Relaxation Oscillator - Understanding OpAmps

Hello,

I'm trying to construct and understand the Relaxation Oscillator
circuit in the Art of Electronics. The circuit simply has a voltage
divider connected to the positive terminal (with equal R's), and the
negative terminal is connected to a feedback resistor to the output
and a capacitor to ground.

Now I'm gonna have to start by saying that I constructed this circuit
but it didn't work. I could see no oscillations (except at the low
millivolt level). The only difference from AoE is that I used a single
supply instead of a dual supply. I used a 1k for the feedback R, 2x10k
for both R's in the voltage divider, and both 10uF and 10pF caps (it
didn't work with either). I used a TLE2144 for the OpAmp.

In analyzing this circuit, I could go two ways. One is to use the
OpAmp equation where the gain A is very large and Vout = A(V+ - V-).
This analysis (at the end of the message for those interested) gives
me the result that the output should be an exponential function of
time (certainly until saturation). However doing a static analysis, it
seems that at steady state, it's very easy for this circuit to
stabilize at Vout = 0, V+=0, V-=0. In such a case I don't really know
what's the incentive for the OpAmp to get out of this stable state and
start charging the capacitor, and I'm guessing that this stable state
is exactly what I'm seeing.

So what am I missing here? Does this not work with a single supply?
Does the response time of the Amplifier has to do with that? Do I have
to add some external excitation for the oscillation to start (I tried
that at different points in the circuit with no luck).

Plenty of questions. I would appreciate if you give me some
clarification to what I am doing (or understanding) wrong

Thank you.

My analysis for oscillator:

Rf is the feedback resistor to v-

Vout = A (V+ - V-)
V+ = Vout / 2 (from the divider)
hence, Vout = A(Vout/2 - V-)
so, V- = Vout(1/2-1/A)
and for very large A => V- = Vout/2 ----------- (1)
so now we established that both terminals have the same voltage.
Now, at the negative terminal:
(Vout - V-)/Rf = CdV-/dt
substituting from (1):
(Vout - Vout/2)/Rf = 1/2*CdVout/dt
so, CdVout/dt = Vout/Rf
and hence, dVout/Vout = 1/CRf dt
integrating this equation leads to exponential rise of Vout vs. t
.

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