Sensor/LED circuit-more help needed please-multiV's and dimmer

Discussion in 'Electronic Basics' started by Gman, Aug 19, 2005.

1. GmanGuest

Hi there,

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.

Gman

2. John FieldsGuest

---
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.

3. John FieldsGuest

---
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 28V
| E |
| | O
Q2 C | |
2N4401 B-----+--[R2]--+ |
E | | O
| [R3] Q3 C |
| | 2N4401 B--[4k7]--+--> >--O--> |
| | E | S1
| | | [100K]
| | | |
GND-------+-------+--------+-----------+

4. GmanGuest

Hi John,

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.

Thanks again, it really is appreciated.

Gman

5. GmanGuest

Also,

Do you have any recommendations on the higher rated transistor?

Mucho gracias
Gman

6. John FieldsGuest

---
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.

7. GmanGuest

Thanks again John,

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?

Thanks again,
Gman

8. John FieldsGuest

---
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:

+VIN>-----+-------+
| |R6
| [430]
|R1 |A
[10K] [LED]
| |
| C Q1
+-----B 2N4401 14V O--> |<--O 28V
| E |
| | O
Q2 C | |
2N4401 B-----+ |
E | |
| +--[R2]--+ |
| | | O
| [R3] Q3 C R4 |
| | 2N4401 B--[4k7]--+--> >--O--> |
| | E |R5 S1
| | | [100K]
| | | |
GND-------+-------+--------+-----------+

---
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.

9. GmanGuest

Hi John,

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?

Thanks,
Gman

10. John FieldsGuest

---
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.
---
---
With that 430 ohm resistor in series with the LED, the dissipation
in the transistor went down to about 350mW, and the case temperature
rose to 42° with an ambient temp of 25°C so that ought to be fine.

Fairchild spec's the junction-to-ambient thermal resistance of the
2N4401 at 200°C per watt, so with a dissipation of 350mW that puts
the junction at:

200°C * 0.35W
Tj = --------------- + Ta = 95°C
1W

Since the part is rated for a maximum junction temp of 150°C, I'd go
with it but, if you want something a little stronger I'd recommend a
Zetex ZTX451.

11. ehsjrGuest

This circuit will work at either 14 or 28 volts Vcc
from the sensors, and includes switchable dimming:

Vcc---[R1]---+---[R2]---+
| |
[Zd] [LED]
| |
Gnd----------+---[R3]---+
| |
+---o o---+
\
Bright/Dim switch

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.

Ed

12. GmanGuest

Hi Ed,

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

Best regards,
Gman

13. GmanGuest

Hi Ed,

I found that the sensors operate by providing ground instead of power.
How does this effect the suggested circuit above, if at all?

Thanks and regards,
Gman

14. ChrisGuest

Hi, Gman. Just to restate the bidding here:
* 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.

` NPN Current Sink PNP Current Source`
`
` VCC VCC
` + +
` | |
` | | +
` | .-.
` R | |
` | |
` | | -
` || |Vdrive + 0.7V
` || Vb|<
` |V Vdrive >---|
` |Ic |\
` | | |
` Vb|/ | |
` Vdrive>----| | V
` |> | Ic
` |Vdrive - 0.7V |
` | +
` R | |
` | | |
` '-' |
` | - |
` | |
` === ===
` GND GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

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\$

` VCC
` +
` |
` |
` V ~
` - ~
` |
` |
` Vb|/
` Vdrive>---| Q1 - Q8
` |>
` |
` | +
` .-.
` 100| |
` | |
` '-'
` | -
` |
` ===
` GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

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
2.7V 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:

` VCC VCC VCC VCC
` + + + +
` | | | |
` | | | |
` | DZ2 .-. | DZ1 .-.
` /-/ | | /-/ | |
` ^ | | ^ | |
` | '-' | '-'
` | | | |
` | | | |
` | |< | |<
` o-----| o----|
` | |\ | |\
` .-. | .-. |
` | | | | | |
` | | | | | |
` SW1 '-' | '-' |
` _/ | | | |
` .--o/ o--' | === |
` | | GND |
` | | | Vdrive
` === '------------------o------->
` GND |
` .-.
` | |47 ohms
` | |
` '-'
` |
` |
` ===
` GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

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.

Good luck
Chris

15. ChrisGuest

Sorry -- Try again with the drive voltage section...

VCC VCC VCC VCC
+ + + +
| | | |
| | | |
| DZ2 .-. | DZ1 .-.
/-/ | |270 /-/ | |110
^ | | ^ | |
| '-' | '-'
| | | |
| | | |
| |< | |<
o-----| Q10 o----| Q9
| |\ | |\
.-. | .-. |
| | | | | |
| | | | | |
SW1 '-' | '-' |
_/ | | | |
.--o/ o--' | === |
| | GND |
| | | Vdrive
=== '------------------o------->
GND |
.-.
| |47 ohms
| |
'-'
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Left out the labels. Q1-8 could be TIP31 or equivalent, and Q9-10
could be TIP41 or equivalent.

Q9 should get a small heat sink for greater reliability. It will
dissipate a little over 1 watt.

Chris

16. ehsjrGuest

No change to the circuit - just wire the Gnd line to the
sensor and the Vcc line to the supply.

Ed

17. ehsjrGuest

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.

-----
Vcc---in|LM317|out---+--------------+-------+----//----+
----- | | | |
adj [R1] [Rled] [Rled] [Rled]
| | | | |
+---------+ [LED1] [LED2] [LED8]
| | | |
| Sensor1 Sensor2 Sensor8
| dim
+---[R2]---o
| | o bright
[R3] o
| |
+--------[POT]
|
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)

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.

Ed

18. GmanGuest

Thanks guys,

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

19. ehsjrGuest

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.

Ed

20. GmanGuest

Thanks Ed,

I breadboarded the circuit with only 3 leds and accidently put in 47
ohm RLEDs, of course the LM317 overheated and shut down, but before it
did, you could work the circuit exactly as described. The LM317 came
back on after cooling down and I changed to 330R for the LEDs, but I
only have 3 LEDs in the breadboard, and can not adjust the brightness
at all. Is this because there isn't enough load? Going to load it up
and see if that changes anything.

Thanks,