[ARC5] Collins xfmr ratings

Kenneth G. Gordon kgordon2006 at frontier.com
Fri Aug 15 16:10:22 EDT 2014


On 15 Aug 2014 at 12:00, John Saxon wrote:

> 
> Great discussion guys, especially since I am now building some 
> power supplies for my ARC-5 stuff.  OK...critical inductance:
> 
> In the 1943 ARRL Handbook, in the discussion of critical inductance 
> it says:
> 
> "The function of the choke is ... to prevent the d.c. output voltage 
> from rising above the average value of the a.c. voltage applied to the 
> rectifier."

Although in point of fact, this statement is correct as it is written, what 
actually happens if the critical inductance is not reached is that the output 
voltage reaches the a.c. voltage's PEAK, not average. Now this doesn't 
happen instantly, meaning that if the choke is close to the critical inductance, 
but still below it, the output voltage will be higher, but not to the peak value 
yet. If the true peak value was reached, that would be 600*1.414=848.4V.

> I have a 1200 v center tapped transformer, full wave rectifier (CT 
> grounded), single choke-input filter.  May be a dumb question,

No...it's not...

> but 
> what is "...the a.c. voltage applied to the rectifier"?  600?  1200? or?

600, because we are talking about the rectifier SYSTEM, in this case, two 
diodes, basically in series (I think).

BUT, what we have to be concerned with when we are dealing with rectifiers, 
especially SS ones, is the peak-to-peak value impressed on them.

For a FW center-tapped system, the P-to-P voltage impressed on them is the 
full secondary voltage, 1200 in this case, times 1.414 (square-root of 2), or 
1696.8 V. So you have to size your rectifiers accordingly.

Here is a very succinct statement on this I found on the web: "The PIV for 
each diode in a center-tapped full-wave rectifier is twice the peak output 
voltage plus one diode drop." A diode drop is about 0.7 V, so for our 
purposes we can ignore it.

Output voltage in your case is ~600. Twice that is 1200. The peak voltage of 
1200 volts is 1200*1.414=1696.8V.

I would use at least 2 rectifiers of 1KV PIV rating in each leg, and three in 
each leg wouldn't be a bad idea. Rectifiers are cheap.

Ken W7EKB


More information about the ARC5 mailing list