Step start supplies and HV keying

[ND8D]

I am beginning to piece together a supply for a transmitter, but am at an impasse when it comes to keying the supply.

For reference:
Transformer: (1KVA, dual 60V secondary) http://www.antekinc.com/content/AN-10458.pdf
Rectifier: 50A 1000V bridge (https://www.digikey.com/product-detail/en/comchip-technology/KBPC5010-G/641-1861-ND/2214448)
Capacitor: 7200 uF spread across 6 350V capacitors.

I understand that inrush is a concern, especially when there is a stout transformer feeding a big capacitor bank. To alleviate that a switched resistor bank is the cure, disengaging after a few hundred milliseconds.

But, if that is to happen every time the TX is keyed, wouldn't that lead to an excessive rise time? Or am I expecting too much from that?

It would be nice to make use of the 12V and 18V secondaries of my transformer to save room, how much of a concern is the resultant spike when the load is removed from the secondary if I were to just switch the DC output? Or should I just put that Idea out of my head?

[W1DAN]

Hi:

I fire up the DC supply only once per radio session. I use the Soft Start pin on the UCC25701 (Pin 14) PDM generator IC through a DPDT transmit switch (the other side keying up the exciter). This ramps up the PDM generator upon transmit. Note you may need to invert the drive to the PDM transistors to get the proper effect.

In listen mode, with the PDM generator cut off, the main supply filter caps float up a bit higher than under load. This is not a worry as it is still under the peak voltage of the RF transistors, and ramping the PDM up brings it down before full carrier is made.

Most others use a step-start relay. Here an AC solid state relay may be used.

Dan

[Opcom]

How about just sizing the peak inrush to the 'full power' rating of the supply or maybe 150% of it?

120V & 1KVA = 8.3A,  about 14 Ohms?
for faster charging, 10 Ohms if you can deal with that. The outlet can take 12A, right?

nice transformer like that, the ol RC time constant might give an answer.
if it is C input off a 120V winding, then DC would be about 150V

R=10 C=0.0072F

1RC time constant=.072 seconds to 63% of the 150V voltage; 94VDC

2RC = 0.141 seconds, 63% of the remaining voltage from 94VDC to 150VDC would be 129VDC

and at that time close the switch to get to 150V. Is 140 milliseconds fast enough?

I think this makes sense, corrections welcome.

I completely admit the electronic soft start using the PDM is much more elegant.

[VE3ELQ]

There are many way to do this but a simple solution (that I'm currently using), employs a surge protector 570-1071-ND in series with the transformer primary and a 48V 10A relay  255-2813-ND across the device, in fact it is mounted on the relay, powered through a series resistor from the B+ set to pull in the relay at about 80% of the B+.  The HV to the modulator is hot all the time, the modulator bridge driver enables the HV to the RF deck only on PTT for TX.

73s  Nigel

[steve_qix]

There are a number of solutions.  The best solution disconnects the AC input to the power transformer between transmissions.  This prevents soaring of the rectified DC between transmissions.  It is also safer.

To do this, a step start solution is definitely the way to go.  The step start limits the inrush current to whatever value you specify by the value of the step start resistor.

For instance, for a 120V AC line and a 400 watt transmitter, using an 8 ohm step start resistor will limit the in-rush current to no more than 15A and is reasonable for the power level (the capacitors will charge up to almost full voltage during the step start period).

I usually use about a 1/2 second delay on the step start, where the resistor is shorted out after 1/2 second.  You need 2 relays - one to key the main line voltage coming into the H.V. transformer primary, and a 2nd relay to short out the step start resistor after about 1/2 second.

The step start resistor has to be able to handle the full inrush current, so be sure to use a reasonably sized resistor.