately has dual meaning.
It is the name given to an amplifier that could perform some math operation
provided it has been surrounded by circuitry that let's it make its two inputs equal. That's how it "operates".
It has fallen into common usage as the name given to a circuit containing
both such an amplifier
and its surrounding circuitry when used with negative feedback to make its inputs equal. It performs a math operation, usually but not always a linear one.
The way to tell whether the latter inverts is to work it in your head. i talk my way through them.
Looking at yours
First question is can this operate ? That is, can it hold its + and - input terminals equal?
Let's see what we know:
we know input- is held firmly at 0.
we know input+ will be at some voltage that lies in between Vin and Vout, because the 3k and 9k resistors make a voltage divider.
we know that making input+ more positive makes Vout more positive because that's what the + means.
So at first glance it appears
applying a tiny positive voltage to Vin will make Vout go positive
so input+ will lie between Vin which is positive and Vout which is also positive
Now - the condition for operation was it must make its inputs equal, that is make input+ equal zero.
Can zero lie between two positive numbers? I think not.
So i think your textbook has thrown you a trick question - that circuit isn't a linear operational amplifier circuit, even though it has an operational amplifier in its heart.
Now there do exist the math operators for comparison, > for greater than and < for less than.
Your circuit can do that function.
Let's apply some arithmetic to your circuit.
Voltage between Vin and Vout is of course Vin-Vout
so voltage across the 3k is ¼(Vin-Vout)
and voltage across the 9k is ¾(Vin-Vout)
Write KVL around this closed loop::: Vin to common to input+ through 3k back to Vin
and you get
Vin - (input+) - ¼(Vin-Vout) = 0
re-arrange
input+ = Vin - ¼(Vin-Vout)
input+ = ¾Vin + ¼Vout
so in order for input+ to be zero, Vin and Vout must have opposite polarity. That's what is meant by inverting opamp circuit.
In the circuit above they don't, but if you swap the amplifier's input pins they will.
So as is, that circuit doesn't invert.
If this was a trick question, the circuit indeed doesn't invert so it must be noninverting. But i wouldn't call it an operational amplifier circuit
Since the opamp has supply of +/- 15 volts that's the highest its Vout can go.
What if Vout were at that limit, let's just say postive 15 volts?
could we drive input+ to zero by applying negative Vin ? If we went just a little further would we tip the balance and make Vout go to negative 15? Of course we could.
So let Vout be +15
what Vin will drive input+ to zero?
input+ = ¾Vin + ¼Vout
0 = ¾Vin + ¼(+15)
¾Vin =
- ¼(+15)
Vin = -5
So that circuit flips its output state whenever input edit: reaches -Vout/3, as shown in svein's delightful post #8 scope trace .
crosses 5 volts with polarity opposite Vout
which would be described as a comparator with hysteresis
also known as a "Schmitt Trigger"
and it does a logical math operation, the inequality "greater than" with memory of last state
and since output polarity follows input polarity
i'd call it a "non-inverting comparator with 5 volt hysteresis"
KVL is great for figuring out op-amps.
Ask teacher if that was a trick question.
Have fun. Talk to your circuits, it'll amuse your roommates no end.
old jim
