circuit help

[Karl W4KRL]

Mike,

A probable flaw in my scheme, Rob's, and yours is that the OFF LEDs will be
back biased by up to 24V due to a reverse path through each load impedance.
Most LEDs won't tolerate that. Off hand, I do not see any way to overcome
this defect in any scheme.

Karl

-----Original Message-----
From: tacos-bounces+w4krl=arrl.net at amrad.org
[mailto:tacos-bounces+w4krl=arrl.net at amrad.org] On Behalf Of Karl W4KRL
Sent: Tuesday, July 08, 2014 4:25 PM
To: 'Mike O'Dell'; tacos at amrad.org
Subject: RE: circuit help

Mike,

Use an emitter follower to RAISE the voltage on the common side of the LEDs.
The potentiometer R1 with resistor R2 form a voltage divider with the low
end of the pot at half the supply voltage = 12V. When the pot is at the low
end the emitter will be at 12.6V and 11.4V (24 - 12.6) will be available to
light the LEDs. When the pot is at the top end Q1 will be full ON and the
LEDs will be dark. The brightness varies from dim to bright according to the
position of R1. You could put a small resistor in series with the top end of
R1 to limit the minimum dimness and you could raise the value of R2 to limit
the maximum brightness. Since the emitter voltage follows the base voltage
plus one diode drop more or less independently of load current the voltage
to the LEDs will not be affected by the number that are lit. 

R3 is necessary to provide a current path to ground for the LEDs. The value
depends upon the number of switches you have. For example if ONE LED in ONE
switch draws 20mA at 12V then R3 must supply 60mA at 12V if you have three
switches. R3 = 12V / 0.060A = 200 Ohms. At full DIM it will see 24V so it
must be rated for 24^2 / 200 = 2.9W. Use a 5W.

73 Karl W4KRL

PS the schematic is drawn in ExpressPCB.

-----Original Message-----
From: tacos-bounces+w4krl=arrl.net at amrad.org
[mailto:tacos-bounces+w4krl=arrl.net at amrad.org] On Behalf Of Mike O'Dell
Sent: Tuesday, July 08, 2014 3:30 PM
To: tacos at amrad.org
Subject: circuit help

I have a design problem that's been driving me batty because it seems light
there should be a reasonably simple way to solve it but I can't see it.
Sigh.

The Problem:
There are a number of LED-illuminated SPDT ON-OFF-ON switches, green for one
position, red for the other. The LEDs are fed from the two switched
terminals via a dropping resistor per LED sized to work switching 12VDC. The
switch is designed to switch the + side of the circuit between the two
positions and there is a common - connection for the bottom of the LEDs.

Normal use case: +12VDC to the moving terminal, ground the common LED pin,
and take switched 12VDC off the two switched positions. Easy.

My use case:
Part 1: Similar to the normal case except switching +24VDC.
	can't change the dropping resistor inside the switch
	but can easily add an external dropping resistor to ground.
	Easy.

Part 2: The lighting needs to be *DIMMABLE*. This is for something
	in a boat dash and it *really* needs to dim down to almost
	nothing to preserve night vision. Simple minded-solution is 
	to just add a common rail for the LED ground terminals that goes
	to ground through a variable resistor (big pot or a pass
	transistor).

Now here's the tricky bit:

Part 3: We also need to keep the brightness constant independent of 
	the number of switches in either on position or off because
	we don't want want to turn off a number of them and have the
	rest of them bloom bright and kill night vision.

What has occured to me so far is to use a 3-terminal variable voltage
regulator that can withstand going to +24VDC on the input.

Feed the *input* of the regulator from the LED common rail and feed the
output to ground through a resistor sized to not exceed the current rating
of the regulator with +24VDC on the input and set for max output voltage. 

So to adjust brightness: 
Turn up the voltage setpoint on the regulator to make it brighter. 
Turn down the voltage setpoint to make it dimmer.

As LEDs get turned on or off, the effective resistance of the various
paralled LEDs (and dropping resistors) varies, but the 3-terminal regulator
maintains a constant voltage drop between its input and ground through the
load resistor. Therefore the LED common rail stays at a constant voltage
above ground in spite of how many LEDs are in parallel at any one moment. 

This gives a constant voltage drop across each LED producing a uniform
current through each LED and uniform brightness. (Ignoring red-vs-green
brightness.  One would hope the internal dropping resistors took that into
account.)


so does this make sense?

	Thanks!

	-mo




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