Transformer voltage creeps up and up

Discussion in 'The Projects Forum' started by SharpyWarpy, Aug 6, 2007.

  1. SharpyWarpy

    Thread Starter New Member

    Aug 5, 2007
    5
    0
    Hi, everybody. This is my first post, hope it's not dumb.
    I cut the secondary windings off a microwave oven transformer and wound enough for my DIY uninteruptible power supply, I want between 13 and 14 volts. But with the transformer wound for 13 or 13.5 volts ( I can't remember the exact voltage, I changed the windings a lot ) when I hook it up to the battery/inverter the voltage jumps up to 14, 15, 16 volts. It made me think of a little voltage doubler circuit I played with a few weeks back, I think it's called a charge pump, so I removed the two 15000 microfarad caps I had in there for smoothing but this does no good. What is this phenomenon that has me pulling my hair out?
    Incidentally, I tried it without the inverter with the same result. Thanks ahead of time.
     
  2. spar59

    Active Member

    Aug 4, 2007
    51
    0
    Presumably the primary side of the transformer is connected to the mains and the mains voltage is stable.

    Are the measured secondary voltages a.c. , d.c. or both ?

    What I am getting at is :- is the problem a varying voltage i.e. not stable with the measurements being at different times or a stable but unexpected voltage with one measurement being the a.c. input and the other the d.c. output ?

    As you are probably aware if you convert a.c. to d.c. through a rectifier and smoothing capacitor arrangement the output d.c. voltage at no load is equal to the peak input a.c. voltage minus the rectifier forward voltage drops.

    So for a bridge rectified supply with 13V a.c. input the output would be approximately 13 * 1.414 - 0.7 -0.7 (assuming 0.7V forward drop in each conducting diode) = 16.98V d.c.

    Even with the smoothing capacitors removed there may be sufficient stray capacitance to get the voltage near this value if the only load is the test meter.

    Hope I have not stated the obvious.

    Steve.
     
  3. SharpyWarpy

    Thread Starter New Member

    Aug 5, 2007
    5
    0
    I didn't know about the formula you stated,
    13 * 1.414 -0.7 - 0.4
    particularly the "* 1.414" part. I'm not an electronics student, I've only just begun about a year ago on my own.
    Do I understand correctly that the capacitance, however there, is creating the voltage jump? Like a charge pump?
    Thanks very much for your reply, this could explain a lot of things for me that I have experienced lately.
     
  4. spar59

    Active Member

    Aug 4, 2007
    51
    0
    I'll try and explain - please post back if you require more info.

    The mains alternating current (a.c.) supply is a sine wave and has a frequency of typically 50Hertz (cycles per second) in the uk or 60Hertz in the usa. This means in the former case the voltage starts at zero gradually increases to a peak voltage following a sine curve in 1/200th of a second then drops back to zero in the next 200th of a second it then goes to a peak negative value in the next 200th of a second and then back to zero in the final 200th second of the cycle.

    So its value is always changing and this gives the problem of how do you define it. It was decided to do this by defining it as a voltage that would do the same amount of work as an equivalent d.c. voltage. This is called the root mean square or r.m.s. voltage if you connected a resistor across a 6V direct current (d.c.) battery or a 6V r.m.s. a.c. supply the same amount of heat would be generated in it.

    If you are familiar with ohms law then Volts (V) = Amps (I) x Resistance (R) also power in Watts (W) = V x I a bit of manipulation gives P = V^2 / R so doubling the voltage quadrouples the power.

    For a d.c. supply V is constant so no problem working out the power for a given resistance. For a.c. with a constantly changing value consider a series of time slices and work out the instantaneous power in each slice and then take the average but since power is proportional to v^2 the calculation needs to be the average of the voltages squared x resistance - this is where r.m.s (Root Mean Square) comes from.

    So far so good - go get more coffee.

    As I mentioned earlier the a.c. waveform is a sine wave and maths manipulation yields that the peak voltage is root 2 x the r.m.s. voltage.

    So for your transformer that reads 13V a.c. on your meter the voltage is actually changing from 0 volts to a positive peak of 18.38V back to 0V then to a negative peak of -18.38 volts and then back to 0V.

    Assuming you then connect this to a bridge rectifier which contains 4 diodes you get d.c. out of the rectifier but the current has to travel from the transformer, through a rectifier diode, to the load, back through a second rectifier diode and finally back to the transformer. These diodes are typically of silcon construction and rather than being perfect conductors they have a voltage drop of around 0.7V each in the forward direction, hence the series circuit via 2 of these drops 1.4V of the available (peak) voltage from the transformer.

    So now we have an output from the rectifier that reaches 18.38V - 1.4V or approximately 17V peak, it still keeps dropping to zero twice every cycle.

    Now consider the capacitor as a battery that can be charged by the transformer / rectifier and discharged by the load (virtually nothing if it is just a test meter) the capacitor is charged to the peak available voltage of 17V, as the voltage from the transformer drops the capacitor holds the charge and the rectifier diodes prevent it leaking back into the transformer.

    If a load is connected the capacitor will discharge until the rectifier output voltage exceeds the capacitor voltage at which time charging will re-commence. Hence the output voltage sags a bit then recovers twice every cycle, this fluctuation is referred to as the ripple voltage.

    Hope that helps.

    Steve.
     
  5. SharpyWarpy

    Thread Starter New Member

    Aug 5, 2007
    5
    0
    Thank you very much, Steve for the clear explanation. There are a couple of things in there I was not aware of and which I'll do more studying. Thanks again.
     
  6. spar59

    Active Member

    Aug 4, 2007
    51
    0
    Glad to be of help - just a couple of oops in my reply:-

    1) typo should be silicon not silcon

    2) Description of root mean square
    is the value at each time slice squared (the square bit) then added together and divided by the number of slices (the mean bit) then the square root of the mean (the root bit) - I forgot the latter - I must read more carefully before clicking post.

    Steve
     
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