|
|
bob mackey wrote:
I am purchasing a 12V power supply that takes
6-24V input and gives 12V out at up to 14 A continuous.
It's a 95% efficient switcher with a quiescent current
around 10mA.
I hope you are ready for a nasty surprise when you actually get one and try it. I have designed over a hundred switching power supplies from a few hundred milliamp to several 10's of amps. The only ones that were in the range of 95% were step down supplies with no reverse input protection, a relatively fixed input voltage and a fixed lower output voltage. When you add input protection for reversed voltage and variable input voltage, the efficiency starts to go down. When you make it step up (output voltage is higher than the input voltage) the efficiency goes down quite a bit more. When you make it a step-up/step-down, the efficiency goes to hell in a hurry.
During cranking, the bus voltage drops
to around 10 V, and the avionics brown out.
Leaving the avionics off is the easiest solution. Going to a 24V system would be easier than what you are contemplating. If the avionics is so critical that you have to have it on during engine start, you really can not afford the risk of having a component of unknown reliability, like this DC-DC, supplying its power. Yes, you can add diodes and cross-feeds to handle the failure, but by the time you are done, you have given up either efficiency, or reliability or likely both. You also need to consider that unlike a resistive load where the current goes down as the voltage decreases, the a DC-DC draws more current as its input voltage decreases. If you are delivering say 10A to a load at 12V and have an 80% efficiency (a more realistic number), you are looking at drawing 15A from the input. Do you really think your battery voltage will stay at 10V if you throw in an additional 15A of load to it? It will go down and as result the current draw will increase even more. How will the starter like the lower voltage?
I want that
problem to go away, and also want to keep the radio working
for a while if the alternator shuts down and the battery drains
below 10V.
How long do you think your 12V battery is going to last once it gets down to 10V as you draw the aforementioned 15A from it? An additional minute? Maybe two if it is your lucky day. You can get far more minutes than that for less money and weight by going to the next larger size battery. That will be significantly more reliable.
Let me end with a little math for a quick sanity check. At 12V and 14A, you are delivering 168W to the output. At 95% efficiency, you would be drawing an average of 29.5A from the 6V input. Lets assume that the parasitic resistance in the current path is a miraculously low 0.010 ohms (including wires, traces, 2 sets of MOSFETs in series, at least one inductor plus other stuff). This means that just the resistive power loss would be 29.5*29.5*0.010=8.7W or 5.2% In real life, you will not be able to get the resistance through the system anywhere near that low or even twice that. Then there is the input protection device, the capacitor ESR losses for both the input and the output caps and MOSFET switching losses which get bigger as the MOSFET resistance goes down. The above analysis also assumes that the power supply is a multi-phase supply optimized for this voltage. Such a supply will have a lower efficiency at other voltage. I can go on with lots of second order effects that can contribute up to 50% in the overall loss, but I do not want to make this into a lecture on power supply design.
Regards,
Hamid
|
|