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PostPosted: 01 Feb 2018, 19:07 
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Location: Australia
A CCS here is off thread. The whole idea of this build is no OPT and built into a slim base so to show off the tubes.

The mention of CCS is OK but not an extended discussion. It is off-thread.

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PostPosted: 01 Feb 2018, 20:36 
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I agree with Mark. Please try keep posts on topic. If you want to diverge from the topic, please feel free to start a new thread.

Gio

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PostPosted: 02 Feb 2018, 00:14 
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Sunc wanted to see a schematic on CCS and I posted one, others had questions so I answered them, I don't see anything wrong.


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PostPosted: 02 Feb 2018, 00:36 
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Time to move on.

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PostPosted: 02 Feb 2018, 05:40 
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mwhouston wrote:
Maybe a 6C33C with CCS should have its own thread and leave this thread to the original concept. Might be time to replant it.

Guess some people (cough! cough! :blush: ) got carried away. Over to you Houston (Or as a Swedish radio station pronounced it - Husston).

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PostPosted: 02 Feb 2018, 06:57 
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No problem there.

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PostPosted: 02 Feb 2018, 06:59 
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Looking for a good voltage doubler cct. I have one but wonder if there are better ones out there.

I need this to get the voltage up for my build.

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PostPosted: 04 Feb 2018, 17:05 
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http://www.learningaboutelectronics.com ... ircuit.php
The size of c1 should equal to c2. Going from full bridge rectifier to a voltage doubler means you get a half wave rectifer.

This means the size of your capacitor needs to doubble for same ammount of ripple compared to a full bridge. Further more its effectivly quad cap size for same ripple because you need two capacitors to doubble voltage.

https://en.wikipedia.org/wiki/Ripple_(electrical) 1.1 filtering
voltage ripple equals to Current/ (Freq*cap size)
in full bridge the rate at which the caps can charge is doubbled, the transfomer is 50hz and effectily charging at 100hz.
Half wave means it charges at 50hz.

Because we now have two caps they will both charge and discharge at same time, the fourmula only monitors the votlage ripple accross one capacitor not two.
Because the supply draws equal power from both caps they discharge equally, and because each cap holds half of supply voltage, the voltage riple seen is doubble.
Imagine a voltage supply with 1 voltage ripple you put two in series for doubble volts now you get 2 voltage ripple, same in what is currently happening.
This means the actuall ripple current is 2*Current/ (Freq*cap size)
We know the current and frequency, now we can find a suitable cap size.


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PostPosted: 04 Feb 2018, 17:20 
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That's actually a decent doubler that I use all the time.

One thing to remember is to look for a cap with high ripple current AT 100/120Hz (SMPS caps are almost always rated at 100KHz). Doubler caps will get warm and are made to do so. Ability to dissipate that heat is a key to doubler quality and why high ripple caps are BIG compared to regular ones of the same V and uF.

Not shown in the above circuit are equalization/bleeder resistors. I have found that a draw of 5 to 10mA (1mA if a little LED supply) at the loaded voltage is good enough to EQ imbalances between them.

Peak and power-up surge currents in the circuit will be very high. If math says a UF4007 will do, use a UF5408 or my fav, MUR4100 :D

Doubling is a forgotten technique that HK, Eico, Heath and others used very successfully ^_^

Cheers!

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PostPosted: 04 Feb 2018, 17:21 
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KochiyaYamato wrote:
http://www.learningaboutelectronics.com/Articles/Voltage-doubler-circuit.php
The size of c1 should equal to c2. Going from full bridge rectifier to a voltage doubler means you get a half wave rectifer.

This means the size of your capacitor needs to doubble for same ammount of ripple compared to a full bridge. Further more its effectivly quad cap size for same ripple because you need two capacitors to doubble voltage.

https://en.wikipedia.org/wiki/Ripple_(electrical) 1.1 filtering
voltage ripple equals to Current/ (Freq*cap size)
in full bridge the rate at which the caps can charge is doubbled, the transfomer is 50hz and effectily charging at 100hz.
Half wave means it charges at 50hz.

Because we now have two caps they will both charge and discharge at same time, the fourmula only monitors the votlage ripple accross one capacitor not two.
Because the supply draws equal power from both caps they discharge equally, and because each cap holds half of supply voltage, the voltage riple seen is doubble.
Imagine a voltage supply with 1 voltage ripple you put two in series for doubble volts now you get 2 voltage ripple, same in what is currently happening.
This means the actuall ripple current is 2*Current/ (Freq*cap size)
We know the current and frequency, now we can find a suitable cap size.

Thanks

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Projects: "Sanctum" - 12AU7 and 6AS7 direct coupled headphone amp | "retro-Oatley 6J6" - 6J6 push-pull headphone amp with OPTs | "retro-Hiraga" - Jean Hiraga Le Monster
Website: retro-thermionic


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