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schematic

simulate this circuit – Schematic created using CircuitLab

I would like to place a non-polarized capacitor in between V1 and R1. Is it possible to replace the non-polarized capacitor with two polarized (electrolytic capacitors) C1 and C2 shown in the diagram above?

Kurst
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    Your description does not match the schematic, and it doesn't even match itself. First you want to *place* a non-polarized capacitor. Fine. In the next sentence you want to *replace* a non-existing polarized capacitor with two polarized. Ok.. You then show us a circuit (above, not below) with non-polarized capacitors, connected with diodes, seemingly redundant? Why two? Why diodes? There are so many mistakes that it's impossible to figure out what you want or what you need. – pipe Aug 24 '16 at 07:18
  • @pipe The reason why I put the diodes there is because electrolytic capacitors cannot take a DC voltage of the reverse polarity. As such, I'm hoping the diode in reversed biased mode will protect the electrolytic capacitors. – Kurst Aug 24 '16 at 07:28
  • @pipe I see your confusion. Sorry I couldn't find the symbol for polarized capacitors in CircuitLab so I just used the capacitor symbol in the app. Just imagine C1 and C2 are polarized. – Kurst Aug 24 '16 at 07:31
  • Two things. 1. Why ? 2. You can get bipolar electrolytics – D-on Aug 24 '16 at 07:38
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    @Kurst its not obvious why you have 2 diodes and 2 capacitors. Did you intend D1 and D2 to be reverse polarity? – Steve G Aug 24 '16 at 07:39
  • @SteveG Yes, sorry I mean to have them in reverse. – Kurst Aug 24 '16 at 08:07
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    Makes me wonder what else you "oops forgot". We know that the textual description is wrong, the diodes are the wrong way, the capacitors in the schematic is the wrong type, C2 doesn't exist.. What else, before everyone wastes more time trying to solve the puzzle? – pipe Aug 24 '16 at 09:49
  • What ceramic capacitor? Do you really have two ceramic capacitors of 470uF each that you want to replace? Hard to believe. – user207421 Aug 24 '16 at 10:59
  • You may find [this answer](http://electronics.stackexchange.com/a/90672/25328) useful. From CDE's [Application Guide, Aluminum Electrolytic Capacitors](http://www.cde.com/resources/catalogs/AEappGUIDE.pdf): "If two, same-value, aluminum electrolytic capacitors are connected in series, back-to-back with the positive terminals or the negative terminals connected, the resulting single capacitor is a non-polar capacitor with half the capacitance." – Tut Aug 24 '16 at 13:49

2 Answers2

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You can simulate the effect of a non-polar capacitor with polar capacitors and diodes, if you connect them correctly.

schematic

simulate this circuit – Schematic created using CircuitLab

As an AC voltage is applied across this combination, each diode will rectify, and charge up the centre node. The reverse voltage across either capacitor will not exceed one diode drop, which is probably OK for aluminium electrolytics.

Note that in use, the effective capacitance in this case is 500nF, the correct addition of two capacitors in series. I have seen people argue that each diode 'shorts out' its respective capacitor on each half cycle, and that the capacitance is therefore the value of just one capacitor. This is only true during the startup charging phase when the diodes conduct, the effective capacitance will be that of just one capacitor, 1uF.

If the pair has been exposed to one AC voltage for a while, so the centre node has charged to that voltage, and the voltage is suddenly increased, there will be a further phase of charging, and a change in effective capacitance while it charges.

If it's expected that the applied voltage will change, it's better to use a high value resistor to the centre node as well, to bias it to a higher voltage than either end will see. This will ensure that both capacitors stay correctly biassed for the whole time.

Neil_UK
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  • What would be a high value for a resistor in this case? – Kurst Aug 24 '16 at 08:11
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    A 'high value' resistor is determined by the application. It needs to be low enough to charge the centre node quickly (don't forget this sees both caps in parallel, so it's charging 2uF) so it's ready to use before you need it, but high enough not to disturb your application, not to put too much shunt resistance on the signal path. You draw in the rest of the circuit, and then a suggestion might be forthcoming. – Neil_UK Aug 24 '16 at 08:33
  • I was trying to run an AC signal through the diode and then a resistor, I found that a small amount of negative current still passes through to the resistor. In this case, will my rectifying bridge even work since I will probably end up running a small AC current in the opposite direction of a polarized capacitor. – Kurst Aug 25 '16 at 00:47
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As for the correct way to do it, have a look at Neil's answer. I just wrote this as a little supplementary answer if OP was creating an RC differentiator

In response to your question, simple answer, no you can't (in that configuration).

Before, with a single ceramic capacitor this would have been an RC differentiator circuit. [RC Differentiator explanation]

The problem with the way you've added diodes will mean that on the falling edge of your square wave, the 5V across your capacitor will have nowhere to go, I've tried to explain it below (although it might not be 100% correct)

schematic

simulate this circuit – Schematic created using CircuitLab

Community
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