MarkT:
But its clearly a homework problem you're passing on to us - if you at least had a go at answering
it first it wouldn't be so galling - people like to help, but cheating isn't helping, its avoiding understanding.
It is a challenge from the teacher ..
He said you can get help from anyone .. But I need you to understand the solution ..
I guess nobody knows how to solve it : (
Nice try and I will give you bonus points for being persistent.
Why don't you check out any op-amp introductory primer, it should be simple to figure out the answer.
Lefty recognized the trick part right off - this is a problem in electronics, not mechatronics.
You're in the wrong classroom.
Also, your instructor tricked you - it's a really a trivial problem. As everyone here obviously
knows [except OP], you go to the basics for opAmp operation, and then solve for the special
case.
oric_dan(333):
Lefty recognized the trick part right off - this is a problem in electronics, not mechatronics.
You're in the wrong classroom.
Also, your instructor tricked you - it's a really a trivial problem. As everyone here obviously
knows [except OP], you go to the basics for opAmp operation, and then solve for the special
case.
I really hope I can solve it ..
We all hope you can solve it too.
You all know the answer?!?!?! I feel like an idiot!
The thing you have to remember that there are two different things a circuit can amplify, voltage or current. So the question is exactly how good is the person setting the question. Do they know that fact?
I asked this question in another forum ..
Here is one of the replies ...
"""""
The output voltage is:
Vo = Vi / ( 1- 1/A)
where A is the open loop gain of the op amp. so A>>1
you can disregard 1/A then you have.
Vo=Vi
That is a voltage follower. In theory the same as the standard one with negative feedback.
But in real life the output will smack against one rail or the other and stay there. Any difference between the two inputs is amplified and increases the difference rather than reducing it.
""""""
So what he is saying ..
If the Op-amp is ideal .. ( I+ = I = 0 ) and ( V+ = V- ) .. Then , Vin = Vout in this circuit ..
and if it is not ideal ,, the op-amp will be saturated and Vout will be equal to the value of the external power supply ..
Yes, Vout=Vin. Value of resistor doesn't matter. You can even put 0R.
But in real life the output will smack against one rail or the other and stay there.
That's true for all opamps. -- not just this circuit.
For the positive cycle, Vout cannot be greater than V+ (where V+ is your supply voltage). So if your Vin >> V+, Vout = V+
(and the converse is also true for V-, if you're using a split power supply).
In real life, Vout max is usually some V+ minus (some voltage value, say 0.7V or something, depends on the opamp).
Yes but as I mentioned before the input impedance is high and the output impedance is low therefore there is a current gain even if there is no voltage gain.
there is a current gain even if there is no voltage gain.
An opamp by definition is a voltage amplifying device, not a current amplifying device.
Don't confuse more output current drive capability with current gain.
In fact, current output capability of any opamp can be further increased by strapping on discrete NPN/PNP booster circuit at it's output... but it's still a voltage amplifier.
What about an emitter follower, that has a current gain but no voltage gain. Using an op amp to make a voltage follower is the same thing.
Just because the current gain is saturated does not make it any the less of a gain. In exactly the same way as a common emitter amplifier being used as a switch has a saturated voltage gain, no proportionality there but there is still gain.