It’s circuit from a client, it’s supposed to generate a sine wave.U5 pin 1 has no DC path for the bias current so its output will end up at one of the rails.
Since U4B has unconnected wires I have no idea what it does.
Where did you get that circuit?
That makes sense, thanks!
LOL I like how you put it, thanks for the reply!If you take U4b and turn it through 180° along with the wire on its output, it will drop in place and make an amplifier with a gain of 11. A rather pointless amplifier as it is followed by an attenuator which reduces the gain by a factor of 55.
That leaves us with a low pass filter with a cutoff frequency of 143Hz.
You don’t say the frequency of the squarewave, but the output won’t be a sine wave, it will be a squarewave with the edges rounded a bit.
If your client thinks it will make a sine wave, then he’s the sort of client to avoid.
With positive feedback as shown in that attachment shown in post #7, U4b will provide a fair squared wave, as it goes from one saturated state to the other.That makes sense, thanks!
Yes, the 4018 will work rather well, BUT it is not a binary counter and so the resistors must all be the same value. That makes it easier to set up and also educes the glitches that would be created by a binary ripple counter, such as the 4020, 4024, or 4040. So really the 4018 is the better choice. What will be important is to properly terminate all of the unused inputs.Swap the 4017 for a 4018 and with four resistors to a summing op-amp, followed by a filter it will make a very nice sine wave.
2:1 gives Butterworth, equal values give critically damped.The antique TDA2003 output amplifier has been obsolete and not available for a number of years
The Q of the lowpass filter is too high. When the opamp has a gain of 1 then the feedback capacitor value should be double the capacitor to ground capacitor value.
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