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  #11  
Old 02-28-2012, 11:08 PM
roee roee is offline
 
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Quote:
Originally Posted by paul330 View Post
... I believe that all that is needed now is a resistor on the positive side of the capacitor circuit to limit the charging current. BUT, won't that drop the voltage to the transponder when main power is removed? So the next question is - would the resistor have the desired effect if placed on the earth side of the capacitor?

CB
|
Diode
|
| ------ GTX327 ------ Earth
|
Capacitor
|
Resistor
|
Earth

Somehow, I don't think that works
...
Paul,

Yes, the resistor will limit the inrush current. The peak inrush current through the capacitor would equal Vbuss / R, where Vbuss is the buss voltage and R is the value of the resistor. Note that the value of the capacitor doesn't factor in to the peak inrush current. The value of the capacitor only affects the time duration for the inrush current to taper off.

Yes, the resistor will also incur a voltage drop to the transponder when drawing current from the capacitor when the buss drops out. The voltage drop across the resistor is equal to Iload * R, where Iload is the transponder load current and R is the value of the resistor.

It makes no difference on which side of the capacitor you put the resistor. They're in series, so all current flowing through one has to also flow through the other.

Since you seem to be enjoying this , have you done the math to select the value and power rating for the resistor?

Anyway, for educational purposes only.

I agree, if you do keep your existing buss architecture, don't bother adding this complexity just to keep the transponder from dropping out. Just live with having to reboot the transponder. No big deal.

Good luck,
-Roee
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  #12  
Old 02-29-2012, 12:57 AM
paul330 paul330 is offline
 
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OK, I think this works



1000ohm-1/4W resistor

Initial charge current = 14/1000 = .014A giving .014x14 = 0.2W

On-line calculator shows initial charge time to be about 1 minute

And you are right, Roee - I am enjoying this. Fills in the time between 3am and breakfast nicely when you are in a hotel far from home.......
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  #13  
Old 02-29-2012, 03:03 AM
roee roee is offline
 
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Quote:
Originally Posted by paul330 View Post
OK, I think this works
Yep, I think your circuit works. Kudos, Paul!

Two footnotes to your circuit:

1) With the diode you added, there will still be a voltage drop from the capacitor to the transponder when powered from the capacitor. But unlike with a path through a resistor, now the voltage drop will be fixed at approx 0.4V (assuming the diode is a Schottky) more-or-less irrespective of the load current.

2) Step back and look at the circuit you drew. You've independently arrived right back at the "diode OR'ed" circuit topology, i.e. two power sources (the ESS buss and the capacitor) permanently connected to a load (the transponder) through a pair of diodes. That's an often seen topology, for good reason

Quote:
Originally Posted by paul330 View Post
And you are right, Roee - I am enjoying this. Fills in the time between 3am and breakfast nicely when you are in a hotel far from home.......
Now, I do believe you're enjoying this mental exercise, so let's continue to improve the circuit.

One downside of your circuit is that the transponder will effectively see two diode drops (i.e. 2 x 0.4V) relative to nominal buss voltage when powered from the capacitor. That's because there's the obvious one diode drop for current flowing from the capacitor to the transponder. But there's also the one diode drop for the current flowing from the buss to the capacitor in the first place, so the capacitor will only be charged to the buss voltage minus 0.4V.

So can you think of any simple way to eliminate the second diode drop when powered from the cap?

Have fun!
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  #14  
Old 03-01-2012, 02:49 AM
paul330 paul330 is offline
 
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Quote:
Originally Posted by roee View Post
One downside of your circuit is that the transponder will effectively see two diode drops (i.e. 2 x 0.4V) relative to nominal buss voltage when powered from the capacitor. That's because there's the obvious one diode drop for current flowing from the capacitor to the transponder. But there's also the one diode drop for the current flowing from the buss to the capacitor in the first place, so the capacitor will only be charged to the buss voltage minus 0.4V.

So can you think of any simple way to eliminate the second diode drop when powered from the cap?

Have fun!
Yes I can! Worked it out with my First Officer on the flight home - don't think we missed any ATC calls

Move the diode in the main power circuit to the right of the junction for the resistor ie take it out of the charge circuit. However, this will result in a voltage leakage back through the resistor during the transient - less than 0.4V though. This can be further alleviated by increasing the value to (say) 5k. This reduces the leakage to virtually nil. Charge time would be about 4 minutes.

