Opamp voltages V(P), V(N) and V(out)
Vcc=+15 V (for t >0)
VCC=-15 V (for t>200µsec)
2.0V
V(out)
1.0V
V(P), V(N)
0V
V(N)
V(P)
-1.0V
-2.0V
-3.0V
-4.0V
0s
50us
V(N) V(P) V(A)
100us
150us
200us
250us
Time
300us
350us
400us
450us
500us
Detailed view of the voltages at t=200 µsec
(switch-on of the negative supply)
0.9V
V(P)
0.0V
V(N)
-1.0V
V(out)
-2.0V
-3.0V
-4.0V
199.617us
200.000us
V(N) V(P) V(A)
200.500us
201.000us
Time
201.500us
202.000us
202.478us
Description/Explanation
In the following, it is assumed that both power rails are not switched-on
simultaneously (time difference 200µsec). This is in accordance with real
conditions.
1.) t<200 µsec
The opamp is not well biased because of single-supply (no linear operation).
V(out) is positive and, thus, also V(N) and V(P) are positive. The voltage V(P) is
rising (capacitor charging) and then discharging (parallel resistor). Thus,
V(N)>V(P) at t=200µsec.
2.) t>200 msec (switch-on of negative supply)
After the second power rail is connected the opamp returns to linear operation.
Because of V(out)>0 and V(N)=V(out)/3>V(P) the ouput voltage V(out) jumps to a
negative value (in our case –4 V). This causes the voltage V(N) also to become
negative. Now, because of V(N)<0 the output voltages V(out) and V(N) return
very quickly back to a positive values.
Therefore, we have a rather large but very short negative impulse, which causes
the WIEN network and V(P) to react with the impulse response of a passive RC
bandpass.
3.) Response of a 2nd order RC bandpass (WIEN network)
The graph shows the impulse response of V(P). Because V(out) has the same
shape the WIEN network will be excited in a closed loop (as a 2nd run) with this
response function. This is repeated again and again. As an example, the 2nd
curve in the graph is the output after the 4th run. (The various runs are
approximated by a cascade of identical WIEN circuits isolated with ideal buffers
(gain of 3).
4.) Result
The sinusoidal output voltage of the osciilator is generated because of the
positive feedback effect that causes a superposition of the various contributions.
50mV
0mV
-50mV
4th run of V(P)
-100mV
Impulse response V(P)
-150mV
-200mV
-250mV
0s
20us
V(C6:2) V(R16:1)
40us
60us
80us
100us
Time
120us
140us
160us
180us
200us