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 
> <[email protected]> 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 
> <[email protected]> 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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