0
\$\begingroup\$

I need to add two 32MHz square wave. It is not possible using all op amps because of the bad frequency response. My question is is there any ic device or some technique to do so???

Update: Actually I am building a music synth. I thought of using a high source of frequency, higher the better, and divide it using counters to get the musical notes. The reason I wanted to add two high frequency wave is to get an octave kind of sound. But I just realised that I cannot divide the added signal using counter because it is not square. So, this question is pointless. I can make another counter circuit for the second wave and then add. I would not have a problem of the bandwidth also except the massive work of designing dividers for each notes- 12notes. Sorry guys for wasting your time. Thank you very much for replying.

\$\endgroup\$
3
  • \$\begingroup\$ What do you expect the output to look like? Two resistors will give you a scaled sum. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented May 9, 2015 at 14:35
  • \$\begingroup\$ Did you try looking for an op amp with an appropriate GBW? \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented May 9, 2015 at 15:07
  • 1
    \$\begingroup\$ A 32 MHz squarewave (any sq wave in fact) has harmonics rising to infinity. What is reasonable to assume is the highest harmonic your sq waves have? \$\endgroup\$
    – Andy aka
    Commented May 9, 2015 at 15:38

1 Answer 1

Reset to default
1
\$\begingroup\$

Sure it's possible - easy, even. Let's assume your source impedance (which you didn't mention) is < 10 ohms, and lets assume that your signal amplitude (which you also didn't mention) is < 5 volts, and that your load impedance (which you didn't mention) is > 1k. Then

schematic

simulate this circuit – Schematic created using CircuitLab

will work fine. You may (will) need decoupling caps and possibly a feedback cap for stability, but otherwise you should be fine.

\$\endgroup\$
6
  • \$\begingroup\$ I didn't read the whole datasheet, but on the front page the LT1226 only promises to be stable at A=25. \$\endgroup\$
    – The Photon
    Commented May 9, 2015 at 15:17
  • \$\begingroup\$ Oops. You're right. Check the edit. Thanks for picking up on that. \$\endgroup\$
    – WhatRoughBeast
    Commented May 9, 2015 at 16:25
  • \$\begingroup\$ Now you have gain of about 1.05 instead of 2. \$\endgroup\$
    – The Photon
    Commented May 9, 2015 at 16:27
  • 1
    \$\begingroup\$ That's what you get for doing engineering on the weekend. \$\endgroup\$
    – The Photon
    Commented May 9, 2015 at 16:28
  • \$\begingroup\$ Heh. Same to you, I think. With both inputs at 5 volts, divider voltage is 5 x (20/520) or .1923 volts. Noise gain is 1 + (1000/19.6) or 52. 52 x .1923 is 10. Or is my brain suffering from weekenditis? \$\endgroup\$
    – WhatRoughBeast
    Commented May 9, 2015 at 16:37

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.