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The operational amplifier is simplu a very high gain amplifier. Its almost ideal in terms of the high gain, high input impedance, low output impedance, but the nice thing is that its gain is very high that is saturates the output. So its not usually used with the open loop configuration as its of now use.
The reason behind the presence of the non-inverting and inverting terminals is that its implemented using a set of differential amplifiers, which are amplifiers that amplify the difference between the two signals.
The very interesting fact is that the inverting term causes the ability to form circuits with negative-feedback configuration which is one of the main configuration the Operational Amplifier (or simply the OpAmp) is used in. The use of negative feedback is simply a trade of the very high gain for the application you want.
where I only use uppercase so subscripts are easily read, and
Vo = output voltage
A = open loop gain
Vp = voltage at plus (I'd call it "non-inv terminal, but you don't know why yet) input
Vm = voltage at minus input
Now the question you may have is, how does that equation differ from the comparator? It doesn't. The two are the same.
So how do you get an op amp from this? You'll see.
Now, let's say, using our equation above, that A=1,000,000 (actually higher but, let's use it), Vp=1mV, and Vm=0V (it's grounded). What do you get?
Did you see the sign change? Now notice where we put our 1mV input each time. This is why the plus input is called non-inverting and the minus input is called inverting. Notice that in both instances we used a positive input, 1mV. When we applied the positive to the one input we got a positive result. But when we applied it to the other input, we got a negative result, an inverted result.
Now that you get that, what about the 1000V?
If you ran an op amp/comparator from a 1000v (plus some) supply, then you'd get a 1000V output. However, in reality, the supply of an opamp/comparator is no where near 1000V. Hence, the opamp/comparator does the best it can to give you 1000V with the supply it has. Instead of 1000V, it just slams the supply rail, either negative or positive, whichever one you wanted, due to your +/- input selection. This slamming of the rail is the comparator's output.
To prevent the slamming of either of the rails, we use feedback, and thus is born the "op amp circuit." An op amp is a comparator with feedback.
Do you understand now? (I won't go into op amps if you haven't followed to this point. However, even if you do get it to this point, review the VDR if necessary, because if you want to know more, you're going to need it.)
However, I'm not saying anythng further if you (or someone else) do(es)n't ask, because, like an op amp, I need feedback. ;-)
In my experience there are two kinds of comparator; so let's be clear. There is the digital, which is the Exclusive-OR gate. It says "I'll give you a high if you give me unlike inputs." Then there's this comparator, the analog. This one says, "if + is higher than -, then I'll give you a positive output, or if - is higher then +, then I'll give you a negative output.
However, with either comparator, the output is the result of a comparison of the inputs, hence the name.
VDR -- This is your best buddy in electronics, the Voltage Divider Rule:
Vx = (Rx/Rt)Vt
Rx = the resistance for which you wish the voltage drop
Rt = the total resistance in the series (1, 2, 3, or infinite number of resistors)
Vt = the supply across Rt
Vx = the voltage drop of Rx
The VDR is your best friend. It makes understanding circuits at a glance so much easier. It is nothing mysterious however, for it comes from Ohm's Law. Simply figure out the voltage drop via Ohm's Law, and you'll see the VDR come out.
I would prove it to you, but I'm out of time again.
P.S. Then once you have the VDR, then the reciprocal of the fraction is the CDR, the Current Divider Rule, for parallel resistances, and using It and Ix instead.
Once again, Vtap = Vm, and Vp is still at Gnd, so we get
Vo
=
A [0 -
(
Rin(Vo-Vin)
--------------
Rin + Rf
+
Vin(Rin + Rf)
---------------
Rin + Rf
)]
=
A(
-Rin(Vo-Vin)
--------------
Rin + Rf
-
Vin(Rin + Rf)
---------------
Rin + Rf
)
Now expand on the right to get
Vo
=
A(
-RinVo+RinVin - VinRin - VinRf
------------------------------------
Rin + Rf
)
Then simplify a little to get
Vo
=
A(
-RinVo - VinRf
-----------------
Rin + Rf
)
Now lose the denominator.
