Rectifying an AC signal

Discussion in 'Supporting Physics' started by JFKtech, Feb 3, 2020.

  1. JFKtech

    JFKtech and then there were two - now JFK Techs :)

    Sorry, another physics question :oops:

    AQA GCSE P5.1 playing with AC and DC and oscilloscopes. 'Your teacher may now demonstrate half-wave rectification of an a.c. supply, in which a diode is used to stop current flowing in one direction.'

    Well their teacher may, if we could make it work ! I tried with the recommended 1N4001 diode, power pack on 1V, variable resistor bringing current to minimum. Plugged it in, momentarily saw a beautiful trace, then the diode fizzled and the trace went back to a straight line. Tried again with a bigger diode (1N5401) and I just get a jagged version of the original AC waveform. Tried a (new) 1N4001 with the sig gen instead of a power pack, still just a squiffy version of the normal waveform.

    Please can someone tell me what diode I need, and what else to put in the circuit ?! Or is it another example of someone writing a practical that sounds good on paper, but not actually checking that it works ?
     
  2. Nick Mitchener

    Nick Mitchener COMMITTEE

    Won't you need a current limiting resistor to protect the diode? I would use a fixed resistor for that, far too easy to take a variable to zero. BTW we teach AQA GCSE and nobody has ever asked me for anything to do with rectification. We just provide AC hand gen and a battery to compare the two on oscilloscope.
     
  3. As Nick, you will have to make sure the current through the 1N4001 is less than 1Amp (or it will die)
    The 1N5401 has a higher current rating 3A IIRC, but will still need a current limiting resistor.
    If you use a SigGen be sure to connect the ground of the SigGen and the ground of the Scope. As they are probably connected to Earth at some point.
    T
     
  4. JFKtech

    JFKtech and then there were two - now JFK Techs :)

    oh, I forgot to say there was a 100 ohm resistor attached to the diode. I just put the variable resistor in because the sheet said I needed one, although it didn't actually tell me what to do with it.
    If you have the kerboodle resources, the bit I quoted is at the very end of the student method for P5.1 Only some of the teachers ever ask for it, and in the past we've always just told them it doesn't work - but I'm gradually trying to make all the 'doesn't-works' work, and it was requested again this week so I thought I'd try again.
     
  5. STEM1

    STEM1 Dave

    You could use a lamp to illustrate the reduced brightness with the diode in circuit. Ensure the lamp draws less than the diode maximum rating. Put the oscilloscope across the lamp. A full-wave rectifier gives a good trace too.
    A diode on it's own won't limit the current so you need a resistance or lamp in series to protect it.
     
  6. JFKtech

    JFKtech and then there were two - now JFK Techs :)

    I'm back to this again ! Please could you explain how I can work out what resistance I need to reduce the current from the 8.5A supplied by the power pack to the 1A required to not blow up the diode ? I've looked it up online, but can't find an answer :(
     
  7. STEM1

    STEM1 Dave

    Try a 12v lamp of less than 10 watts or so. Or a resistor, say 15 or 22 ohms rated at more than 10 watts. Wire it in series with the diode and look at the waveform across the lamp/resistor with the oscilloscope.
     
  8. Ignore that the powere supply can supply 8.5A, this is the maximum that the supply can happily give. It wont supply it unless the load resistance is small.
    Take the output voltage and divide it by the current you want to pass through the diode (less than 1A)
    For example 4V divided by 0.5A gives 8ohms (nearest higher std value resistance 10ohms) The resistor will dissipate 8 times 0.5 equals 4W
    As stem1 a bigger15 or 22 ohm lamp will reduce the current to 4 divided by 22 equals 0.18A.
     
  9. JFKtech

    JFKtech and then there were two - now JFK Techs :)

    This is the bit I was missing, thank you !!! I didn't know that, I thought the 8.5A was somehow set. So I had V=IR, but two values of I, and got thoroughly confused :oops: Why didn't they teach me anything useful in A level physics ?!
     
  10. I have put a capacitor across the CRO, if I remember rightly it cleaned up the incoming signal quite nicely