How does the ideal op-amp model explain the need for feedback?

[kimmy510]

In an inverting opamp let the output be feedback to inverting input through a feedback resistor Rf. We write the direction of flow of current from input terminal to output terminal in the feedback resistor Rf. But if this direction is correct then how can we say that input is fed back?? it seems like the input current flows through Rf and reaches output terminal.

 

[waht]

Since the '+' terminal of the op-amp is grounded, then the '-' terminal will also be grounded.

Therefore for any Vout votlage, current will flow into '-'

$$i = \frac{V_{out}-0}{R_f}$$


http://www.elexp.com/tips/invert.gif

 

[Bob S]

No current can flow either into, or out of, the "-" input. Where does it go?

Bob S

 

[waht]

No current can flow either into, or out of, the "-" input. Where does it go?

Bob S

Yes you are right, there is no actual current flowing into '-' or '+' in an ideal model, but rather, a virtual ground is generated at '-' due to negative feedback, or balancing action of the inverting op-amp setup.

When Vin is POSITIVE, one can track the input current flowing from Vin, through Ri, and Rf and to the Vout.

When Vin is NEGATIVE, the current will be flowing from Vout, through Rf, and Ri, and to Vin.

 

[jsgruszynski]

No current can flow either into, or out of, the "-" input. Where does it go?

Bob S

Actually a tiny amount does go into the terminal. It's called "Bias Current" and is spec'ed on the Op Amp data sheet. As you intuition suggests, you can't get things for "free". The reason you can ignore it (up to a point) and use this analysis technique is that the ratio of the bias current to your circuit currents is usually safely assumable "sufficiently small" as to be ignorable.

If you look at the circuit diagram for the innards of the Op Amp you are using (they are usually in the datasheet), you'll see a bipolar, JFET or MOSFET differential pair are at the + and - inputs. BJTs of course have base current and this is where bias current comes from in a BJT Op Amp. JFETs have reverse diode leakage which is where JFET Op Amp bias current comes from. And MOSFETs have gate leakage current so that's where MOSFET Op Amp bias current comes from. You might think that this would make MOSFET Op Amps best for low voltage/low current applications, but because the gm (transconductance or "gain") of MOSFET is worse than JFETs, this isn't the case. The best low bias current Op Amps are JFET-based.

The general rule in electrical engineering is if some quantity is 10x or more than another similar quantity, the smaller is "swamped out" by the larger and you can approximate the analysis by ignoring the smaller. Obviously this is not absolutely true but that's how engineers get things faster - by making thoughtful and fully aware approximations.

So when you do need to worry about the bias current is when you are doing a low voltage, low current or low power Op Amp design - when you design current are within 10x of the bias current or whatever resolution/accuracy you are seeking. At that point you have to worry about the current that flows into the terminal because they may be larger than what you are trying to measure or what is induced parasitically in your circuit layout.

 

[LowEParticle]

Unlike the small signal model of a BJT, a current-oriented device, which probably preceeded your current study of op-amps, the op amp is a voltage-oriented device. This may explain some of the confusion you're feeling!

In the ideal (perfect) op-amp model the internal gain, often called the Open Loop Gain, is infinite, and the input impedance of both inputs is also infinite. Since the input impedance is infinite, no current can enter or leave either input terminal. Ignoring the bias current (described in an earlier post) the op-amp may be thought of as instantaneously sampling the voltage at the two inputs, subtracting V- from V+, multiplying that difference by the Open Loop Gain, and outputting the product as a voltage on the output terminal.

Since the Open Loop Gain of the op-amp device itself is assumed to be infinite in the ideal model, then you can pretty easily see why you usually need some sort of feedback arrangement outside the device! Otherwise, the output of the op-amp would only have 3 possible states:
If V+ > V-, Vout is at the positive supply rail
If V+ < V-, Vout is at the negative supply rail
If V+ = V- (exactly), Vout is 0.