Vo (Rin + Rf)
=
A(-RinVo - VinRf)
Expand on the left and right.
VoRin + VoRf
=
-ARinVo - AVinRf
Group
VoRin + VoRf + ARinVo = -AVinRf
Factor out Vo.
Vo(Rin + Rf + ARin) = -AVinRf
and solve for Vo
Vo
=
-VinARf
-------------------
Rin + Rf + ARin
Now, make the right more understandable by multiplying by A^-1/A^-1=1
Vo
=
-VinRf
----------------------
(Rin+ Rf)/A + Rin
This is an important result.
It says that, were A not infinite, the closed-loop gain (the absolute value, actually) would go down, because th denominator would increase by (Rin + Rf)/A. This is important when you start to push the bandwidth of an op amp to where the open- loop gain is no longer very large.
However, if you make sure you're working substantially below the bandwidth edge of the op amp, you can make the assumption that Vo/A = 0, simplifying things a bit, as was shown previously.
(In another thread, a questioner was wonderng why his high-pass filter, using a 741, was a bandpass, if I remember correctly. It's all about the open-loop gain falling as frequency increases, which reduces the closed-loop gain of the circuit.)
Note: Send A to infinity again, and you'll see the ideal equation, the one in the previous post, come out of this one.
Now we'll continue to the non-inverting op amp circuit. ...when I get some more time ;-)
...I hope I haven't bored people, especially the author of this thread...
The opamp is an amplifier where it amplifies the difference of the two inputs. it suppresses the common mode signals like noise in the inputs and amplifies only the differential signal. it has a high open loop gain which is unpredictable and unstable. it is used in the closed loop mode since the closed loop gain is predictable and stable.
Now to the non-inverting op amp circuit. With this one we'll send the input signal to the non-inv input but still set up a voltage divider from the output to the inverting input, so as to reign in the open loop gain (A) again.
The Non-Inverting Op Amp Circuit
----------------------------------------
Gnd ------- R1 --------- Vtap --------- Rf ---------- Vo
Op Amp: +in = Vin; -in = Vtap; out = Vo
Onc again, we use our friend the VDR, which says
Vtap
=
R1(Vo)
---------
R1 + Rf
Next, use the general eq for the op amp/comparator
Vo = A(Vp - Vn)
Now, this time, make Vp = Vin and Vn = Vtap and substitute.
(Notice we didn't take the simplification that Vo/A=0 this time, as we saw that we'll get both the ideal and real at the end anyway.)
Now factor A from the numerator
Vo
=
VinA(R1 + Rf)
------------------
R1 + Rf + AR1
Then, to make it easier to understand, do our trick of multiplying by 1=A^-1/A^-1.
Vo
=
Vin(R1 + Rf)
------------------
(R1 + Rf)A + R1
This is the general equation for the non-inverting op amp circuit.
Once again, we see that, with A not equal to infinity, as the ideal assumption goes, the output decreases because the denominator increases by (R1 + Rf)/A, just as was the case with the inverting op amp.
Likewise, setting A = infinity, the general equation goes to the familiar ideal equation:
Vo
=
Vin(R1 + Rf)
---------------
R1
Finally, note that it is impossible to set a closed loop gain of less than one, since the eq can be adjusted to
Vo
=
Vin(
1 + Rf/R1
)
Also not one last thing. What if Rf = 0 (a short) anr R1 = infinity (an open)? Can you see that the equation then just becomes
Vo = Vin (1) = Vin
This is called the voltage follower configuration. Its advantage is its use as a high-impedance buffer. It presents a high impedance to the input source and a low impedance (a characteristic of an op amp circuit, generally) to the load.
There ya go, both circuits
The VDR rocks! Many more complex op amp circuits can be understood and designed simply by knowing the VDR and the general op amp/comparator eq, Vo = A(Vp - Vn).
In other words, it doesn't matter how many resistors there are or that Vin is unknown, you can still find the current in each branch if the total current (It) is known and the total resistance (Rt) is known.
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