Thanks for the help on the test button question. Got another non-electronics technician problem.
I am still working on this Sensor Circuit with LED indicators. The box I'm building, I want it to be able to be used on either system voltage of 14Vdc or 28Vdc with a spdt switch setting the box for the appropriate voltage.
So I have eight sensors, eight LEDs (some with different ratings, see below), two voltages and two states of DAYTIME (Bright) and NIGHTTIME (Dim).
The LEDs are supplied directly by the sensors. This means that each LED has it's own power supply so speak. But, all supplies will either be
14Vdc at one time, or 28Vdc at one time (you will not have both voltages at the same time).
Because of the different LED colors I have to deal with a 2V and a 3V LED at the two voltages of 14Vdc and 28Vdc, so my resistances are 550,
600, 1.25K and 1.3K at 20mA.
Its easy enough just to put a 600 ohm R inline with each LED, but how do I switch so that ALL LEDs have a 1.3K R in front of them without running all the power through my voltage selector switch (which would defeat the purpose of using the power from the sensors and having an indicator light up when power is supplied by the sensor)?
My next problem will be to DIM all the LEDs for the NIGHTTIME setting, yet a third resistive value????
I'm trying to use more passive components rather than ICs and trying to keep it simple as possible all the way, any suggestions are hugely appreciated.
--- I think the esaiest way to do it would be with 8 constant-current sources, like this:
+VIN>-----+-------+---------------------------------+ | |A | [10K] [LED] | | | | | C Q1 | +-----B 2N4401 | | E | | | | Q2 C | | 2N4401 B-----+--[R2]--+ | E | | O | [R3] Q3 C | | | 2N4401 B--[4k7]--+--> >--O--> | | | E | S1 | | | [100K] | | | | GND-------+-------+--------+-----------+
Vin is the output voltage from your sensor, and Q1 and Q2 form a constant current source where R3 determines the current through the LED. I just built one to see how well it would work, and with 39 ohms for R3 I got 17.07mA through the LED with Vin = 14V, and
17.83mA with Vin = 28V. Not bad!
R2 and Q3 do the dimming, and when Q3 is turned on by closing S1, (the common BRIGHT/DIM switch) R2 will be placed in parallel with R3, requiring more current to flow in order to bring the circuit into regulation. That means that "BRIGHT" will be when the switch is closed, and "DIM" will be when it's open.
After I tried the basic circuit I added on the dimmer, and with R2 and R3 both equal to 39 0hms, here's what I got:
Vin Idim Ibrt V mA mA 14 16.4 29.8 28 16.0 32.9
You'll have to fiddle with R2 and R3 to get the ratio of bright to dim to be what you want but, basically, it works.
Also, I forgot to show it on the schematic, but I bypassed the supply, at the circuit, with 0.1µF.
--
P.S.
2N4401\'s are rated for 625mW max, and Q1 will be dissipating about
500mW with a 28V supply, so if you decide to use this circuit you
may want to get a transistor that can handle a little more power or
put a dropping resistor in series with the LED or even glue all the
Q1\'s to a heat sink...
And, BTW, the BRIGHT/DIM switch needs to be wired to the 14V/28V
POWER switch, not to the output of a sensor, as I showed before.
Like this:
+VIN>-----+-------+
| |A
[10K] [LED]
| |
| C Q1
+-----B 2N4401 14V O--> | >--O--> |
| | E | S1
| | | [100K]
| | | |
GND-------+-------+--------+-----------+
Thanks. With how much posting you do, I really appreciate you taking the time for my simple questions. Can you tell me what the 1uF cap is for, noise, filter? I'm not skilled in electronics so I'm just curious.
I just breadboarded the circuit, if I am understanding the circuit (it came a little out of whack in alignment), R2 should be between the Base of Q2 and Collector of Q3? Also, SW1 should be supplied system voltage not supplied from the sensor?
I'm gonna go out on a limb and say that I got it right (far limb), when I pull R3 the "dimness" doesn't change, changing the value of R3 seems to have no effect on "dimness" if you will. Which asks the question (at least in my feable mind), why is it there? Should I rather mess with R2?
--- It's to lower the impedance of the supply. What happens is that the wiring between the supply (in this case, the outputs of your sensors) and the load (the constant current LED drivers) often can cause problems because of the indutance and capacitance it can add to the circuit, which can result in oscillation. Placing a capacitor across the supply, at the load, will generally stop any foolisness from happening since it will provide a nice low impedance path to ground for the potentially offending signals. It cuts them off at the knees, so to speak, before they have a chance to grow and do any damage.
That's especially important with devices like your sensors, which are supplying "courtesy" outputs capable of sourcing fairly decent amounts of current, but which aren't really power supplies.
Sorry for being such a simpleton, I have only found this forum in the last few days. You say you posted the picture on "abse", I feel like an idiot, but what is "abse"?
