Hello.
I've started to get into radio circuits and a variety of capacitors are needed so I bought an assortment of non polar capacitors. To my misfortune, many of the capacitance levels are too low.
After doing some research, I came across this thread.

http://www.diyaudio.com/forums/parts/107159-polarized-non-polar-capacitor-wiring.html

From what I understand from the above thread is that if I place two capacitors this fashion: + - - + (or - + + -), where each capacitor has a value of 1uF, the equivalent of that would be a non polar capacitor that I could use for radio circuits with a capacitance of .5uF?

Is this true?


Also, if I made a radio circuit and let's say my antenna is resonating at 1Ghz. If there is no signal in the air of that frequency will I hear static?
I made a radio circuit and picked arbitrary values for my tank circuit and I wasn't able to hear anything . Need to know if the error was on my part or on my components.


Thank You for the help.
I really appreciate it.

 

That method works. I have done it before. By having the two electrolytics in series, you get 1/2 the capacitance rating and 2x the voltage rating if they are the same value. Works well.

 

^^ And many times the physical size of a single ceramic capacitor.

 

Hello,

Not completely.
The capacitance value will be half the value with equal capacitors.
The voltage allowed will be the max of ONE capacitor, as a capacitor will not like a reversed voltage.
(there is a kind of diode effect).

Bertus

 

Hello,

Not completely.
The capacitance value will be half the value with equal capacitors.
The voltage allowed will be the max of ONE capacitor, as a capacitor will not like a reversed voltage.
(there is a kind of diode effect).

Bertus

Thank you all for the input.
Is there any where I can read about this? Can't seemed to find any literature/data.

 

From what I understand from the above thread is that if I place two capacitors this fashion: + - - + (or - + + -), where each capacitor has a value of 1uF, the equivalent of that would be a non polar capacitor that I could use for radio circuits with a capacitance of .5uF?

The short answer is yes, but be very careful to not exceed maximum ratings. The long answer: Read this: http://electronics.stackexchange.co...tic-capacitor-out-of-two-regular-electrolytic

As Bertus pointed out, you do not have double the voltage, nor do you have half the capacitance.

TheComet

 

Thank you all for the input.
Is there any where I can read about this? Can't seemed to find any literature/data.

http://electrochem.cwru.edu/encycl/misc/c04-appguide.pdf

 

If you put two capacitors of equal value in series and apply a DC voltage across the combination, what is the voltage across each capacitor?

 

If you put two capacitors of equal value in series and apply a DC voltage across the combination, what is the voltage across each capacitor?

The charge of every capacitor in series is equal to the total charge of the capacitors.

\(Q_{total}=Q_1=Q_2\)

Given the DC voltage across both capacitors in series, one can solve for the individual voltages:

\(U_{total}=10V\)

\(C_{total}=\frac{1}{\frac{1}{C_{1}}+\frac{1}{C_{2}}}\)

\(U_{1}=\frac{U_{total}*C_{total}}{C_1}\)

\(U_{2}=\frac{U_{total}*C_{total}}{C_2}\)

So the answer would be 5V on both capacitors (half the supply voltage).

However, due to leakage currents and imperfections capacitors have, this value will slowly begin to drift, so there's no guarantee the two voltages will remain as they are for long periods of time. This is why supercaps need resistors in parallel, to make sure the voltage doesn't rise too much and destroy it.

TheComet

 

Hello,

Here is a page with info on reversing the polarity of an electrolitic capacitor on it:

http://www.capacitorguide.com/electrolytic-capacitor/

Look for polarity and safety.

Bertus

 

The charge of every capacitor in series is equal to the total charge of the capacitors.

\(Q_{total}=Q_1=Q_2\)
...

Not in this case. With two polarised electro caps one cap is reversed and acts very much like a short circuit. the reversed cap does not "charge". The total capacitance of two 10uF electros in reversed series (in most applications) is still about 10uF.

And the voltage rating is about the same as one cap also.

 

I have some papers on caps here http://electrochem.cwru.edu/encycl/misc/c04-appguide.pdf and here http://ebooksgo.org/engineering-technology/ElectrolyticCapacitors.pdf. This is what the firt paper say about making a bipolar electrolyte cap

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
to either of the original pair. The two capacitors rectify
the applied voltage and act as if they had been bypassed
by diodes. When voltage is applied, the correct-polarity
capacitor gets the full voltage.
In non-polar aluminum electrolytic capacitors and motorstart
aluminum electrolytic capacitors a second anode foil
substitutes for the cathode foil to achieve a non-polar
capacitor in a single case. While non-polar aluminum electrolytics
are available for momentary-duty AC applications
like motor starting and voltage-reversing applications, the
high DF of aluminum electrolytic capacitors – from 2% to
150% – causes excess heating and short life in most AC​

applications.

 

Not in this case. With two polarised electro caps one cap is reversed and acts very much like a short circuit. the reversed cap does not "charge". The total capacitance of two 10uF electros in reversed series (in most applications) is still about 10uF.

Obviously, yes. I thought he meant two non-polarized caps.

