[time-nuts] Using a common power supply among few time standards
Charles Steinmetz
csteinmetz at yandex.com
Thu Feb 13 02:48:48 UTC 2020
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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