0
\$\begingroup\$

Square to saw converter

I am designing a square wave to saw wave converter by first integrating the square wave into a triangle wave, then toggling between the triangle wave and it's inverse, using 2 CMOS 4066 switches triggered by the original square wave.

The output is an incomplete saw wave because of the triangle wave going too negative relative to the 4066 supply voltage.

scope image

I know I could fix this by scaling and adding a DC offset to the triangle waves at the input of each 4066 switch. But for a number of reasons I want to keep the waveforms balanced with no DC offset. I also want to keep the circuit as simple as possible because it will be repeated many times in this project.

Can I power the with 4066 and 4069 with -12v to +12v so as not to cause this problem? If so what is a simple way to shift the logic level of the input to the 4069 to the right level?

\$\endgroup\$
  • \$\begingroup\$ There are FET analog switches that will take +/-15V. Or, run the 4066 off +/-6V and level-shift its control inputs. \$\endgroup\$ – Brian Drummond Sep 13 '17 at 13:32
  • \$\begingroup\$ Can I power the with 4066 and 4069 with -12v to +12v Look at their datasheets, for those made by TI it is a no as the maximum supply voltage is 20 V. Why not simply run your circuit from +/- 10 V, then you can supply all the ICs from the same supply lines and avoid the problem you have. For the CMOS logic ICs then a 0 will be -10 V and a 1 will be +10 V. \$\endgroup\$ – Bimpelrekkie Sep 13 '17 at 13:49
0
\$\begingroup\$

You can run the 4000 series parts off a bipolar supply but the voltage is limited (IIRC the limit is 20V total which works out to ±10V) and you need to level shift the input signal, this is perfectly doable with a comparator and a few resistors but it's still an extra part.

Or you can use a modern part like the ADG1219. It's a 2:1 mux rather than a plain switch so you don't need an inverter and it has a seperate pin for the logic ground so you can run it off a bipolar supply while controlling it with a unipolar signal.

\$\endgroup\$
  • \$\begingroup\$ I will try reducing the supply voltage. The ADG1219 is a lot more expensive than the 4066. I will am making 180 of these circuits for a polyphonic synthesizer. \$\endgroup\$ – John Spence Sep 13 '17 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.