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Q light system modifications


neilalexrose

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I throw myself on the collective wisdom of the Blue Room once again on matters just outside my area of expertise.

 

I'm looking at making some new out stations for our simple Q light system. It currently uses 12v filament lamps that I am looking to replace with LED fittings, specifically these from CPC, rated at 12v 20ma.

 

So, where I struggle is knowing which resistor to put in series with the LED lamps, so they don't blow up every time they are used.

 

It's an old fashioned q light system with a 24vDC power supply, and the master station and out station 12v lamps are wired in series, to complete the 24v circuit.

 

What I think I need is a resistor that will resist the current down to the 20ma the LED requires, so my calculations are as follows:

The LED requires a voltage of 12v down from 24v, so the voltage drop divided by the current is 12v / 0.02a = 600ohms.

 

Is this right? Am I miss-leading myself by thinking the voltage drop is 12v, when the two lamps are 12v in series?

 

Should I be looking at putting a 12v Zener diode in series with the resistor, to keep the current low for the LED lamp, and providing a bit of protection? Or would this muck up the voltage for the entire circuit?

 

All help great fully received I this electronics matter!

 

Thanks

 

Neil

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OP unsure if you mean to relamp I) just the outstations or ii) both the master and the outstations

 

 

I) will be hard work,

 

, to make the master station work you must maintain the current appropriate to its lamp in the whole circuit, you can work out the current from 12v and the power of the lamp W /V = A

you then need a resistor that is in parallel to the LED outstation lamp such that the extra current required above the LED will flow.

 

example

LED 10mA master lamp 60mA then you will need 50mA lost in the parallel resistor

find the resistor, Ohm's Law V = I R >> 12/ 0.050 = R = 240 The resistor will dissipate W = V x A 12 x 0.050 = 0.6W so to be sure I would use a 1 W wire wound

 

It would be best to put the resistor as physically close to the LED as possible, the lamps were always wired in series so if one failed the DSM would see a non illumination and transmit the cue another way

It was also traditional for the RED (standby) to have a push button break switch so the talent when given a standby would press the switch for a few flashes so the DSM would know the talent was ready.

 

You can also consider that the resistor is using up all the electricity that we have saved by going LED - lamps in series should be similar!

 

 

ii) Easy Just change both lamps in each circuit and off you go

 

 

the lamps you mention already have a suitable resistor inside to make them work on 12V so thats easy they are 12 V lamps

 

 

Should lamps of different current requirement be in unprotected series you have a potentiometer chain and the voltage across each will vary inverse to its current, so the LED will be overvolted and go bang then nothing will work.

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Hi.

 

Thanks for the replies, both very helpful. The purpose of this is to make extra out stations to add to an existing system, so adding a resistor in parallel is the way to go. Which in fact tallies with some existing LED q lights we already have, that we opened up today and discovered a 120ohm resistor in parallel with those LED fittings (not the same as we want to use here.)

The existing filament lamps operate at 100ma, so the current drop is 80ma. Therefore the maths is: 12v / 0.08a = 150 ohms, and 12v x 0.08a = 0.96w. So I've picked a 150ohm 2w resistor to go in parallel with the LED fitting.

 

Much as we would love to re-wire the entire system and convert to LED (as per whiskers's option 2), we just don't have the time or manpower to do it in the turnaround between the next 2 shows. We just need a couple of extra out stations to hopefully cover a 3 show rep.

 

The next question I have is about making the red lamp flash, and then turn solid once the PTB switch is pressed. I have an old circuit diagram that shows a diode in parallel with the red lamp, and in practise this makes the red lamp flash. (I know this works, as a wireman I know made all the q lights for the venue I used to work at). My question is, would this work if the 1n4001 diode would do the same thing if wired in parallel with the LED lamp. (Diode found from a bit of motorcycle forum research, it's how they sometimes fit aftermarket indicators)

 

Schematic of the proposed out station:

http://i1338.photobucket.com/albums/o700/neilalexrose/image.jpg1_zpsbqlksx63.jpg

 

 

Thanks

 

Neil

 

 

EDIT to add proposed schematic.

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Sorry NO the flashing will not work like that

 

The main unit in the case of your old venue is more complex and sends AC

 

it "multiplexes" the wires by using the diode to separate the AC into two DCs one lights the lamp and the other sets a relay in the main unit.

 

If you do not have time to change the main unit indicators you certainly do not have the time to introduce all the other whistles and bells

 

In truth I have always felt that extra complexity introduced more fault opportunities for quite a small gain in functionality.

 

KISS = Keep It Simple Stupid

 

On the bright side the sums look good, just dump the diode, you will know if the master is common - or common + and could reverse the diodes if required

 

or if the time fairy really bites just more incandescents and no more resistor shenanigans

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