50 Hz oscillator based on a LM324N IC

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Just a bit more about unused sections of chips: if you just leave them unconnected, as cowboybob says they can misbehave and do any of the following.
(1) do nothing.
(2) float around randomly at a low frequency.
(3) oscillate, often at a high frequency and do no harm
(4) oscillate at a high frequency and take excessive supply current
(5) oscillate and destroy themselves
(6) oscillate and destroy the entire chip.

In a company that I worked for, one pin left open circuit on a board in a big system, cost £25,000UK and delayed the program two months by the time we found the fault.

The general rule is that all inputs on a chip must be put into a safe, stable and defined state. Outputs can nearly always be left open circuit though.

End of lecture.

spec
 
Hy iods,

Neat schematic and the LT1007 is a nice opamp, but the LT1007 will not work with a single 12V supply line; the LT1007 needs +-15V (30V)

An ideal chip would be an OPA192 or OPA197

R5 needs to be 100K, rather than 200K, for 50Hz frequency, .

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well I would have used LM324 but I couldn't find a close opamp in LTspice , and about the R5 , if I use 100k I'll end up with 100Hz no ?

btw , thanks
 
isnt F=1/t ? .. in that case we get F= 1/0.02 = 50Hz ? no ?

The formula for the frequency with that type of oscillator (integrating oscillator) is complex because it is affected by the level that the opamp output voltage swings to, but the LM324 output swings pretty close to the supply rails.

The best type of opamp for an integrating oscillator is a rail-to-rail output (RRO) type, like an OPA192.

I am too lazy to work out the exact timing so I just use:

period = 1/f, which for 50Hz is 0.02 seconds (20ms). A half period is thus 0.01 seconds (10ms) so make C *R = 0.01 seconds.

And on the circuit, 100K * 100nF = 0.01 seconds (10ms). And then adjust the resistor to get the exact frequency that you want from the circuit when it is built.

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Thanks man
 
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