 

I have some papers on caps here http://electrochem.cwru.edu/encycl/misc/c04-appguide.pdf and here http://ebooksgo.org/engineering-technology/ElectrolyticCapacitors.pdf. This is what the firt paper say about making a bipolar electrolyte cap

" the resulting single
capacitor is a non-polar capacitor with half the capacitance
to either of the original pair."

" The two capacitors rectify
the applied voltage and act as if they had been bypassed
by diodes. "

Yes they act like the second (reversed) cap is bypassed.

I'd like to see the justification for saying they have half the capacitance of one cap?

Imagine one of these caps made with two 10uF electros, charging from 2v to 10v, then back from 10v to 2v. The second cap is bypassed the whole time as it has reversed voltage across it. The second cap acts a lot like a short circuit or very low resistance.

So cap A alone charges from 2v to 10v, then 10v to 2v. Cap A is 10uF, the whole thing charges and discharges to a capacity of 10uF.

How would it supposedly be 5uF?

 

Yes they act like the second (reversed) cap is bypassed.

I'd like to see the justification for saying they have half the capacitance of one cap?

Imagine one of these caps made with two 10uF electros, charging from 2v to 10v, then back from 10v to 2v. The second cap is bypassed the whole time as it has reversed voltage across it. The second cap acts a lot like a short circuit or very low resistance.

So cap A alone charges from 2v to 10v, then 10v to 2v. Cap A is 10uF, the whole thing charges and discharges to a capacity of 10uF.

How would it supposedly be 5uF?

I am only referring to what I have read several places. Of course some less thrustworthy than others. So I presented two sources that I regard as thrustworthy(see page 141 in the EBOOK). It is also a discussion here you may like to read. http://electronics.stackexchange.co...tic-capacitor-out-of-two-regular-electrolytic
This is something that I have not tested my self. And can not bring any experience to the table.

 

Thanks t06afre, I hope I did not sound argumentative, it's just that it goes against common sense to say it is half the capacitance of one cap.

Maybe someone writing that got confused and was thinking of two caps in series, not a bypassed (shorted) cap in series with one charging cap?

It should be very easy for someone with a simulator to test. They could charge a polarised electro from 10v to 20v through a resistor, and see the time on the simulator scope trace.

Then compare to a reversed series electro pair, also charging from 10v to 20v through the same resistor.

If the "pair" acts like a cap of half the value it should charge twice as quickly from 10v to 20v. Personally I can't see it happening.

 

It has been my experience that it is true. I even put it in my library because we have to tell new folks this occasionally.

I've been using this trick a very long time, I learned it almost 50 years ago, from electronics magazines. No resistors needed.

It still has all the high frequency problems electrolytes have, so it tends to be only for lower frequency audio stuff.

Anyone with a LC bridge can prove it. I have one, for example. The odd cap does not short out, but the stress from the voltage on the plates is taken on by the other cap.

 

It has been my experience that it is true. I even put it in my library because we have to tell new folks this occasionally.

I've been using this trick a very long time, I learned it almost 50 years ago, from electronics magazines. No resistors needed.

It still has all the high frequency problems electrolytes have, so it tends to be only for lower frequency audio stuff.

Anyone with a LC bridge can prove it. I have one, for example. The odd cap does not short out, but the stress from the voltage on the plates is taken on by the other cap.

I agree. A commonly used trick.

Ramesh

 

I'm aware of the two electros trick Ramesh and have used it decades ago when repairing valve amps, although always worked on the principle that Ct = C not Ct = 0.5C.

Bill, I just went out to the workshop and did some tests, and sorry to say your library is wrong, although it can be right in some specific cases.

I tested the RC charge time (DC) of a number of polarised electros using my digital storage scope. Once the reversed electro reaches the "diode" condition at about 5v (8v on some caps) the entire remaining charge occurs solely in the non-reversed electro. Charge time matches C of the non-reversed cap, as it should.

However, in cases where the votlage is sufficiently low (ie <5v) no electro "dioding" effect takes place, and the two caps charge equally in time and voltage, exactly as if they were two non-polarised caps working normally. So in that case, Ct = 0.5C exactly as it would in non-polarised caps.

My memories of using this trick were mainly from valve amps where the caps work at hundreds of volts, so the "dioding" effect is very much the case, and is a very small percentage of the cap working voltage. So this holds true; Ct = C.

Since the discussion is specifically for polarised electros and the reversed dioding effect was mentioned and assumed to occur, the actual formula for two reversed polarised electros should be Ct = C (assuming perfect electros).

Now obviously in the real world the capacitance will be 0.5C for low voltage waveforms and approaches C as the waveform voltage rises significantly above the dioding voltage.

So I can see why people would make that mistake and think Ct = 0.5C, as many cursory low voltage tests would give that result (even on my cap meter!) but that result only occurs because the electros DO NOT at that time exhibit any reverse "shorted" or "diode" effect so what the test measures is NOT two reversed polarised electros, it's measuring two non-polarised caps.

 

So I can see why people would make that mistake and think Ct = 0.5C, as many cursory low voltage tests would give that result (even on my cap meter!) but that result only occurs because the electros DO NOT at that time exhibit any reverse "shorted" or "diode" effect so what the test measures is NOT two reversed polarised electros, it's measuring two non-polarised caps.

Good to know. So the correct answer is. The Ct will depend on the caps electrochemical properties more or less.