Also, if we get the circuit working the way it should, I probably will want to go for a higher rated transistor, any suggestions on that?
--- You should have your newsreader set to display text in a non-proportional font like Courier or Courier New and then the schematic will line up.
That end of R2 should be connected to the junction of Q2 base, Q1 emitter, and the ungrounded end of R3. Yes, system voltage goes to S1. Here's a better schematic, with R6 added to cut down on the power dissipated by Q1:
--- Something's wrong. R3 is used to set the dimness with S1 open, and R2 is used to set the brightness when S1 is closed.
I've posted a picture of the working circuit on abse so you can see what it looks like and check your wiring against it. I just noticed (after I posted it, of course) that there are two R5's on the photo. Aaarghh! The top one is supposed to be R6.
--
Don\'t feel bad, this is sci.electronics.basics (seb) where there are
no stupid questions. Abse is a newsgroup
(alt.binaries.schematics.electronic) where you can post binaries,
unlike this one (seb) where you can only post text.
The zener is 12 volts and R1 is 470 ohms, both at 1 watt. The the LED and R2 "see" 12 volts regardless of the input, so R2 is computed for 20 mA at 12 volts, or 500 ohms for the 2V LED, and 450 for the 3 volt LED. The bright/dim switch shorts out R3 in the bright position. You'll need to choose R3 experimentally for the degree of dimness you want.
A note regarding R2. There is not much difference in current if you use the same resistance for both the
2V and the 3V leds. If you pick a value of 470 ohms, the 2V LED draws ~21.27 mA. The 3V LED draws ~19.14 mA. I don't think you will be able to see any difference.
Thanks a bunch. I will do a breadboard and look at that one to. That has even less complexity over the other circuit, do you guys see advantages/disadvantes between the two?
You've want a circuit whose power supply can be switched between
14VDC and 28VDC.
You've got a sensor which turns sinks current to GND (low resistance to GND) when the light is bright.
You want to drive 8 LEDs in such a manner that, for either voltage selected, they have higher current (say, 20mA) during the day, and a lower current (say, 12mA) at night.
For some reason you want to do the job with only discrete components
-- no ICs.
The responses you've been getting from others that you should be driving the LEDs with a constant current source are good ones. Actually, a transistor current source is a fairly good "one-trick pony" which should do most of the work for your circuit.
The idea goes something like this: If you have an NPN transistor and set it up so the base is driven with a stiff voltage source, the transistor will try to keep it's emitter about 0.7V less than the base by allowing current to pass through the collector. If you put a resistor from the emitter to GND, that will mean the transistor will try to keep the steady voltage across that resistor -- in other words, constant current sink through the load. Likewise, with a PNP transistor, the emitter will be kept about 0.7V higher than the base, meaning you get a constant current source. Here's the simple way to get the eight individual LED drivers (view in fixed font or M$ Notepad):
Your largish NPN transistor will have a voltage applied to the base, and the collector current will be about 0.7V less than that. (This is, of course, an approximation, but it should be sufficient for the needs of the day.) So, if you've got a 100 ohm collector resistor, applying
2=2E7V at the base will mean 2V across the 100 ohm resistor. Since the base current is only a very small fraction of the collector current (usually about 1/40th for TO-200 power transistors), you can assume your LED current will be about 20mA too, whether the supply voltage is
14VDC or 28VDC. If you put about 1.9V at the base, you'll get about
12mA through the collector. You'll need 8 of these -- 1 for each LED.
Now let's talk about supplying the eight NPN transistor bases with a steady 2.7V in daylight (20mA) or 1.9V for darkness (12mA). I'm not sure what you're using as a light sensor, but you give the impression that it acts like a switch -- it has a high resistance to GND if it's dark, and a low resistance to GND if it's light. Of course, you could give more information if this isn't exactly what you have. But for the sake of discussion, let's assume it is a switch (SW1). That would make it fairly easy to provide the constant voltage source to the bases of Q1 - Q8 with two more transistors:
The zener diode shown would be a 1N4733, a 5.1V zener with a test current of 49mA and a knee current of 1mA. Q9 is set up to supply a constant 40 mA to the 47 ohm resistor (that's about 1.88V, for those who are following with their calculators). The base of Q10 will normally not be drawing any current, so it's off when it's dark. But when the switch is turned on (it lights up), the 5.1V appears between the base of Q10 and V+ (whether 14V or 28V), which means Q10 will supply another 16.2mA to the 47 ohm resistor. That will total 56.2mA, which means about 2.64V. Of course, the other transistors will draw typically 0.3mA to 0.5mA each, but that total 2.4mA to 4mA won't affect the voltage that much. If you find it's not quite there, tweak the 110 ohm resistor above Q9 down to 100 ohms, and the 270 ohm resistor down to 240 ohms. That will increase the drive voltages to well over 2.7V (light) and 1.9V (dark).
