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# Need help with signal generator 5kHz and 8.7 kHz

Hi,

I'd like some help. I'm trying to make a simple signal generator that will produce a 5kHz and an 8.7kHz square wave signal. I have some basic (very) electronics skills, and know how to follow a schematic and solder things up. The ones I've found online are outdated, using obsolete IC's.

Any help would be appreciated.

Thanks in advance.

If a 555 will operate at that frequency (which I suspect it will) then the ubiquitous 555 astable mode circuit is a simple answer. If you need extreme precision, thermal stability etc. then you will need some more advanced components- perhaps dividing the output of a high frequency crystal oscillator would do it.

What purpose do you need it for (control electronics, radio stuff, audio?) and what are these obsolete ICs in the designs you have found?

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Hey, thanks.

The 8038 is obsolete.

So, I was looking at signal generators that I could find online. Here’s a project that I found using a 555 timer chip:

http://talkingelectronics.com/html/SquareWaveOsc.html

It has six choices for output frequencies. I only want two. Well, three if you count “off”. :)

As far as doing the math to know the frequencies to generate a 5kHz and 8.7kHz signal, that’s a bit out of my league (though I’m willing to learn). Obviously, I don’t have an oscilloscope. Nor would I know how to use one.

So, I’ve tried to modify the parts list a little. Can you let me know if I’m on the right track? Instead of the six caps they list for the signals, I was thinking of using two. One for 5kHz, and one for 8.7kHz.

The substitution of a 5000 pf cap to generate the 5kHz signal, and a 2300 pf cap for the 8.7kHz. Would that be right?

Trouble is, I can’t seem to find a cap in the 2300 pf range. Do they make such a beast? Can I substitute two to make up the difference? Or how does it work? What can I do? What I need ideally is a schematic.

Sorry to be a newb at this.

Thanks again for the help you have provided, and thanks in advance for any more you’d be willing to offer.

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Sorry to be a newb at this.No trouble at all- you have shown you are willing to try and work things out for yourself, which is more than 90% of the people asking for help in electronics will do (mostly they just ask you to provide schematics for them).

The substitution of a 5000 pf cap to generate the 5kHz signal, and a 2300 pf cap for the 8.7kHz. Would that be right?You're on the right lines- I don't know where you got those numbers from, though. You seem to have realised the fundamental part of this circuit, that doubling the capacitance will halve the frequency and vice versa. From that you can work out that if a 1nF capacitor generates a 10kHz wave, a 2nF capacitor will generate 5kHz. Getting 8700 Hz is a bit more difficult, so for that I used the equation for frequency from the Wikipedia page.

This gave a value of 2062 pF for 5000 Hz, which more or less matches the quick calculation from earlier, and 1185pF for 8700 Hz. You could put these capacitance values into their schematic and get the frequencies that you want out.

Trouble is, I can’t seem to find a cap in the 2300 pf range. Do they make such a beast?Picofarads aren't exactly a "beast", being one thousandth of one billionth of a farad :) You can certainly get roughly that capacitance, but finding the precise value you want can sometimes be tricky. Close substitutions might be ok depending on how exact you need the frequency to be.

You can combine capacitors to make different values, but they are a little bit confusing as they work the opposite way round to resistors- putting two capacitors in series lowers the capacitance, and two in parallel add their capacitances together.

Hope that all helps!

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Wow, you're awesome, thanks so much.

I got the numbers not from a formula, but just from deduction, which is why I wanted to check with someone who knows what they are doing.

I'll try a 1200pF cap. That looks like it would be fairly close, no? I don't think the frequency would be too far off from what I need it for. The other frequency is more important for my application anyway.

Thanks again for your help.

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