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low voltage AC

Discussion in 'Electronic Basics' started by Guest, Oct 20, 2003.

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  1. Guest

    Guest Guest

    I need to have an AC power source for about 6 volt, and 3 W. It doesn't
    have to be perfectly sinusoidal or block, as long as it goes from
    positive to negative.

    To get this I can use a low voltage DC power source (batteries or
    computer power source) and convert to AC. Or would it be easier to use
    220V AC and somehow downtransform? Which is the easiest route?

    Is it possible to achieve this using a simple circuit?
  2. John Fortier

    John Fortier Guest

    You haven't menmtioned frequency, but assuming you are happy with 50 or 60
    Hz and your 6 volts is RMS ratehr than peak to peak, all you actually need
    is a transformer which will downconvert from mains voltage either 110 or
    230, depending on our location. You can get one from Radio Shack or from
    Maplin. The secondary should be capable of providing 500 milliamps, to give
    yo your 3 W.

  3. Guest

    Guest Guest

    yup, it doesn't matter too much what frequency. Is getting a transformer
    the 'easiest' way though (meaning, I thought more like a simple circuit
    that either transforms, or converts DC into AC).

    Isn't there an easy oscillator circuit somewhere?
  4. John Fields

    John Fields Guest

    You said in an earlier post that you could use 220V mains for a source.

    Since the mains are already AC, all you need is a transformer to step
    the voltage down to the 6VAC you need.
  5. Terry

    Terry Guest

    Yes; unless the weight of a transformer would be a consideration,
    it would (at 50 or 60 cycles) probably be the cheapest, simplest
    and most reliable (fewest parts involved) solution. About the
    only extra 'component' I would suggest would be a fuse in the
    input! At 400 hertz (e.g. aircraft) a 3 VA transformer could be
    quite small.

    An ordinary mains transformer with a 6 volt 0.5 amp transformer
    secondary should be pretty small; also the inductance of the
    transformer might provide some protection against sudden voltage
    spikes (transients) that occur on the commercial mains supply.
    You might be able to salvage such a transformer from something
    small such as a desk calculator/adding machine, old printer or
    something similar. Even an old model railway transformer or one
    of those that operates residential front door chimes (although
    those latter often don't like a continuous load and and can burn
    out). Even some wall warts can supply 500 milliamps at around six
    volts. A transformer also nicely isolates the 'load' from the
    main supply; often required and good safety design.

    Only other suggestion; if you really DON'T want to use a
    transformer, and assuming the load being is say close to a pure
    and unvarying resistance, would to put it in series with an AC
    rated capacitor across the mains supply. Fuse this at around one
    This will produce a leading current across the resistance portion
    of the circuit.
    This is a vector diagram situation; but in this case the
    resistance will be small (less than 10% or so) in comparison to
    the capacitive reactance that within practical limits the 'R' can
    be ignored!

    Have done this several times to power the filament heater of a
    single tube rated at 6.3 volts 0.3 amps. (Very similar to the 6
    volt at 0.5 amp requirement here!) from 230 volt mains supply.
    [6.3/0.3 = resistance of tube heater, hot = 21 0hms].

    Using the formula Xc (capacitive reactance) = 1/(2pi x frequency
    x capacitance);
    That is Xr = One divided by the total of (two pi times frequency
    in cycles, times capacity in farads).
    Thus at say 60 cycles a one microfarad capacitor (10 to the minus
    6 farads) will work out to;

    Xc Ohms = 1/(2pi x 60 x 10^-6) = 10^6/ 375 = 2667 ohms or approx.
    2700 ohms. [One microfarad].
    So a 6 microfarad cap will have one sixth of the reactance and
    will 'pass';
    At say 230 volts then; 230/450 = 0.5 amps.
    At 115 volts then double the capcitance to 12 mfd for 115/225 =
    0.5 amps.
    It's rough and ready but it works!

    Disadvantages of the series capacitor method:
    1) One side of the mains supply is connected to the load circuit.
    2) The size of capacitors (and possibly their weight) to
    withstand these voltages and the cost if buying new), will be
    greater than that of a simple cheap transformer.
    2) If the capacitor ever breaks down full line voltage will
    impressed on the load circuit. Pose a possible fire and safety
    hazard and will most likely destroy the load. Whatever it happens
    to be?

    I.e. KISS. (Keep it simple, Simon). Why complicate it?

    Good luck, have fun. Terry.
  6. Baphomet

    Baphomet Guest

    It's easier, simpler, cheaper and much more reliable to use just a step down
    transformer. For example uses 2
    transistors, 5 resistors, 2 diodes, plus the need for a transformer in any
    event. While this isn't the exact circuit you need, I think you can get an
    idea of the added complexity of going the inverter route.
  7. John Fields

    John Fields Guest

    Well, there's a _little_ more to it than what you've described.

