# The way an op amp goes about doing its magic ???

Discussion in 'Electronic Basics' started by [email protected], Aug 6, 2008.

1. ### Guest

I have a few questions about feedback. For discussion consider an
inverting op amp configuration with R2 feedback @ 4K ohms and R1
(connected to V-) @ 1K at 1K ohms. The open loop gain to be 200,000 .

Assume that Vin is at 2 volts, according to my calculations using Vout
= (V+ - V-)*200,000 and V- = (Vin-Vout)*R2/(R1+R2)+Vout, V- should be
around .0003999 volts and Vout should be around -7.9980005 volts.

My question is in what fashion does Vout get to -7.9980005? I know it
happens but it seems like magic. I have looked at the schematic
diagram of a 741 for example. It will probably take decades before I
would be able to understand something like that. Is it instantaneous
or does the op amp spend some time hunting for this value? If it
hunts, does it swing low to the negative rail and move towards the
upper rail or do something else?

I did some experimenting using excel trying to iterate between Vout =
(V+ - V-)*200,000 and V- = (Vin-Vout)*R2/(R1+R2)+Vout to see if the
thing will somehow converge. Here are some examples:

Assume that the op amp guessed the output was -8.1 making V- = -.02
(Voltage Divider), but this would mean that Vout should be 400
(amplifying the difference between V+ and V- , ....

Vout V-
-8.1 -0.02
400 81.6
-1632000 -326398.4
6527968000 1305593602
-2.61119E+13 -5.22237E+12
1.04447E+17 2.08895E+16
-4.1779E+20 -8.3558E+19

Assume that the op am guessed the output was -7.9

Vout V-
-7.9 0.02
-400 -78.4
1568000 313601.6
-6272032000 -1254406398
2.50881E+13 5.01763E+12
-1.00353E+17 -2.00705E+16
4.0141E+20 8.0282E+19
-1.60564E+24 -3.21128E+23
6.42256E+27 1.28451E+27
-2.56902E+31 -5.13805E+30
1.02761E+35 2.05522E+34

Vout for both of these does not seem to converge on anything??? I
don't think that clipping off Vout to stay within the positive and
negative rails would help.

How does an op amp do its magic?

Any help would be greatly appreciated. Thanks

2. ### DJ DelorieGuest

For R1=1k and R2=4k inverting...

V- = Vin + (Vout-Vin)/5

V- = (Vin*4 + Vout)/5
= Vin*4/5 + Vout/5

The op-amp will try to make V- = 0, resulting in:

Vout/5 = -Vin*4/5
Vout = -Vin*4

Starting from "off", though, the instant you apply +2v to Vin, this
happens:

Vout = 0
Vin = +2
V- = (2*4+0)/5 = 1.6v

The op amp will try to do this:

Vout = (V+ - V-) * 200,000

I.e. Vout = -320,000 volts.

HOWEVER, it cannot do this instantaneously! The output voltage will
gradually (perhaps over a few nano- or microseconds, depending on the
op amp) change from its old value (0v) to the new one (-320,000).
I.e. it will start going negative. In this example, let's use 1V/nS
as the fastest the output can change.

AS IT CHANGES, the V- value changes also. Which changes the op amp's
desired Vout. This is how "feedback" happens. Let's say a nanosecond
later, Vout is at -1v. V- is now (2*4+(-1))/5 or 1.4v, and the op amp
wants Vout to be -280,000 volts. Continuing on the 1v/nS trend:

nS Vin V- Vout(desired) Vout(actual)

-1 0 0.0 0 0
0 2 1.6 -320,000 0
1 2 1.4 -280,000 -1
2 2 1.2 -240,000 -2
3 2 1.0 -200,000 -3
4 2 0.8 -160,000 -4
5 2 0.6 -120,000 -5
6 2 0.4 -80,000 -6
7 2 0.2 -40,000 -7
8 2 ~0.0 -8 -8 (approx)

Now, if there's any stray capacitance or inductance in your circuit
(likely), the output value may overshoot, just like any other signal.
If it overshoots, the op amp will reduce its output accordingly, etc,
until it stabilizes. Hopefully the overshoot will be tiny and
stabilize quickly.

The open loop gain tells you how close to ideal the op amp is. For an
open loop gain of 100, for example, the V- would have to be 0.08 to
get a Vout of -8.0. You'd need a slightly more positive Vout to do
this, so the actual Vout is more like -7.62:

Vout = - V- * 100
V- = Vin*4/5 + Vout/5
Vin = 2

V- = -Vout / 100
V- = 1.6 + Vout/5

-Vout/100 = 1.6 + Vout/5
-Vout = 160 + Vout * 20
-Vout * 21 = 160
Vout = - 7.62
V- = 0.076

For a gain of 200,000, the result is Vout = 7.9998 and V- = 0.00004.

3. ### Andrew HolmeGuest

It takes time. Vout can't change instantly. The maximum speed at which
Vout can change is called slew rate and is specified on the op-amp
datasheet. If your Vin changes, the resultant differential input voltage
(V+ - V-) causes Vout to start moving in the right direction. It actually
overshoots before finally settling at -8V or whatever. The dynamics depend
on factors such as bandwidth and phase shift as well as slew rate.

4. ### Phil AllisonGuest

** You can imagine an op-amp (with negative feedback applied) to be a
creature with a one track mind. It has just one goal which is to keep the
voltage levels at the V+ & V- inputs exactly the same. It has only one way
to do this which is to vary the voltage on its output.

In your example, the op-amp was initially in a stable condition with input
and output sitting at 0 volts. Then you suddenly apply +2 volts to the 1k
resistor and instantly the voltage at the V- input rises to +1.6 volts,
since the 1k and the 4k act as a simple voltage divider.

The V+ input remains at 0, so there is a big drive signal to the V- input to
the op-amp causing its output voltage to swing as fast as it can in the
negative direction. The speed of this swing ( in volts per micro-second) is
limited by the need to charge a small capacitance inside the op-amp with a
restricted source of current.

As the output swings negative, the voltage at the V- input reduces from +1.6
volts towards the desired value of 0 volts. When the output reaches -8
volts, the V- input will be restored to 0 volts, matching the V+ input and
the op-amp is stable again.

Using your figure of 200,000 for the DC gain of the op-amp, the actual
voltage at the V- input will then be 8 / 2exp5 = 40 micro-volts.

...... Phil

5. ### EeyoreGuest

Open loop gains are approximate, average, not accurate values.

It's called negative feedack and works with all amplifiers. That's what
you need to study up on.

Graham

6. ### neon

1,325
0
Oct 21, 2006
a 741 whatever open gain it has [not all are the same ] the gain=A=input resistance devided by the F/B resistance. that is the basic the + input need to see the same resistance to eliminate some unbalance on the input. the you may not use 100mega homs as F/B why the offset current will saturate the sucker. as point out before there is a slew rate to contend with v/msec. and there is also storage delay if sat. all these things are in the spec read and understand  