# low voltage AC

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

1. ### GuestGuest

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 FortierGuest

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.

John

3. ### GuestGuest

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 FieldsGuest

---
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. ### TerryGuest

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
amp?
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. ### BaphometGuest

It's easier, simpler, cheaper and much more reliable to use just a step down
transformer. For example http://www.electronics.50g.com/cdcac.htm 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 FieldsGuest

---
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
circuit.

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
load.

^^^^^
6.032E-06F

9. ### TerryGuest

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
circuit.
Over design is to be avoided? And the requirement as described
sounded like a pretty straightforward and unregulated voltage was
needed.
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 GriseGuest

Can you say "Wall Wart?"

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

Have Fun!
Rich

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