    First, you've not provided any means to keep line transients from
    affecting the load and, in fact, just the initial turn-on surge could
    damage the load if it couldn't handle about a 311 volt negative or
    positive spike coming from the OP's 220V mains. Both problems can be
    mitigated by putting a bidirectional Zener TVS across the load. Ideally,
    a fuse should also be placed in series with the load. I prefer to fuse
    in front of the cap so that mains hot will be disconnected from _all_ of
    the circuit when the fuse blows. Unfortunately, it's not always
    possible to ensure that the circuit's return won't be connected to mains
    hot, so it remains a dangerous circuit if any part of it can come in
    contact with any part of a human!^)

    Second, it's not just the reactance of the capacitor which determines
    the current through the circuit, it's the impedance of the entire

    For example, assuming the OP's load is resistive and dissipates 3W, it
    will have a resistance of 12 ohms and draw 0.5A from the mains with 6V
    across it. Since the current is the same in all parts of a series
    circuit, and the entire circuit will have 220V across it, its impedance
    must be Z = E/I = 220V/0.5A = 440 ohms in order to allow 0.5A to flow.

    Now, since Z = sqrt(R²+Xc²), we can rearrange to solve for Xc:

    Xc = sqrt(Z²-R²) = sqrt(440²-12²) = 439.8 ~ 440 ohms

    Now, since Xc = 1/2pifC, we can rearrange to solve for the capacitance:

    C = 1/2pifXc, and if the mains frequency is 60Hz,

    C = 1/6.28*60*440 = 6.032 ~ 6µF,

    So your answer was pretty close because of the low resistance of the
  8. John Fields

    John Fields Guest

  9. Terry

    Terry Guest

    You are correct. I agree on all points. And that was the point;
    the phase angle with very low R is 'almost' 90.
    If the Xc reactance was of same order as R then the phase angle
    would be large etc.
    Could have solved the vector diagram. But for practical purposes,
    with R so small and component tolerances the rough and ready
    approach will work.
    The surge/voltage spike question is valid; and my tube heater was
    not specifically a fixed resistive load; most likely its
    resistance before warm up, for a few cycles anyway, would be
    lower than at rated voltage of 6.3v/0.3a. (Or another instance
    12.6v/0.15a), so voltage across it, at start-up, would be less.
    But was trying to make the point that if someone wants a quick
    and dirty (and cheap) way to provide six volts why make it so
    complicated? Especially if its for a one-time experimental
    Over design is to be avoided? And the requirement as described
    sounded like a pretty straightforward and unregulated voltage was
    Providing six volts AC at 0.5 amps (at say 50 or 60 Hertz) is
    such a common and basic requirement that one wonders why it even
    needs to be asked. I must have a half dozen scrapped transformers
    (and a couple of Wall warts) in my basement that have a winding
    that will provide six volts with a mains input. Either that or
    I'll bung in a variable voltage from a Variac to the primary of
    one of them until I get exactly six volts from a suitable
    secondary winding.
    But I carry around in my head that one mfd. at 50 hertz is
    'roughly' 3000 ohms!
    Thanks for the discussion.
    Cheers. Terry.
    PS. Metal cased waxed paper ex-telephone industry capacitors
    rated at 48 volts DC, tested during production to 2000 volts DC
    (between terminals and to metal case), have worked, as above, OK
    at 115 volts AC line voltage; one is currently in use to provide
    a phase difference between two windings of an ex radio
    transmitter equipment cabinet ventilation fan. I used fan to
    ventilate my attic roof space during the summer. That cap (0.5
    mfd) I think, manufactured in the 1960s, has been in use in that
    manner, for at least the past ten years. The ex-telephone
    switching equipment caps. are in a variety of values from 0.02 to
    4 mfd. So a six, say, could be made up of a a four+two. The
    dimension were similar in at least two directions, thickness can
    vary with capacitance but generally there are two or three
    'standard' sizes. North American (Western Electric/Bell System)
    caps with rounded corners were/are similar in size and values to
    the rectangular style caps made by various UK manufacturers, into
    the 1950s and 60s. In 1955 I used a group of them out of scrapped
    equipment as 16 mfd in a DC power supply at 350-400 volts. Still
    got that power supply somewhere; I wouldn't be surprised if the
    paper caps are OK; but the 'big' 450 volt electrolytic is
    probably dried up and a goner?
  10. Rich Grise

    Rich Grise Guest

    Can you say "Wall Wart?"

    6 VAC RMS, 3 VA - One Part, about five bucks.

    Have Fun!
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