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Author Topic: F-16 fuselage  (Read 74678 times)

Offline simpit

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Re: F-16 fuselage
« Reply #375 on: November 08, 2017, 06:46:28 AM »
Very nice.

I like the switching box. I think I am following your lead on that. Do you have pictures of other parts of your powering circuit, especially how are you branching out from the main supply wire.

Cheers

Tulkas

Hi Tulkas,

see the picture above in my post from 3. September 2017 (http://www.viperpits.org/smf/index.php?topic=10598.msg162073#msg162073). there you perfectly see the methode of wiring. Rebuild it on your own risk  ;) - im not an expert in electric skills (ask henkie!!!).

@henkie:

After connecting serveral real panels I have a strong impact of voltage at the end of my 1,5m - 2 m long wire (1,5mm diameter), due to the resistance of the wires and the low voltage of 5V. Every panel leeches about 1 Amp current. I think there will be no other way as to install for every panel a separate wire. What do you think?

Have a nice day

Hans Dieter

Offline henkie

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Re: F-16 fuselage
« Reply #376 on: November 08, 2017, 03:58:11 PM »
Indeed, easy wiring if going from one panel to the next panel.
But easy and good is not always going well together  :whistle:

Assume you have five "consumers", and each consumer draws 1 Amp.
The wires to the last consumer only have to transport 1 Amps, but the wires going to the one-but-last needs to transport the current for the one-but-last *plus* the last consumer, thus 2 Amps.
So, if you have five consumers, the wires from the power source to the first consumer must be rated for 5 Amps (at least).

Also, always keep in mind that high currents *require* low resistance in its connection(s).
Ohm's law makes this clear. In the following formulas U stand for Volts, I stands for current (Amps) and R stands for resistance (Ohms). P stands for power or energy.
Ohm's law says U = I * R  (1)    and  P = U * I   (2)
So, if you you have a connector transition resistance of "just" 0.1 Ohm your voltage drop across that connection will be according to formula (1): U = 5 Amps * 0.1 Ohm = 0.5 Volt.
You can see that as current increases the voltage drop (loss) will also increase.

But there is an other aspect that might be even more important, and that is energy (read: heat due to losses). Using formula (2) and substituting U using formula (1) you get
P = (I * R) * I = I^2 * R. Thus, energy loss increases linearly when the resistance increases, but the loss increases with the square of the current.
In the above example, current is 5 Amps and resistance is 0.1 Ohm. The energy loss will be 5^2 * 0.1 = 2.5 Watt. That 2.5 Watt will cause the connection to become "hand-warm".
If the current is higher you could risk some charring or even a fire!

Conclusion.
Try to keep currents low. That minimizes loss (voltage drop) and energy loss (heat development).
So, it would be better to wire each panel directly to the power source. I know, that is more wiring, but as I said easy and good not go well together  ;D

Especially with *real* panels. Those bulbs for backlighting do suck some *real* juice!
Measurements of a pit with all panels real show a total current of 30+ Amps at 5V!  ::)
Obviously thát current requires (very) good handling ...

Offline Rufus

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Re: F-16 fuselage
« Reply #377 on: November 08, 2017, 09:04:23 PM »
Well explained, Henkie... :thumbsup: ...this is info we can all use!
- Rufus