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Old Mar 11, 2012, 11:05 AM
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Zener diode question

Hi and thanks for looking.

I have installed two LiFe (6.6v, 2100mAh) batteries in my 33% Edge, each with it's own switch and feed to a Futaba R6014HS Rx.

I have modified the switches to include a LED (on the output side) to show me that each pack is online.

Because Futaba Rx's use a bus for power to all channels, when I switch one battery pack on, both LED's light because of the 6v being supplied from the Rx.

I understand installing a Zener diode (5W, 6.8v) will resolve the problem but my question is on how to install it.

Do I simply put it inline on the +ve feed to the Rx (between switch and Rx)?

Many thanks

Goose
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Old Mar 11, 2012, 12:36 PM
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Waterloo, Belgium
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Quote:
Originally Posted by V6Goose View Post
Hi and thanks for looking.

I have installed two LiFe (6.6v, 2100mAh) batteries in my 33% Edge, each with it's own switch and feed to a Futaba R6014HS Rx.

I have modified the switches to include a LED (on the output side) to show me that each pack is online.

Because Futaba Rx's use a bus for power to all channels, when I switch one battery pack on, both LED's light because of the 6v being supplied from the Rx.

I understand installing a Zener diode (5W, 6.8v) will resolve the problem but my question is on how to install it.

Do I simply put it inline on the +ve feed to the Rx (between switch and Rx)?

Many thanks

Goose
No zener diode, a schottky diode between switch and Rx, twice.
Led just after switch.
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Old Mar 11, 2012, 01:50 PM
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Thanks Ramboman...
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Old Mar 12, 2012, 02:30 PM
Dave the Rave
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One other thing, make sure the current rating of the diodes you use is large enough to handle the current load. I notice your plane is a big one, a 33% Edge, so consider what all your servos might draw in a situation where you're moving them all at once, like an extreme 3D maneuver, you'll want the diode(s) to be able to handle the total current load. If they fail, it will be like turning the switches off, not a good thing at all.
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Old Mar 13, 2012, 01:14 AM
Airlady
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Very True
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Old Mar 13, 2012, 01:28 AM
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Not really.

When diodes fail from overcurrent they become permanent short circuits, conducting in either direction with virtually no voltage drop. If this happens, the worst consequence will be that "Goose" will have the same situation he has now - both LEDs will light when either switch is on.

The current level needed to actually burn a diode in half and create an open circuit is much higher than the level that causes it to short circuit.
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Old Mar 13, 2012, 01:34 AM
Airlady
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Maybe
But ive had diodes just crap out maybe due to manufacturing Brand - a assy line run flaw etc.... even repaired quite a few computer motherboards ( wont name ) because of this problem
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Old Mar 13, 2012, 01:40 AM
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I'm not so sure that diodes are the best idea anyway. It depends upon what sort of problem "Goose" is trying to detect. If he just wants to check the LEDs to make sure he remembered to turn both switches on, it might be better to use double-pole switches with the LED circuits wired to one side and the plane's electrical system wired to the other side.
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Old Mar 13, 2012, 02:02 AM
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Quote:
Originally Posted by V6Goose View Post
................
I understand installing a Zener diode (5W, 6.8v) will resolve the problem but my question is on how to install it.

...................
Just what is the problem ?
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Old Mar 13, 2012, 09:03 AM
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There is no need for zener diode. To isolate the two batteries (and so the two leds) you can use two MBR 745 high current schottky diodes, as shown in the attached diagramm. The same thing like power supplying RX and servos of a twin using the BECs of both speed controls.
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Old Mar 13, 2012, 11:14 AM
Oxford Panic
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Don't forget the resistors needed in series with the LEDs to limit the current.

A.
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Old Mar 13, 2012, 12:05 PM
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Originally Posted by Jimbello View Post
Just what is the problem ?
Exactly.

You can't identify the best solution to a problem until you know what the problem is.
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Old Mar 13, 2012, 12:55 PM
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Originally Posted by AndyOne View Post
Don't forget the resistors needed in series with the LEDs to limit the current.

A.
HOOPS!!! Yes you are right. I forgot the resistors in series to each led.
For 6.6 volts max 6.8 (under load not during charging) of a LiFePO battery a 220-330 ohm resistor in series to each of the leds will be fine.
I have just added the resistors to the diargamm.
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Old Mar 14, 2012, 01:27 PM
Dave the Rave
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Quote:
Originally Posted by Miami Mike View Post
Exactly.

You can't identify the best solution to a problem until you know what the problem is.
He told us what the problem is in the original post. He's trying to use two batteries and two switches (for redundancy), he's added an LED to each switch, but when he plugs either battery into the receiver, both LEDs light up. Of course that's because the power leads in a Futaba receiver are on a common buss. He wants to isolate each battery/switch/LED from the other.

The circuit bladeskiller posted works fine, with the added advantage that you can connect it all up with a servo wye plugged into a single channel on the receiver, no need to use up a spare channel for the other battery/switch.
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Old Mar 14, 2012, 04:05 PM
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Originally Posted by dmccormick001 View Post
He told us what the problem is in the original post.
No, not that problem. That may be what Jimbello meant, but I'm asking what the problem is that he's trying to detect. What sort of failure is he trying to identify and prevent with this LED circuit? See my post #8.
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