1
\$\begingroup\$

I'm working through an old Popular Electronics article, trying to apply and learn some of the math behind the project. Math for the clock pulse generator frequency has me stumped; wondering if anyone can provide a hint or might know what I'm doing wrong.

According to the description, the clock pulse generator should generate a square wave with a period of 8.1 microseconds or 122.8 khz. It uses part of a 4049 CMOS hex inverter, and the project runs on 9VDC; this feeds a counter chain which provides other multiples of the clock. In the diagram below, C1 is 100pf, R1 is 100k, R2 is 15K, and R24 5K variable/pot. Datasheets and other online discussion points to the formula as freq = 1 / 2.2RC, which I find gives me about twice the frequency I'm expecting here (2.2 x 17.5k x (100 x 10^-12)?). Am I trying to apply the wrong formula? Thanks.

Edit: I have not built the project, just working on paper.

Partial circuit diagram - oscillator

\$\endgroup\$
  • \$\begingroup\$ What formula are you trying to apply? \$\endgroup\$ – user207421 May 24 '15 at 23:28
  • \$\begingroup\$ Formula I tried: frequency = 1 / 2.2 x R x C \$\endgroup\$ – lrom May 25 '15 at 0:58
0
\$\begingroup\$

I think it's closer to 2.53*RC than 2.2*RC (Ref RCA ICAN-6267 - note that the complex formula (2) in that app note is wrong, but the test results are usable), but the formula you have should be basically correct for low frequencies, relatively high resistance and R1 > 2*(R2+R24).

The resistance is probably still low in comparison with the output resistance of the 4049 (probably < 100\$\Omega\$), but the input capacitance of IC1A is 15pF typically, so there's another RC in there (charge and discharge) with the 100K which will cause it to oscillate much more slowly.

Basically the equation falls apart at high frequencies. If you re-did the test at ~1kHz with C = 100nF you'd have much closer agreement with the simplified formula.

\$\endgroup\$
  • \$\begingroup\$ Thanks for the reply - sorry I can't vote this response up yet. I'll look for a copy of that application note online; thanks! \$\endgroup\$ – lrom May 25 '15 at 1:33

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.