I usually put (+) to (+) but in reality it should not make any difference......but.........
For small capacitors - say, 1 uF or less - a non-electrolytic type will very
likely be satisfactory if its size - these are usually much larger - is not a
problem.
There are several approaches to using normal polarized electrolytic capacitors
to construct a non-polarized type.
None of these is really great and obtaining a proper replacement would
be best. In the discussion below, it is assumed that a 1000 uF, 25 V
non-polarized capacitor is needed.
Here are three simple approaches:
* Connect two electrolytic capacitors of twice the uF rating and at least
equal voltage rating back-back in series:
- + + -
o----------)|-----------|(-----------o
2,000 uF 2,000 uF
25 V 25 V
It doesn't matter which sign (+ or -) is together as long as they match.
The increased leakage in the reverse direction will tend to charge up the
center node so that the caps will be biased with the proper polarity.
However, some reverse voltage will still be unavoidable at times. For
signal circuits, this is probably acceptable but use with caution in
power supply and high power applications.
* Connect two electrolytic capacitors of twice the uF rating and at least
equal voltage rating back-back in series. To minimize any significant
reverse voltage on the capacitors, add a pair of diodes:
+---|>|----+----|<|----+
| - + | + - |
o-----+----)|----+-----|(----+------o
2,000 uF 2,000 uF
25 V 25 V
Note that initially, the source will see a capacitance equal to the full
capacitance (not half). However, the diodes will cause the center node
to charge to a positive voltage (in this example) at which point the diodes
will not conduct in the steady state.
However, there will be some non-linearity into the circuit under transient
conditions (and due to leakage which will tend to discharge the capacitors)
so use with care. The diodes must be capable of passing the peak current
without damage.
* Connect two capacitors of twice the uF rating in series and bias the center
point from a positive or negative DC source greater than the maximum signal
expected for the circuit:
+12 V
o
|
/
\ 1K
/
- + | + -
o----------)|-----+-----|(-----------o
2,000 uF 2,000 uF
35 V 35 V
The resistor value should be high compared to the impedance of the driving
circuit but low compared to the leakage of the capacitors. Of course, the
voltage ratings of the capacitors need to be greater than the bias plus the
peak value of the signal in the opposite direction.