So that's 8 power NPNs, 2 power PNPs, 2 1N4733 zener diodes, and 13 1/4 watt resistors. I don't think you're going to reduce that parts count too much.
Again, I'm kind of wondering what you'll use as a light sensor. I get a feeling it might be a little more difficult than a switch to GND. Feel free to post again with more information.
Chris replied, and at first I wondered why he offered a complicated circuit, until I realized he's thinking about making all the LEDS either bright or dim with one switch, while my circuit, although simple, requires individual switches to dim each LED. Therefore, I'll offer a different simple circuit to dim all the LEDS with one switch, and add adjustable dimming.
R1 = 240 ohm 1/4 w R2 = 47 ohm 1/4 w R3 = 1500 ohm 1/4 w POT = 5K linear taper potentiometer Rled = 330 (will work for both 2V and 3V leds)
How it works: The LM317 is configured to supply ~9.06 volts when the switch is in the bright position, regardless of whether Vcc is 14 or 28 volts. When the switch is in the bright position, Rled limits the current to a 2 volt LED to ~21 mA; for a 3 volt LED Rled limits it to ~18 mA.
When the switch is moved to the dim position, R2 and the pot are placed in parallel with R3. The resistance from the adj pin to ground can be varied by the pot from ~1252 ohms to ~45.5 ohms, meaning that the LM317 will supply from ~7.77 volts down to ~1.48 volts. That means you can vary the current to the LED between ~17mA and 0 for the 2 volt leds, and between ~14 mA and 0 for 3 volt LEDs
You will need a heatsink for the LM317. Under worst case conditions (Vcc = 28 volts, all leds lit, switch in bright position), it will need to dissipate ~3.34 watts.
Whew! For a babe in the woods I have to read that about 20 times to "get it".
Here is the gist again.
I have both magnetic and proximity sensors. I have a total of 8 sensors, 4 of which are magnetic 4 are proximity. When the sensors close (sense), they provide a ground, not power.
That is the sensor side of things that you were wondering about.
Separate Issue: Supply will be either 14 volts or 28 volts and only want one unit that will handle both voltages. Separate Issue: I will need to be able to have 3 levels of dimming for all LEDs Separate Issue ***New***: I have had to add clusters of backlighting LEDs on the front panel. Groups consist of 4 to 6 SMT LEDS in series. Need to be able to dim these at the same time as the indicator LEDs
I didn't want to go with ICs because of static, extra driver circuits and components. But I see that that might not be an option. If necessary I may tell the powers that be that we should just go back to incandescent indicators. I want something that is almost mechanical (not literally, but you know what I mean).
I know this is simple to you guys, but I'm not above admitting I may be in over my head on this. I thought I had the schematic wrapped up and about the same time I got the reply I realized that all of the grounds and the dimming circuit were tied together and that it wouldnt work that way.
One more thing. I need to try and pack all this onto a 3.036" square.
Thanks again guys (did I say anything that made sense?) Gman
Up to this point, all of the above requirements have been met by the designs posted.
3 levels of dimming is a new requirement - and is easy to meet. See below.
We can work that out as a separate but related issue. First we need to know: What is the voltage for these leds? How many groups?
Second: Your groups will be 4 LEDs each if they are 2 volt leds, and 3 LEDs each if they are 3 volt leds.
The ground that the sensor supplies to the circuit *MUST* be common with the 14 volt or 28 volt supply ground. The circuits posted after you told us that the output of the sensor is ground will work fine.
3 LEVEL DIMMING SCHEMATIC:
-------------------------
----- Vcc---in|LM317|out---+--------------+-------+----//----+ ----- | | | | adj [R1] [Rled] [Rled] [Rled] | | | | | +---------+ [LED1] [LED2] [LED8] | | | | | Sensor1 Sensor2 Sensor8 | (Sensors connect to ground when active) | | | dim bright dim off +---[R2]---o o +---o o | \\ sw1 | \\ sw2 [R3] o----------+ o | | | | P P +----------->O 1 +---->O 2 | T | T | | +-----------------------+ Gnd
R1 = 240 ohm 1/4 w R2 = 47 ohm 1/4 w R3 = 1500 ohm 1/4 w POT = 5K linear taper potentiometer Rled = 330 (will work for both 2V and 3V leds)
When sw1 is in the bright position, a LED will light at full brightness when its sensor activates and goes to ground. When sw1 is in the dim position and sw2 is in the off position, pot 1 will control the brightness. Set it for medium. When both sw1 and sw2 are in the dim position, use pot 2 to set the brightness to the low level. Once you have set the pots for the brightness levels you want, you do not need to touch them again. The circuit works on either 14 volts or 28 volts.
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