"Lewis Carson" wrote in message
...
Thank you all for the responses. But to clarify my original question,
is there anything wrong with the circuit below or not? Is the design
flawed or incorrectly implemented?

Thanks,

Lewis


On Sat, 31 Mar 2007 07:30:53 +1000, Lewis Carson
wrote:

I found this circuit online.

http://members.ozemail.com.au/~bisagoma/step_sine.jpg
I found this circuit online.

http://members.ozemail.com.au/~bisagoma/step_sine.jpg

And re-drew it as follows.

http://members.ozemail.com.au/~bisagoma/step_gen.jpg

But ... it does not appear to work as described. The input is a 512 Hz
clock pulse. The output on pin 3 of the CD4051 is a slightly noisey
plain squarewave divided by 32 (16Hz). IOW no steps.

Can anyone point out a design fault in this circuit or other reason
why it might not work?

Lewis


Someone already told you how to fix your drawing. You need to put the opamp, or
at least a load resistance, on the output of the mux. Otherwise, the resistor
chain is open-ended, no current is being passed through them, therefore, you're
seeing only the sum of the 4040 outputs, which are all CMOS level square waves.
Remember, the 4040 is an analog switch. It only makes a path from input to
output, it doesn't source or sink anything at the output except what is applied
to the inputs.

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer to the end, the faster it goes.