Peter said:
I'm using an ideal op-amp in MultiSim, 0.022uF cap and a 100kohm resistor
with the non-inverting input grounded.
I have three questions about how an intergrater works.
First question: If DC voltage can't pass through a cap, how come the
op-amp
out put doesn't hit the rail instantly? It slowly decends negative (per
the
intergrater formula) and once it hits the rail (in my case -21volts), the
RC on the input starts to rise at a rate of T=RC.
Second question: Why does the RC start once the op-amp output hit the
rail?
Third Question: If I put a 50pF capacitor on the inverting input to ground
(between the 100k and the 0.022uF), this changes the ramp constant very
slightly.... how is that 50pF calculated into the circuit and what is this
providing for the circuit's function?
When you apply a small positive voltage to the input of the resistor, that
doesn't mean the input to the op-ap 'sees' the same voltage. There is a
voltage drop across the resistor because there is current flowing through
it.
Now, an *ideal* op-amp has an infinite input impedance. So there is no
current flowing between the inverting input and the junction point
(resistor, capacitor and inverting input to the opamp). But there *is*
current flowing to the capacitor.
Pretend for a moment the opamp isn't there and the output side of the cap is
just tied to ground. When you apply a signal voltage to the input of the
resistor, current begins to flow through the resistor to charge up the
capacitor. Simple series RC circuit. As the capacitor charges, the voltage
at the point between the resistor and capacitor would slowly rise and
eventually equal the input voltage.
But now put the opamp back into things. As the voltage at the
resistor-capacitor tie *tries* to rise above ground, the op-amp, sensing a
rising voltage on its inverting input will slew its output negative. So as
the capacitor charges, and the voltage across it rises, the output side goes
negative by exactly the same amount as it charges. This keeps the inverting
input 'virtually' at ground potential. (in the ideal case anyway, in real
opamps it will shift very slightly).
As more charge builds up on the capacitor (owing to a fixed amount of
current flowing in from the applied input signal and resistor), the opamp
keeps slewing its output negative more and more in order to maintain the
inverting input at 'virtual ground'.
All this time, the voltage across the capacitor is rising at a *linear* rate
(not the exponential rate often associated with RC networks). Once the
output 'hits the rail', the cap continues to charge. But now the voltage at
the inverting input can no longer be maintained at 'virtual ground' and it
begins to rise. Also now that we've 'hit the rail', the charging current
into the cap begins to decay (the voltage between resistor input and the
junction point is slowly decaying).
With just this simple a circuit, once the integrator 'hits the rail', and
the inverting input voltage rises above 'virtual ground', the circuit is
sometimes said to be 'winding up'. Reversing the input signal's polarity
will *not* cause the output to start to integrate right away. It will
remain at the negative 'rail' for some time while the charge is gradually
bled off the cap. Once the charge on the cap has bled down to the point
where the junction is at ground potential, *then* the output will move from
the 'rail' voltage as the cap's charge continues to bleed down.
Various designs have been used to prevent or at least minimize this delay in
reacting to a change in input signal. In the process-control world, the
circuits are referred to as 'anti-windup' circuits. One simple method is to
place diodes across the two opamp inputs. This prevents the cap from
charging excessively and keeps the inverting input within 1/2 volt (okay,
one 'diode drop') of the ground potential.
hope this helps
daestrom
