0
\$\begingroup\$

I'm looking at coupling the output of two computer PSUs, driving a fairly stable high power project (around 300-400W peak) most likely using all output rails (12V, 5V, 3.3V) all well within the rated limits of a single PSU. Ignoring issues balancing load between rails on a single PSU, I want to use a second supply as a backup to take the load if the first one fails.

I have two questions for the electrical engineers out there:

1) can I simply use two high current power diodes as simple backflow prevention to allow failover to occur reasonably transparently, by using a common ground and tapping off +12V between the diodes below?

+Vin (1)   ~ o----->|----o----|<-----o ~   +Vin (2)
                         +Vout

I'm thinking of using one or more bridge rectifiers to do this, purely from a cost perspective - they're cheap and ready to use (rated ~50A, maybe more). I'd be applying +V from each PSU to the each of the two AC inputs and taking the output off the + terminal. Is there any problem inherently with this design?

2) I'm having trouble understanding the datasheets, I'm trying to work out (a) what the forward voltage drop will be if I feed in 12V, 5V or 3.3V, hence what the heart dissipation requirements will be, and (b) determine if the voltage drop or heat dissipation requirements are going to be too high for my requirements, in which case I'll work out a relay-based solution instead. Take this one for example:

http://www.farnell.com/datasheets/2341003.pdf

It lists "Forward voltage per leg" as 1.2V, is this the voltage drop? At what temperature? Or is it the minimum forward voltage?

Should I be looking at Schottky diodes instead, as a 1.2V drop is significant on a 3.3V line? Are there any other options I could consider?

\$\endgroup\$
2
  • \$\begingroup\$ shottkies are better for voltage drop than silicon. Best of all are FETs, but you will need to drive them on, so more complicated, and more expensive than simple diodes. \$\endgroup\$
    – Neil_UK
    Commented Mar 8, 2018 at 10:06
  • \$\begingroup\$ Yeah, I was trying to keep this side simple, but FETs are another option, you're right. \$\endgroup\$
    – pcdev
    Commented Mar 8, 2018 at 10:35

2 Answers 2

1
\$\begingroup\$

1) Is there any problem inherently with this design? - To maintain optimal cross regulation, the feedback voltage reference for the SMPS control loop is derived generally from one or more secondary outputs which have the lowest voltage/s and the highest power drawn. Having said this, your scheme could work well if the load connected to the output rails are 'not' voltage sensitive and can tolerate the series diode drops (in your suggested scheme, one diode drop per rail irrespective of the diodes selected). Adding Schottky diodes (or similar low Vf Diodes) could likely affect the circuitry powered; especially on the 3.3V rail. Note that the forward voltage of Schottky diodes increase with higher currents drawn. Since your application is meant as a 'back-up' (i.e. Output load is within the power rating of a single Power supply), you needn't ensure that voltages on each of the regulated outputs (especially on the high current rails) being diode-'OR-ed' are closely matched.

2) "Forward voltage per leg" as 1.2V - This implies that while drawing 25A through each diode instantaneously, this single diode would drop an absolute maximum of 1.2VDC; as per your proposed scheme of using 2 of 4 diodes in each Bridge.

\$\endgroup\$
1
\$\begingroup\$

If you use Schottky diodes then the voltage drop will be significantly less, perhaps 600mV rather than 1200. This will be significant for a 3v3 rail, which will drop to 2v7. Running power supplies in parallel like this is quite safe provided the output voltages are within a diode drop of each other, which is very likely the case.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.