[time-nuts] Using a common power supply among few time standards

[Bob kb8tq]

Hi

Unless there is something wrong with the linear supply, it will produce heat
proportional to the output load. They all have very low “parasitic” current in the
control circuits. 

A unit with a damaged transformer or bad caps *can* indeed heat up with no 
load. In that case, it needs to be repaired. 

Bob

> On Feb 12, 2020, at 10:34 PM, Taka Kamiya via time-nuts <time-nuts at lists.febo.com> wrote:
> 
> Yes, I did consider use of a second linear supply for 12V.  But decided against it.  This particular project was to use ONE 24V supply only.  This is because of my previous project.  Also the issue with another 12V linear is that these open frame supplies are awfully inefficient.  They generate enormous amount of heat for little current.  Since this is a rubidium box, heat management was important.
> 
> On previous project, I used 24V linear supply, and 12V switcher supply.  AND 5V DC/DC off 24V, 5V 3 terminal off 12v, and 3.3V internal to one device.  Each was routed all over the place.  The end result was a good working system.  But ground situation became very complicated.  I was very concerned because EFC can easily be affected.
> 
> This is still a work in progress.  There is already a plan for 3rd iteration with GPDSO, and 4th iteration with OCXO.  I'd expect similar problems.  I'm still debating a good design as OCXO requires +/-5V supply that has to be very clean and stable.  
> 
> At one point, I plan to remake the first project.  Besides the fact it works, I didn't practice good engineering process.  Lack of foresight and slapping on STUFF as needed led to this awful configuration.  There got to be a much better way, and I got to get there for my own satisfaction.
> --------------------------------------- 
> (Mr.) Taka Kamiya
> KB4EMF / ex JF2DKG
> 
> 
>    On Wednesday, February 12, 2020, 9:49:42 PM EST, Charles Steinmetz <csteinmetz at yandex.com> wrote:  
> 
> Taka Kamiya wrote:
> 
>> I started with 24V linear power supply (open frame type).  This is the only power supply in the box.  From there, it branches to Rb, a small dist amp, and a simple interface circuit.  Dist amp is 12V rated and interface is 5V.  I modified the dist amp by including a 3 terminal regulator.  *  *  *  I crated a 5V source right at Rb's interface board to be used as EFC power.  This is the only purpose of this branch.  I also created a separate 5V for the simple interface circuit.
>> 
>>   *  *  *  EFC circuit was fine but digital interface board killed the regulator. Put a heat sink on it and issue is resolved.  A bit surprising but it shouldn't be....  The circuit only uses 25mA or so but voltage drop is huge.  *  *  *  I will have to consider use of small smp board for large voltage drops.
> 
> Good idea to use two, 5 V supplies in this situation.
> 
> The EFC circuit presumably draws no more than 0.5 mA, so regulator power 
> dissipation is probably at most 19 V x 0.5 mA, or <10 mW.  As you found, 
> no problem.
> 
> At ~ 25 mA, the interface board dissipates ~ 19 V x 25 mA, or ~ 500 mW. 
>   As you found, nothing a moderate heatsink can't handle.  But you don't 
> have to dissipate all 500 mW in the regulator.  For example, you could 
> feed the 5 V regulator from the 24 V supply through a 400 ohm, 1 W 
> resistor (putting a capacitor of, say, 470 uF to ground at the input of 
> the 5 V reg).  This would share the ~500 mW dissipation equally between 
> the 5 V regulator and the resistor.  (A 10v, 1 W  zener diode would also 
> work.)
> 
> Another option would be to use two, mains-powered linear power supplies 
> -- one 24v supply, and one 12v supply.  Then sub-regulate the two 5v 
> supplies from the 12v supply rather than from 24v.
> 
>> My only remaining concern is if regulator fails short, then what happens??  It will kill the particular device, which I'm fine with, but what else?  Perhaps simple fuse is in order for every branch.
> 
> A fuse by itself will not necessarily open in this situation.  If you 
> are worried about failing short, use a simple zener diode crowbar with 
> the protective fuse.  Use a shunt zener with a breakdown voltage about 
> 2v greater than the regulated power supply voltage across the load, fed 
> through the protective fuse.  The fuse should be rated at about 200% of 
> the load current.  The zener must be sufficiently robust to withstand 
> the fault current until the fuse opens.
> 
> With a zener rated 2v greater than the operating voltage, there is a 
> good chance the downstream load will survive without damage.
> 
> Best regards,
> 
> Charles
> 
> 
> 
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