# Op-amp behavior in open loop

I'm trying to understand how op-amp works, but I'm bit stumped by its open loop behavior.

In different sources I found that $$\v_u=A ⋅ v_d\$$, where $$\v_d=v_+ - v_-\$$; since $$\A=∞\$$, the explanations I found concludes that $$\v_d=\frac{v_u}{A}=0\$$, with no limitation to closed loop (which I found only in Wikipedia: "In a closed loop the output attempts to do whatever is necessary to make the voltage difference between the inputs zero", https://en.wikipedia.org/wiki/Operational_amplifier#Ideal_op_amps).

I understand why $$\v_d=0\$$ is true in closed loop, but I'm not sure why it is in open loop.

On an MIT paper I found that (note: I changed $$\V_i\$$ to $$\v_d\$$ and $$\V_o\$$ to $$\v_u\$$ to have the same notation as above):

Note that when using the ideal op-amp rules we should remember that they are limits and so we must perform our analysis by considering them as limits. For example if we consider the equation

$$\v_u=Av_d ⇒ v_d=\frac{v_d}{A}\$$

Which in turn implies that $$\v_d → 0\$$ as $$\A → ∞\$$. However, this does not mean that $$\v_u → 0\$$ but rather that as $$\A → ∞\$$, $$\v_d → 0\$$ in such a way that their product $$\Av_d=v_u ≠ 0\$$.

I get why "tends to 0" is different to "equals zero", but "as $$\A → ∞\$$, $$\v_d → 0\$$ in such a way that their product $$\Av_d=v_u ≠ 0\$$" sounds like a closed loop behavior, since it states that the input changes according to the output.

Moreover, also with the idea that $$\v_d\$$ takes a close-to-zero value so that $$\v_u\$$ gets to a non-zero value, $$\v_d=v_+ - v_-\$$ is an input, so I can give them whatever value I want; if I use an open loop op-amp as comparator, $$\v_+\$$ and $$\v_-\$$ can have significantly different values, so $$\v_d=v_+ - v_-\$$ wouldn't be equal (or tending) to zero.

I understand I'm missing something in my basic understanding of how an op-amp works, but despite looking on several sources I wasn't able to find what I'm missing.

• I understand why vd=0 is true in closed loop, but I'm not sure why it is in open loop. In open loop, vd isn't zero. The closed loop (with negative feedback) makes vd = 0. Without a loop (open loop) that will not be the case. if I use an open loop op-amp as comparator Be precise in your wording, an opamp by itself has no feedback, there is no loop. So you can simply say: "if I use an opamp as a comparator". Any loop is made by the circuit. The opamp simply (ond only) does: $v_u=A ⋅ v_d$ that's it! (I'm ignoring limited output voltage range etc.). Sep 23 at 14:12
• if I use an open loop op-amp as comparator, v+ and v− can have significantly different values, so vd=v+−v− wouldn't be equal (or tending) to zero. Yes, that is correct. In order to have vd = 0 (more precise: almost zero) you will need to add a feedback loop with negative feedback around the opamp. Sep 23 at 14:18
• electronics.stackexchange.com/questions/441184/…
– G36
Sep 23 at 14:21
• All-in-all I think that you do understand correctly how the opamp works: $v_u=A ⋅ v_d$ just realize that the differences occur when you're using that opamp in a circuit. Negative feedback, positive feedback (search for: "Schmitt-trigger") or no feedback (comparator): it doesn't change the behavior of the opamp itself. The behavior of the complete circuit does change! Sep 23 at 14:24
• Mauro, you should always have in mind, that Vd=0 is an assumption/simplification only, which allows us to use simplified formulas with an error that is very small in comparison to all other uncertainties (parts tolerances, etc.). In reality, Vd is always finite (often in the µ-Volt range).
– LvW
Sep 23 at 14:42

I understand I'm missing something in my basic understanding of how an op-amp works, but despite looking on several sources I wasn't able to find what I'm missing.

I believe you are just confusing yourself. The simplification that you set the difference voltage, $$\v_d\$$ to zero, then solve for the output voltage, $$\v_u\$$, is only for stable, closed-loop circuits. If you don't know ahead of time that your circuit is closed loop, and believe ahead of time that your circuit will be stable, then it's not a simplification -- it's an obfuscation and a complication.

For the idealized model of an op-amp where you take the amplification factor $$\A\$$ as being (or tending to) infinity, the open loop behavior is just

$$v_u = \begin{cases} V_{SS} & v_d > 0 \\ V_{DD} & v_d < 0 \\ \text{¯\_(ツ)_/¯} & v_d = 0 \end{cases}$$

NOTE!!

The whole $$\v_d = 0\$$ assumption is, as you are finding, very narrow. It is exceedingly handy if you are working on simple circuits. I've been designing circuits for 40 years now and I use it all the time. But using it is like dancing on a mile-high bridge without guardrails. It doesn't predict stability, it doesn't work if your precision requirements exceed the capabilities of the op-amp, it doesn't help you determine if your op-amp's gain bandwidth product is big enough, etc., etc.

So -- use it where you can, understand it (and its limitations), but if it's confusing you you're probably using it in the wrong situation.

• Maybe I understood what I was missing: what just struck me is, we know $v_d$ is an input in an open circuit, so I can't read $v_u=Av_d ⇒ v_d=\frac{v_u}{A}=0$ as "Since $A=∞$, then $v_d=0$", but rather as "If I want a finite (but undefined) $v_u$ I have to input a $v_d$ equal/tending to 0". I can give $v_d≠0$, that just means $v_u→∞$ (or, rather, to $V_{DD}$/$V_{SS}$ due to saturation). I think what I was missing is the "finite output" part: $v_d=0$ isn't a consequence of $A=∞$, it is a requirement for a finite non-saturated $v_u$. Is my understading right? Sep 23 at 15:55
• It's right from a theoretical perspective. In practice it's meaningless, because there's no practical use of an op-amp in open loop except when you're using it as a comparator (which may or may not work, depending on the particular part). In fact, when someone goes to measure an op-amp's "open loop" parameters, they do it by putting the thing in a closed loop circuit, applying a signal, then measuring outputs and inputs (which are non-zero, because for a real op-amp $A$ is finite, and there's offset, etc., etc.) Sep 23 at 17:59
• Sure, I understand; I was asking to check my understanding of the theoretical part, I understand that the practical applications differ. Thanks! Sep 23 at 18:10