[ARC5] Collins xfmr ratings

[Kenneth G. Gordon]

On 15 Aug 2014 at 10:23, D C _Mac_ Macdonald wrote:
 
> I'm just WAGing that a bleeder load of 10% or maybe even 5% 
> of the total current would do that, Ken.

It might, Mac. Yes.

But since I have the opportunity, let me try to explain for the rest of the crew, 
if any of us don't know about it.

"Critical Inductance" only really becomes an important issue when the load 
on an AC power supply is varying a lot....like in a CW transmitter.

Dynamotor and regulated power supplies never have this sort of problem.

For AC power supplies in which the load is fairly constant, it is not a concern, 
since the constant load makes this issue moot.

The "critical inductance", defined as that inductance needed to prevent 
output voltage from soaring, needed for any power supply is approximately 
the "load resistance", divided by 1000.

For instance, for an ARC-5 level transmitter, if the DC voltage on the rig is 
500 VDC, and we are loading it to 200 mA at key-down, the load resistance 
is then 500/.2=2500 ohms. Therefore, in this case, critical inductance would 
be 2.5 H.

But, at key up, the load might drop to 20 mA. Then the critical inductance 
would need to be 10 times as much, or 25 H. For this case, load resistance 
has risen to 25000 ohms due to the much lower current being drawn at key 
up. 

Since we have only installed, say, a 4 H choke, the output voltage would rise 
WAY up, in this case to almost 800 VDC, and in addition to the dynamic 
voltage regulation being terrible, contributing to chirp and other distressing 
noises, the voltage might be high enough to severely stress some 
components in the transmitter, like blocking capacitors, etc.

So, we add a bleeder resistor to the power supply which will draw enough 
current at key UP to keep the critical inductance required to no more than 
our installed choke.

In this case, we would want the key-UP load resistance to be 4000 ohms or 
less. So, we would need a 4800 ohm resistor at 100 watts as a bleeder. This 
bleeder would draw 52 watts at key up from our power supply and would 
keep the voltage from soaring.

Of course, you can increase the needed resistance value of the bleeder 
resistor, thereby reducing the power dissipated in it, by increasing the size of 
the choke.

Although I have not done the math, I would think that if you used an 8 H 
choke, you could use an almost 10K bleeder, which would then only have to 
be 50 watts, and the bleeder would be drawing only 25 watts.

Electric Radio Magazine, in their August issue, beginning on page 3, has a 
reasonable discussion of this, although for our purposes, it isn't quite 
complete.

Also, I hope all my math above is correct. If it isn't, I am sure someone will 
correct me. :-)

Ken W7EKB