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Originally Posted by Manoel
The square wave you referred to
does not have a DC offset. You can get technical and define the DC offset as the constant term in the Fourier series below or just look at the waveform generator and check how far you got the "DC Offset" dial from zero 
Anyway, you are really using a 5V pulse train and my initial concerns were pointless. Which leaves me without ideas of what is going on. 
Could you possibly hook the waveform generator to an optoisolator just to make sure there's nothing else interfering? |
A couple of points here. 24 years ago, when I got my EE, a DC offset was defined as you stated. It still is today. In real world applications, a DC offset can be determined a couple different was, running it through the Fourier Series is one of the most time consuming and accurate ways. Looking at the dial on the signal generator is the quickest, but least accurate. Measuring with a scope or DVM is the most accurate and fairly quick.
When I asked about th offset, the reason was that an offset of 2.5vdc can prevent a circuit from ever seeing a transition from high to low or low to high.
Here's why. Lets assume that the interrupt, or digital, input on the mini-rc identifies a signal as a high when it is above 4.25vdc and a low as below .5vdc. Now a 5v p-p signal with a 2.5vdc offset is running from 2.5 to 7.5 vdc. So, it never sees a level below .5vdc and therefore never sees a low. On the other hand, if the offset is -2.5vdc, the signal runs from -2.5vdc to +2.5vdc, so the input never sees a high. In either case, the pulse train never cycles the input high and low so that neither the rising nor falling edge can be detected.
If, in fact, the signal at the input is connected as Mike indicates and the signal is running from 0 vdc to 5vdc, then I too am at a loss as to what is the problem.
BTW, I am not building this, Mike D. is.
Mike,
I was also wondering if you could just run the signal to a small transistor circuit that just short the signal input to gnd, and released it back to 5vdc instead of running the signal generator directly to the input. This would eliminate any problem with the signal generator being able to sink the needed current to pull it low. (This shouldn't be an issue, but I have seen stranger things) It will also isolate it like the opto that Manoel suggested. In fact, the opto isolator would be easier to implement.