Huge capacitance just after the diodes in a SE6550 power supply

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Heads up guys, I have a question that's been in my mind for years.I builts a 6550 SE amp according to ''Audio Express'' sept.'' 2001 issue, I went overboard and instead of using what was used in the article, I changed (amongst MANY other issues) the capacitance coming out of the diodes(in the p.s.) with a1800uF 450 volt 3''x6'' mammoth...So the question is ...did I do the right thing? Yours Truly Al
 
That cap can store 180 joules (0.5 x 0.0018 x 450^2). Thats a lot more than is needed. Lets say 5% ripple voltage is normal (for want of a figure) that's about 10% energy ripple (energy in a cap goes as voltage squared). 10% of 180J is 18J. between successive half-cycles of 50Hz is 0.01 seconds, 18/0.01 = 1.8kW.


So that cap can provide smoothing for several kW of amp used near its rated voltage. I'd worry about the pulse rating of the diodes.


You'll have to be very carefully to fully discharge it before working on the amp, that sort of energy at HV is definitely a killer, and it will
be able to recharge itself to lethal levels too unless strapped to ground while working. Large HV electrolytics are stored with a shorting link between the terminals.
 
Last edited:
Oh course, the same can't be said about push pull amplifiers. I use between 470 and 5000 or so depending to keep the bass tight.

Usually, a vacuum diode requires less than 50uF capacitance input, and also 100R dropping resistors in the plate leads. This makes a sloppy amp if it's PP class AB, but it's usually fine when it's SE because the idle vs full volume current is the same average (class A).
 
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.