1
\$\begingroup\$

I was working on a layout for a guitar effect, and noticed on the schematic that 1 opamp has a 1N4148 diode on V+. IC6 on the schematic below.

Given it has an internal power supply, I'm assuming this is not for polarity protection (guitar effects often run off external power supplies, and diodes are commonly placed on the input of V+)

Q1) What is the purpose of this diode

Q2) Would it matter if it was not there

Thanks for any assistance available.

EDIT: SCHEM UPDATE, PREVIOUS INCORRECT

PHASOR II SCHEMATIC

\$\endgroup\$
11
  • \$\begingroup\$ It looks like there is an error in that schematic - IC6b pin 5 should not go just to a capacitor. That will not work. I suspect the capacitor should go to pin 6 and pin 5 should be grounded. \$\endgroup\$
    – Kevin White
    Commented Apr 8, 2021 at 21:30
  • \$\begingroup\$ @KevinWhite Agreed... except that there is feedback from the output back to IC6A, which I'm still looking over. \$\endgroup\$
    – jonk
    Commented Apr 8, 2021 at 21:33
  • 1
    \$\begingroup\$ Thank you - you are correct. I found another schematic, where 5 is grounded and the cap is across pin 6 & 7 \$\endgroup\$
    – Andrew Hopkins
    Commented Apr 8, 2021 at 21:47
  • 1
    \$\begingroup\$ @jonk - It is never good just have an input just going to a capacitor - it will charge up due to bias current until the opamp saturates. \$\endgroup\$
    – Kevin White
    Commented Apr 8, 2021 at 22:07
  • 1
    \$\begingroup\$ " I found another schematic" - where did you find it? \$\endgroup\$
    – Bruce Abbott
    Commented Apr 8, 2021 at 23:08

1 Answer 1

Reset to default
0
\$\begingroup\$

Q1) The diode is there because it drops the Vdd on IC6 by about a volt (whatever the Vt of the diode is). That means IC6.2 is only feeding a maximum of +8V into IC5.1

Q2) Will it matter? I suspect so, but define "matter". Guitar pedals like to throw diodes into random places because it causes or ameliorates "clipping" which creates non-linear effects on the output audio spectrum.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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