P
Phil Allison
Guest
<jalbers@bsu.edu>
** 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
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?
** 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