# Relaxation Oscillator - Understanding OpAmps

S

#### ShamShoon

Jan 1, 1970
0
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

J

#### John Popelish

Jan 1, 1970
0
ShamShoon said:
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.

That is a big difference as far as the positive feedback
divider goes. In one case (dual supply) the divider reduces
the output voltage by half around a voltage half way between
the two supply rails, so voltage to the + input always stays
in the middle half of the range between the two supplies
(half way between zero and the positive rail to half way
between zero and the negative rail).

With a single supply (and with the negative rail being the
"ground" end of the divider), the divider reduces the output
swing by half, but biases the result in the negative
direction, so the result, though about half of the supply
rails, is offset negatively, from the negative rail to about
half way between the positive and negative rail.
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.

The opamp gain equation applies only when the two inputs are
almost exactly equal. This circuit should spend almost all
of its time with the two inputs at very different voltages
and with the output saturated against one or the other
supply rail.
So what am I missing here? Does this not work with a single supply?

It can, but you have to make one change in the positive
feedback divider. Double the resistance of the resistor
that ties the + input to the negative rail and add another
similar (double value resistor) between the + input and the
positive supply rail. That way, this pair of resistors
across the supply simulate a single resistor to a voltage
half way between the rails, as far as dividing the output
voltage in half for the + input.
Does the response time of the Amplifier has to do with that?

Not at all.

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