Op Amp

I understand pretty well how an op-amp operates at small signal.
You need to bias the op-amp in order to set the transistors at their constant current region, to achieve max output voltage swing, and to eliminate DC voltage/current at output.

Once these are set, you can calculate the small-signal gain of the amplifier.

However, how do you know what is the large-signal gain of the amplifier?

Moreover, how can you tell the DC input and output resistances of the amplifier?

The only thing you need to do is ensure that there is a path for dc current that can be sourced or sinked by the opamp input.
Rin depends on the specific circuit.
And Rout is very low thanks to negative feedback
Rout=Rout_open_loop/(1+Aol*K)
Where
Aol is a open-loop gain of a opamp (see datasheet)
K - feedback factor ( feedback gain ), sometime we use letter "β" for feedback gain.
http://forum.allaboutcircuits.com/showthread.php?t=24971&highlight=loop

And for example this circuit


Rin≈R5; Au=Vout/Vin=1+(R2/R1); F_cut-off=GBW/Au
GBW - Gain bandwidth product (in the datasheet )
And low corner frequency depends on C1 and R5, C3 and R1, C4 and Rload.
And ideal opamp has
Rin=∞; Rout=0; Aol=∞;

alphacat said:

Once these are set, you can calculate the small-signal gain of the amplifier.
However, how do you know what is the large-signal gain of the amplifier?

Click to expand...

What is the difference between small-signal gain and large-signal gain.
In opamp large-signal gain is Aol
http://forum.allaboutcircuits.com/showpost.php?p=173559&postcount=9

Thank you very much Jony.

Rin depends on the specific circuit.

Click to expand...

Well, the small-signal input resistance of a BJT-based Op-amp is 2*rπ.
How do you calculate the large-signal input resistance?

And Rout is very low thanks to negative feedback
Rout=Rout_open_loop/(1+Aol*K)

Click to expand...

How do you reach the large-signal Rout_open_loop?
I know how to calculate the small-signal Rout_open_loop but not the large-signal one?

What is the difference between small-signal gain and large-signal gain.
In opamp large-signal gain is Aol

Click to expand...

You are right, they are the same, my mistake.
Do you calculate the large-signal gain by calculating first the small signal gain and placing ω=0?

When we design circuit with opamp we usually treat them as ideal opamp. And if we want to know Rin or Aol , Rout wee look to the datasheet, we don't need to do any calculations .
For example for old op amp LM741

http://pdf.elenota.pl/pdf/Fairchild/lm741.pdf
Rin=250KΩ (Figure 2.Input Resistance and Input
Capacitance vs Frequency)
Rout_open_loop=100Ω (Figure 1. Output Resistance vs Frequency)
And Aol=15000[V/V] ω=0
http://www.national.com/an/AN/AN-A.pdf
http://users.ece.gatech.edu/~mleach/ece4435/chap02.pdf

And can you tell me the difference between small-signal vs large-signal resistance ?
For me there is no such thing as large-signal resistance.

Thank you very much Jony!
I got it now.
My problem was that i didnt know there was no difference between Rin, Rout, Aol in small signal (for low frequencies) and in large-signal.

Theoretically, in large signal analysis, I can draw a 2*rπ resistor between the op-amp's input pins, in order to calculate DC voltage across the circuit?

alphacat said:

Theoretically, in large signal analysis, I can draw a 2*rπ resistor between the op-amp's input pins, in order to calculate DC voltage across the circuit?

Click to expand...

Well, yes but we never do this. We treat op-amp as black box and assume that:
Rin=∞, so input current is 0A. (No current is flow to the op-amp inputs)
Rout=0Ω
And the open loop gain is infinite large.
And with this assumptions we get that in linear region non-inventive input voltage "V+" is equal inverting input voltages "V-" and we call it "virtual short". And that is why is this diagrams V+=V-




http://focus.ti.com/lit/an/slod006b/slod006b.pdf

Got it!

Thank you very much!

You taught me very important things.