can't solve this voltage measurement problem


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  1. #1
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    You haven't mentioned the frequency of the ripple. Is it AC mains? If it is, 0.1uF is too small a capacitor to smooth that ripple. I would say, a safe starting value should be in the 10s of uF. You also mention having a half wave rectifier prior to the measurement, so, you should be able to smooth it real good if you increase that capacitor.

    Can you post a schematic? It might help understand the problem better.

  2. #2
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    The AC ripple on the DC is about 5VAC and I have little experience with solving AC ripple problems.
    Christopher4187 the attached ripple chart could help you in choosing the right capacitor for your specific problem. (Mind the chart is for 60 Hz, for a different frequency you have to adjust it.).

    Al.
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    Last edited by aratti; - 11th October 2009 at 17:10.
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  3. #3
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    Jerson,

    I've attached a quick drawing of the circuit. The problem is, I don't have the piece of equipment with me but if I remember correctly, the frequency is 500Hz but it is very erratic. Not only that, not all pieces of equipment have this problem because some are new chargers and some are old. It's not AC mains, it's the output of the battery charger.
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  4. #4
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    Try something like this
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    Thanks Jerson. I realized I drew it wrong because I had the diode on the input to the voltage divider. If not, my measurement would be WAYYYY off!!! I'll try the increased cap size. I didn't put it at the input to the voltage divider because the size would have to be much bigger. How important is it to have the cap before the voltage divider instead of after?
    Last edited by Christopher4187; - 11th October 2009 at 18:44.

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    It also wouldn't hurt to take an average. Since 60 Hz has a period of 16.66 milliseconds, and an A/D conversion takes about 30uSecs, you could average 16 samples (a full cycle) something like this:

    Code:
       ADCAvg = 0            ; Clear the variable
    For X = 1 to 16
       ADCIN,x, ADCresult
       ADCAvg = ADCAvg + ADCresult
       Pauseus 1000  
       Next X
    ADCAvg = ADCAvg >> 4    ; divide by 16
    This should do a pretty good job of smoothing the result.
    Charles Linquist

  7. #7
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    I've tried all kinds of software routines but nothing seemed to help. I think the AC ripple is not a normal sine wave, maybe that's why.

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