We also worked out some solutions using transistors to switch circuits and eliminate the diodes completely.
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  #15  
Old 03-01-2012, 03:42 AM
paul330 paul330 is offline
 
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Default Flame Suit On

I stand accused of firstly designing an electrical system with a single point of failure and secondly modifying one of Guru AeroElectric Bob's circuits without just cause. Here is the case for the defence:

The relay in question is NOT an automotive headlamp dipper relay. It is a high quality item with 30A rating and a quoted life of one million operations. When power is taken off the control circuit, it returns to the NO position by a cantilevered leaf spring. It's a large flat sheet of metal and simply cannot do anything else - no coil springs to break or jam. It would take a huge current to fuse the contacts closed and this could only come from a short on the ESS BUSS. In which case the ESS BUSS isolate switch breakers would trip (rated lower than Avionics Master). If there is a short on the ESS BUSS then it is lost anyway but this system protects the avionics buss. In any case, the relay can be by-passed by the AUTO/OFF/ON switch to manually restore power to the ESS BUSS in the event of some other failure mode. I should also mention that (not shown in the diagrams below) there is a 2A fuse in the control circuit for the relay. In the event of any issues there, the fuse will blow putting the relay back to the NO position and assuring ESS BUSS power.



Now with loss of Main Buss (say Master Relay fail):



In the event of some sort of internal failure of the relay:



I should further point out that I have a dual Skyview system with back-up battery on the main screen and naturally I have split my equipment between the Avionics and ESS Busses. Worst case is a short on the ESS BUSS which trips the Isolation Breaker and the fuse. This would still leave me with SL30 (NAV/COMM) and second Skyview (giving flight/engine instruments and GPS navigation) thus assuring IFR capability (Fail Operational). Main Skyview is available for 60 minutes. Worst case double failure results in total electrical loss but still leaves the main Skyview with flight/engine instruments and VFR GPS for 60 minutes (Fail Safe).

I am open to comment, especially if someone can show me a single failure which causes loss of both ESS and Avionics Busses. In any event, the parts for all this cost a few bucks although the mental gymnastics required to work it all out aren't factored into that! It would be easy to ultimately replace it with the standard always-on/diode protected circuit. Somehow (although I can't prove it) that set-up seems to me more prone to loss through a major short.

The defence rests......
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Last edited by paul330 : 03-01-2012 at 03:52 AM. Reason: Factual correction
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  #16  
Old 03-01-2012, 03:59 AM
roee roee is offline
 
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Quote:
Originally Posted by paul330 View Post
Yes I can! Worked it out with my First Officer on the flight home - don't think we missed any ATC calls

Move the diode in the main power circuit to the right of the junction for the resistor ie take it out of the charge circuit.
Well done, Paul!

Your proposed change to the circuit is exactly what I would do to eliminate the second diode drop from the cap to the transponder.

Quote:
Originally Posted by paul330 View Post
However, this will result in a voltage leakage back through the resistor during the transient - less than 0.4V though. This can be further alleviated by increasing the value to (say) 5k. This reduces the leakage to virtually nil. Charge time would be about 4 minutes.
And you're correct that this does introduce a minor leakage through the resistor, but your explanation of it needs further discussion. There will indeed be a leakage through the resistor, but I'm not sure what you mean exactly by a voltage leakage. The issue is that with the resistor being upstream of the main diode now, when the buss drops out, current will leak from the capacitor through the resistor back to the buss and down to the other loads on the buss. The capacitor is no longer completely isolated to just the transponder. But note that this leakage will have no direct effect on the voltage at the transponder, other than that the current leakage through the resistor to the other buss loads will contribute to discharging the capacitor a little bit faster. But since you can tolerate a long charge time and have therefore chosen a high resistor value, the leakage current will be relatively small compared to the transponder's load current, therefore having no significant effect on the capacitor discharge time. The max leakage through the resistor is limited to Vcap / R, so at V=12V and R=1Kohm you're only talking about 12mA (and in reality will be even less, that is the short circuit worst case). And as you pointed out, the bigger the R, the smaller the leakage. True. But even at R=1Kohm, 12mA leakage is just insignificant. So we're agreed, with the values you've chosen, this leakage issue is a non-issue.

