Ideal op-amp rules

YOU WILL BE ABLE TO

State the ideal op-amp assumptions.
Explain negative feedback and the virtual short.
Apply the two golden rules to analyse a feedback circuit.
Explain why enormous open-loop gain is useful rather than unusable.

ME-3 ended on a bargain: raw gain is cheap, predictability is expensive, and the modern answer is to throw gain away in exchange for behaviour set by external resistors. The operational amplifier is that bargain taken to its limit. Inside its package sit dozens of transistors arranged to produce an open-loop voltage gain of 100,000 or more — and you will essentially never use that gain directly. You will spend it, buying accuracy.

The op-amp: two inputs, one output. The − and + marks are the whole interface — everything else is inside.

The op-amp amplifies the DIFFERENCE between its two inputs: $v_{out} = A(v_+ - v_-)$ , with $A$ enormous. Take that seriously for a moment. With $A = 100{,}000$ and a supply of $\pm 12$ V, an input difference of just $120$ µV drives the output all the way to a rail. On its own the device is almost unusable — a hair-trigger. It becomes useful the instant you connect its output back to its inverting input, which is negative feedback.

Now the payoff. Suppose the circuit has negative feedback — a path from the output back to the inverting input — and suppose the output is NOT slammed against a rail. Then the output must be finite, and since $v_{out} = A(v_+ - v_-)$ with $A$ enormous, the input difference must be essentially zero. That gives the first golden rule, and the ideal input impedance gives the second:

v_+ = v_- \quad\text{(the virtual short)}

i_{+} = i_{-} = 0 \quad\text{(no input current)}

the two golden rules

These two lines analyse almost every op-amp circuit you will meet. Note the logic carefully, because it is easy to misremember: the op-amp does not magically force its inputs equal. The FEEDBACK forces them equal, by driving the output to whatever value makes them equal. Remove the feedback and the rule evaporates — which is precisely what a comparator is, and why Lesson 4 treats it separately. The rule also fails when the output saturates: if the op-amp cannot swing far enough to equalise the inputs, it stops trying and sits at the rail.

WORKED EXAMPLE

The virtual short, in numbers

An op-amp with $A = 10^5$ is in a feedback circuit and its output sits at $5.0$ V.
The required input difference: $v_+ - v_- = v_{out}/A = 5.0/10^5 = 50$ µV.
Against signals of volts, $50$ µV is noise. Treating the inputs as equal introduces an error of about one part in $10^5$ — far smaller than your resistors' tolerance.
That is why the ideal assumption is not a simplification you apologise for; it is more accurate than the components around it.

50

µV — the "short" is virtual but numerically excellent.

WORKED EXAMPLE

The voltage follower

Connect the output directly to the inverting input, and drive the non-inverting input with $v_{in}$ .
Rule 1: $v_- = v_+ = v_{in}$ . But $v_-$ IS the output. So $v_{out} = v_{in}$ .
Gain of exactly one — which sounds useless until you notice what else it does: rule 2 says the input draws no current, and the ideal output impedance is zero.
So a follower converts a high-impedance, easily-loaded signal into a low-impedance one that can drive anything. It is the cure for CA-2's loading trap, and it is the single most common op-amp circuit in existence.

v_{out} = v_{in}

, with the loading problem solved.

PRACTICE

An op-amp has $A = 2 \times 10^5$ and its output rests at $-8$ V in a feedback circuit. What is the voltage difference between its inputs?

State the two golden rules, and the condition that must hold for the first one to be valid.

A follower is driven by a divider made of two $1$ MΩ resistors from $5$ V, and its output feeds a $1$ kΩ load. What voltage appears at the load, and what would happen without the follower?

Why is an open-loop gain of $10^5$ described as something to "spend" rather than to use?