I am using a ne555 configured as an astable multivibrator to measure the "wetness" in soil. The higher the resistance (the soil is dry), the lower the frequency coming out of the 555. This signal then goes to a lm2907 frequency to voltage converter that determines when the water pump should kick in. My problem is that the 555 outputs DC frequency and the 2907 wants AC.. Is there an easy way to convert the signal to AC or is there any other way to do this?
I attached an example from the datasheet that shows what I want to accomplish..
Thankful for any help!

/ Christian

 

The output of the 555 timer will go from nearly 0v to about Vcc-1.3v.

That should be OK for the LM2907 input.

 

The output of the 555 timer will go from nearly 0v to about Vcc-1.3v.

That should be OK for the LM2907 input.

So my output wont be DC? does nearly 0 V mean under 0 V, because to be AC it has to go from under to above GND right?

 

The input is to the noninverting input of a comparator or opamp. If the signal went below ground, it might damage the input.

It should work OK as it is.

 

I will try this setup, thank you for your help!

 

Does the spec sheet list input specs? From block diagram , if + input never goes below - input of "OP" amp, output will always stay high. If circuit does not work might try capacitivley coupling 555 to LM 2907 input via resistor voltage divider. Just founr this: AN162, " anothercircuit whose output may not go below ground. This may be remedied by using ac coupling....Fig 5b. Remember that the minimum input...is only 30mV pp, but must swing +- 15 mV minimum either side of ground." Might try Cap 5X 555 timing cap, in to ground same at timing resistor... all open to suggestions.

 

After looking at the datasheet again, the LM2907/LM2917 are limited to an input voltage of 0v to +28V, but the LM2907-8/LM2917-8 can take +/-28V.

If you are using an LM2907-8, then you can use a small capacitor (0.1uF would work well, NOT polarized electrolytic or tantalum) between the output of the 555 timer and pin 1 of the LM2907-8, and a 10k resistor from pin 1 of the LM2907-8 to ground. This will allow the edges of the signal to pass, but will keep the average voltage on pin 1 near ground.

 

now I am abit confused by all the information. Will the ne555 output get close enough to ground for the lm2907-8 to register?

 

now I am abit confused by all the information. Will the ne555 output get close enough to ground for the lm2907-8 to register?

Here, I edited the schematic you originally posted:

The output from the 555 timer gets connected to the 0.1uF capacitor. The 10k resistor keeps pin 1 centered around 0v. The positive pulses from the 555 timer get level-shifted by the capacitor and resistor network (in this case, known as a differentiator circuit) to positive and negative spikes.

Oh, just a note about the two 5k resistors that are shown in the original schematic: 5k Ohms is not a standard resistance value. However, the exact value is not as important as the two values being close to equal. You could use a pair of 4.7k, or a pair of 5.1k, or a pair of 10k Ohm resistors for that matter. Just as long as the two resistors are about equal in value, and the total value of resistance is somewhere between 2k and 50k, you will be fine.

 

Umm... By all means continue to work on your project but uhh... Isn't this a more simple solution?

http://www.amazon.com/Bulbs-Plant-W...ie=UTF8&s=miscellaneous&qid=1238177581&sr=8-3

 

CoolbeaNUmm... By all means continue to work on your project but uhh... Isn't this a more simple solution?

What's the fun of those things? This is a part of an "intelligent home project".. Plus it's good practice in electronics!

Thank you SgtWookie for your effort! I just found another datasheet where they discribed just that schematic (Ac-coupling right?). I will get back to this thread wether it's with good results or more questions

Thanks! / Christian

 

So.. it has been awile now sice I last wrote on this thread. Stil having some troubles getting the circuit to work..

I hooked the output from the 555 to an oscilloscope. the picture shows me a squarewave pulse going from 180mV to ~ 3,56 V with a frequency of 29,896 kHz.

My circuit (see attachement) has the following values:

C = 10nF
R = 180 ohm
and the capacitor in parallell with R is 100nF

According to the calculation in the datasheet load is energized when
freqin >_ 1 / 2RC

This means that my load should be energized with the pulse from the 555
since my 1/2RC = 22,727 kHz right?

I have uploaded two pictures that illustrate the picture from the oscilloscope, the first one is when the oscilloscope is connected directly to the 555 output, and the other one with the scope connected after the 0,1uF capacitor.

Still nothing happens.. =(

Any thoughts? =)

/ Christian

 

Now, which IC are you using again? Is it the 8-pin or 16-pin version?
[eta]
You're off on your calculation. With R=180 and C=10nF, F=277,777Hz, or 278kHz.
The cheap fix would be to increase R to 1.8k, which would give you about 27.8kHz

Note that capacitor values may vary quite a bit, perhaps 20%. You can adjust R to compensate. Use a 1.3k resistor in series with a 1k 10-turn trimpot to give you a range of roughly +/-30% adjustment.

 

I´m using the 8-pin version. I changed R1 to 2,2kohm which gives me 22,272kHz. So that should make it work, but it doesn't.. My Vcc is 5 volts.. could this have someting to do with it?

 

Look again at the schematic; the allowable range for Vcc is 6v to 24v.
5v does not fall within the allowable range.

 

In my schematic, there is a capacitor in parallell to the resistor R, what value should I put there?