Quote:
Originally Posted by paul330 View Post
We also worked out some solutions using transistors to switch circuits and eliminate the diodes completely.
Yes! You can eliminate the diode drops altogether by using transistors instead. But then the circuit design and analysis will get much more involved. Extra credit if you can pull this off!

And my suggestion would actually be to do it with FETs (field effect transistors) rather than BJTs (bipolar junction transistors, or simply "transistors"). Modern power FETs have incredibly low on-resistance and therefore very high current-carrying ability. And FETs are, generally speaking, much friendlier devices to design with than BJTs. If I were to do a modern clean-sheet design for the electrical system, it would definitely use FETs for the main power switching.

P.S. You're getting good at this! Want to trade jobs for a week? What do you fly?
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  #17  
Old 03-01-2012, 04:46 AM
PaigeHoffart PaigeHoffart is offline
 
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Default Is it aircrews proof?

Your topology looks very similar to a Boeing design...we actually checked the circuit during ground ops, each flight. In about 1500 hours, I saw 2 failures of the relays...which probably had 18,000 hours on them.

One potential gotcha that I see:
Let's say that you are flying along and experience an alternator failure. You are aware of the potential switching transient, and decide to play it safe and turn standby power on before you kill the master switch. When you kill the master, your relay will remain engaged, drawing power through the essential buss...your entire electrical load will flow through the standby power feed.

If there is a good load, a fuse will blow, or a CB will pop. Fuses are generally faster acting than CBs, so I wouldn't be surprised if the 25A battery fuse blows. Now you are stuck with turning your master switch on, and manually shedding loads.

If the load remains low enough to keep from blowing a fuse or CB, you will just waste power (relay holding current and other loads that you probably thought were shed when you killed the master)

So how do you prevent this? You could add a caution to the POH, or you could design that failure mode out by wiring your relay coil through a second pole in your selector switch. (what's the current rating on that switch?)

BTW, with a "diode or" circuit, there are gotcha's also...a transient essential buss short could take out both feeds.



Paige
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  #18  
Old 03-01-2012, 07:06 AM
paul330 paul330 is offline
 
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Quote:
Originally Posted by PaigeHoffart View Post
Your topology looks very similar to a Boeing design...we actually checked the circuit during ground ops, each flight. In about 1500 hours, I saw 2 failures of the relays...which probably had 18,000 hours on them.

One potential gotcha that I see:
Let's say that you are flying along and experience an alternator failure. You are aware of the potential switching transient, and decide to play it safe and turn standby power on before you kill the master switch. When you kill the master, your relay will remain engaged, drawing power through the essential buss...your entire electrical load will flow through the standby power feed.

Paige
Yes, you are right. I haven't shown all the failure modes. The idea with a main ALT failure would be to leave it in AUTO - initially, you have lost nothing. Then looking at your flight time and load requirements, you start the standby generator and accept a drain on the battery or switch off the Batt Master and run on Standby Power indefinitely - yes, the POH would have to be written accordingly. The way I have the circuits configured, the main drains off the Main Buss are the external lighting which can be switched back on for landing.

Funny, I am flying the B747-400 so maybe that is where I got the idea from. A lot of aircraft I have flown have had the AUTO/OFF/ON algorithm and I think it is pretty sound.
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  #19  
Old 03-01-2012, 07:18 AM
paul330 paul330 is offline
 
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Originally Posted by roee View Post
Yes! You can eliminate the diode drops altogether by using transistors instead. But then the circuit design and analysis will get much more involved. Extra credit if you can pull this off!
So, you replace the diode on the discharge circuit with a NOT gate - controlled by a feed from the charge circuit upstream of the resistor.

I am going to be in SFO next week and there is a Radio Shack at the bottom of the street from our hotel. For a few bucks, I am going to buy the bits and build this design (just resistors and diodes) and see if it works.

And sorry, flying B747-400 out of Hong Kong and not prepared to swap....
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  #20  
Old 03-01-2012, 09:40 AM
krw5927 krw5927 is offline
 
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What would happen if the negative terminal screw into the battery were to come loose and you lost that connection? Would either bus work? Would it dump charge into the airframe with a steadily decreasing potential (voltage) between the positive terminal (or alternator, or stby generator) and the airframe?

Wondering if there's not still a single point of failure in our single battery systems.

Of course you'd still have your Skyviews with backup batteries, but would anything else be operational? Or would it not matter one bit and everything would continue to work?
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