Some basic OP Amp Help

[imnotorginal]

Hello members, its probably obvious but if not I am new to the forum. I come here in seek of assistance with some electronics HW and general guidance when it comes to solving common electrical questions. I do not seek just simple answers; I seem advice and common methods for solving the problems. With that being said, let me introduce my first problem. I have the following circuit; I need to figure out how to calculate the output current (iout) as well as the current at i3. I understand the basics of opamps and that i2 = i1 and point A has a virtual ground, there is no current going into the opamp because of an Infinet resistance. This is more a basic circuits 1 question really.

Also can someone offer a little explanation as to what the feedback loop is used for? Thanks for any assistance in advance.


**broken link removed**

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[srizbf]

i3=((R2/R1)*V1)/R3

this is a simple one for current (magnitude)through R3.

srizbf
2ndjuly2010

 

[hameeds01]

V1 = 0 volts ???

 

[srizbf]

i3=0 Amps.

srizbf
2ndjuly2010

 

[keith1200rs]

Also can someone offer a little explanation as to what the feedback loop is used for? Thanks for any assistance in advance.

The output current simply depends on the output voltage and the load resistor. So I3=Vo/R3. In you example you have zero volts in, which is not a useful example - you will end up proving 0=0. Try it with another input voltage.

The feedback is used because the gain of an opamp is very high - far higher than you would normally require. Adding the negative feedback allows the gain to be set at a more practical value. Also, because the opamp 'open loop' gain is so high the actual value usually doesn't matter and the 'closed loop' gain just depends on the resistor ratio.

Keith

 

[Element_115]

First off, if "V1" is = 0V, everything is off . all your answers are 0( 0V or 0A) !

Now if V1 is some value of voltage (AC or DC) we can explain things better!

Note:
U1 pin 3 is tied to Ground, Thus V+ = 0V !
So by Definition of an "OP-AMP", at all times" V-" must be (or will be adjusted
by Vo to be) = 0V (Because V+ ≡ 0V, OP-AMPS require V+ ≡ V- ) !

So with this in mind just think of it as :

What ever voltage wave-form V1 generates, Vo "WILL" create a related wave
form that always Keeps V- ≡ V+ !


Also one practical point that new students miss is:

Now matter how you set the gain of an op-amp and what voltage you put on the
inputs. The voltage rails of the op-amp set the highest voltage you can
produce at Vo !

If U1 pin 7 = +15V & U1 Pin 4 = 0V(Ground)

Vo max will always be +15V !

Thus you "i3" can only be as great as "Vmax"/R3; where Vmax is :
(U1 pin 7 Voltage) - (U1 pin 4 Voltage) = Vmax

 

[imnotorginal]

I just put the input voltage to be 0 to generalize the problem, really it should just say V1 instead of V1 = 0. I wanted to generalize the problem to be able to understand where the solutions came from instead of just getting a number as the solution.

 

[bejoy]

voltage at both terminals will be same by virtual gnd concept so voltage at node A can be considered as 0v.
now apply KCL at node A and B.
op amp draws no current.
By solving the eqns obtained from KCl u can find current voltages etc.

So for most of op amp probs KCl alone required.


now find the gain formula frow the eqns obtained above. we can see that we can control the op amp gain just by altering the external capacitors.
this is a great strenght of op amp.
open loop ideal op amps have very high gain.
usually -ve f/b is used for stable amplifier ckts.
and +ve f/b used generally for schmitt trigger ckts etc.