S

#### ShamShoon

- Jan 1, 1970

- 0

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