# Can someone explaing OPAMPS tasks to me?

1. Aug 30, 2014

### the_man

I am learning about operational amplifiers or so called OPAMPS. And I know how to solve these two examples

EXAMPLE 1 - inverting opamp
EXAMPLE 2 - non-inverting opamp

But how to solve these two:

1. The problem statement, all variables and given/known data

IMAGE 1
1. As you can see on picture No1, it is solved, but I need someone to explain to me the logic behind it, because this scheme doesnt match neither inverting or non-inverting opamp. And why formula looks the way it is. Uin = E / ( (r1+r2) / r2 )? If E is Uinput and E goes to +, that means it non-inverting OPAMP, and the usual formula is (R1+Rf)/R1. So thats why I dont understand.

IMAGE 2
This one is similar. The scheme doesnt match to inverting and non inverting opamp, so I am confused how the solution for Uinput is 12,2V.

2. Relevant equations

So if someone can tell me how I can figure out the formula for the first one or the second example, or for any similar scheme when it comes to opamps, I would be really greatful

2. Aug 30, 2014

### milesyoung

3. Aug 30, 2014

### CWatters

Circuit 1...

This is a inverting comparator..

ε and R1&R2 can be treated as a simple potential divider to calculate a "reference voltage" at V+ of 1.12V.

If V- is > V+ (eg >1.12V) the output will be low (approx. -12V)
If V- is < V+ (eg <1.12V) the output will be high (approx. +12V)

The sketch graph appears correct.

Circuit 2...

This is a non-inverting comparator..

In this case ε provides a 5V "reference voltage" on V-.

The input u is divided down by the divider R1&R2 and presented to V+. The divider ratio is about 0.41.

If V+ (=u*0.41) is > V- (=5V) the output will be high (approx. 12V). For this to be the case u must be greater than 12.2V (= 5/0.41)

If V+ (=u*0.41) is < V- (=5V) the output will be low (approx. -12V). For this to be the case u must be less than 12.2V

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