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2S2P series/parallel capacitor + extra connection

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LAN_403

Signification

Mon Jul 22 2019, 03:36AM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

I have connected four identical photo-flash capacitors (2x2) , each rated: 800uF @ 360Vdc, to get a total rating of 800uF @ 720v. (please forgive the lack of graphics.) Let's call them C1, C2, C3, and C4. The two series connection sets go like this: C1- to C2+ AND C3- to C4+. To put these two series-connected sets in parallel, I connect C1+ to C3+ (+out) AND C2- to C4- (-out). Also, I have 200k@2W resistors across each cap for balancing/bleeding.

THE QUESTION: what is the difference between this configuration and one with an added connection to C1-,C2+ AND C3-,C4+ ?

klugesmith

Mon Jul 22 2019, 04:57AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

Some benefits of the added connection:
1. Need only two instead of four balancing/bleeding resistors.
2. The added connection adds flexibility if you want to match the _measured_ values of C's in series, so the smaller one doesn't get voltage reversed when a rapid discharge happens.

There might be some disadvantages of the added connection when configuring 2S2P batteries. Electric-powered model enthusiasts might know about them. Maybe they apply to your electrolytic capacitor project.

Sulaiman

Mon Jul 22 2019, 08:20AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

Adding the link may give slightly better voltage balancing,
but under fault conditions (a short-circuit capacitor) individual strings of capacitors in series, connected in parallel, may give slightly less explosiveness.
Even if you connect capacitors in parallel, the overall current (hence heat) distribution is better in multiple resistors rather than one single resistor, that could fail open circuit.
So, no need to change your arrangement, but no harm if you do.

DerAlbi

Mon Jul 22 2019, 08:40AM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

Luckily, capacitor manufacturers thought of covering the basics already.
Give some love to TDK:

Besides the fact that one should read everything, your particular problem is addressed in chapter 6, starting on pdf page 26. Have fun.

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Mon Jul 22 2019, 08:58PM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

I am trying to recall this from memory, but a while back I had a similar capacitor data/app book as Albi referred to. There was one page concerning the connection of capacitors in straight parallel. Foe example, suppose there are five capacitors, C1-C5, all lined in single file (C1,C2,...C5) and connected in parallel with two straight conductors. IIRC, the diagram indicated a "Right" and "Wrong" way to connect to this bank. The "Right" way to connect to this parallel bank was to make one connection at C1 and the other polarity at C5.

Apparently, there was some 'disadvantage' in taking the connections at a single capacitor at the end. I don't recall the reason.

2Spoons

Mon Jul 22 2019, 11:28PM

Registered Member #2939 Joined: Fri Jun 25 2010, 04:25AM
Location:
Posts: 615

Its quite simple. Draw the 5 capacitors but include the impedance of every connection. You will see that a connection at c1 and c5 yields equal total impedance to each capacitor. But if you were to connect only at C1, then C1 has the lowest total impedance, C2 is a bit higher, all the way up to C5 being the highest - which means current will not be shared equally among the capacitors.

Signification

Wed Jul 24 2019, 03:15AM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

Many thanks,
This leads to another similar situation: Years ago when the first LED string arrays came out, I remember having one with MANY LED's (each light 'unit' was composed of an LED in series with a resistor set for 12vdc operation). The 'units' were all connected in PARALLEL, and additional LED strings could be plugged into the the end of the previous one for longer arrays. The 12v source (which supplied enough current for all LEDs) was only connected at ONE end of the parallel array.
The brightness was NOT uniform--could this possibly be solved by making the 12v power connections at opposite ends? Much like the proper capacitor array connection in the discussion.

2Spoons

Wed Jul 24 2019, 05:28AM

Registered Member #2939 Joined: Fri Jun 25 2010, 04:25AM
Location:
Posts: 615

Yes, exactly.

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Thu Jul 25 2019, 03:46AM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

OK, one more thing involving connecting capacitors that seems to somewhat fit here...and I have wondered about this one since pre-internet times. When I looked for solutions on youtube, I am finding that many other people, attempting to work this problem 'on paper', get the obvious answer--WHICH IS WRONG--and accept it as true and just keep on going.
The paradox: You have two identical capacitors. One is charged and the other discharged, the two capacitors are connected in parallel (I see sparking when I try it, which I think is key to the wrong answer obtained by the obvious method). It turns out they all come up with the same conclusion, that is, that the initial charge is EQUALLY shared among the two capacitors. For an example, see

This is WRONG! It actually appears that 1/2 of the energy seems to "radiate away" when this experiment is done. Does anyone know of any way to keep more than 1/2 of the total energy at steady state--and an explanation of the energy loss? would this process 'always' be 50% energy efficient?

Sulaiman

Thu Jul 25 2019, 04:55AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

From memory;
There is always some resistance which looses energy
there is inductance, so if you connect two capacitors with different open circuit voltages together,
you get a damped resonance with a very high oscillating current loop - an rf transmitter.
If the circuit resistance is significant then up to half of the energy will be lost to heat with very little rfi.

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