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this close to setting the perfboard on fire, 2013 darlington array

solder followed by anger

Oct 8, 2014
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good afternoon ladies and gentlemen,

I have been banging my head on my workbench for a few days now and figured it was time to ask for some help.
I have been trying to build a little big lamp for a while now and thought I had it all figured out until I transferred the components to the perfboard. I had the pictured circuit in working order on the bread board and to the best of my knowledge everything made it to the perfboard the way that it should have.

The only things I have seen is that when powered up i get no lights, 12.22 vdc going in from the wall wort, through the 555 timer (ic1), but at the nte darlington array (1c2) i get - negative 25 to negative 114 mv relative to the position of the potentiometer. Soldering appears to be complete and un shorted.

That being said i am newb to doing this sort of thing and more than likely messed something up. just following directions don't fully understand why these things do what they do.

Thankyou for your time appreciate any help.

OAjV6dWDf6asPkeq.jpg
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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We'll need to know the part number of IC2.

I suspect you need to connect your load between the outputs and the positive side of your power supply -- is that what you're doing?

Are you seeing an output of the 555 turning on and off as you want (I assume it's slow enough that you can see it happen using a multimeter)

If this is designed to dim some lights then congratulations you are building something I made last weekend, so we're working on the same stuff :-D
 

solder followed by anger

Oct 8, 2014
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Thankyou for your response,

The part number of ic2 is NTE LF133J NTE 2013.

As for the connection between the load and the output the power for ic2 comes from pin 3 of ic1with a 680 ohm resistor in between it and

pin one of ic two, also i have good power to pin 8 of ic1 and good ground on pin one of ic. I have not noticed a change in the amount of

vdc across pin 1 and pin 3 of ic1 just a solid 12.22vdc. Hope this write up isn't to confusing (I am confused). frustrating when it works on

the breadboard (that was a journey in its self) and then fall on its face on the perf board :( . First time using a breadboard let alone

transferring to perfboard mistakes are bound to be made. thank you for your time hope this helps with a conclusion.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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I assume that your LEDs have some sort of current limiting.

What I would do is to disconnect the 680 ohm resistor from pin 3 of the 555 (IC1) and touch it to the +ve supply. This should turn the LEDs on. Disconnecting this, and/or connecting it to the negative rail should turn them off.

If this doesn't work then it may be that there is insufficient voltage to turn on the LEDs.

Can you give more information about the LEDs?
 

solder followed by anger

Oct 8, 2014
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took some time to get back to the computer to answer your question about the leds they are vetco 5mm leds 8000 mcd with a forward

voltage of 3.2 volts they are wired up like this in order to do away with the resistors in series.
6yGTYIh4AKcWghcJ.jpg

I made some progress with the wiring i found a short in the soldering on the negative side of the circuit and got the leds to light up but

they are nowhere near as bright as they were when i had everything on the breadboard. Also they are blinking slow to fast enough that

i cannot see the blinking depending on the position of the potentiometer. When I had every thing on the breadboard the lights were

so bright that they were uncomfortable to look at directly and the pwm was spot on and it exceeded my persistence of vision on all

brightness settings. starting to wonder if i over heated the 2013 when i soldered it in dont know if that would be the cause of the

dimness of the leds as well.
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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As @Colin Mitchell said, you need a current-limiting resistor is series with each group of three series-connected LEDs. The total current supplied from pin 3 of IC1 would then be distributed to the six sets of three LEDs through separate connections for each set to IC2. A current limiting resistor is needed for each set. When you wire it this way, R3 is eliminated. Connect IC1-3 directly to IC2-1,-2,-3,-4,-5 and -6. The current-limiting function of R3 is replaced by six individual resistors, one for each set of three LEDs that are wired in series, connected to IC2-11, -12, -13, -14, -15, and -16.

You absolutely, positively, under no circumstances, subject to losing your experimenter's license, ever try to drive two or more LEDs connected in parallel, through a single current-limiting resistor as you are trying to do with R1. This includes trying to drive in parallel two or more series-connected strings of LEDs: treat each string as it were a single LED and provide that string it with its own current-limiting resistor.

Why do as I am suggesting? Because the forward voltage drop across any LED is not a reliable constant value. It will only be approximately 2 V, so a string of three will only be approximately 6 V. If you place a current-limiting resistor of identical value in each string, then excite the LEDs-with-resistor strings in parallel (which you should do now), you will measure slightly different voltages across each string of LEDs. That means if you were to try to parallel them all together (as you are trying to do now) with just one current-limiting resistor, the LED string with the lowest forward voltage drop will take over, and a disproportionate amount of the current will flow through it. This will cause the other LEDs, with higher forward voltage drops, to be more dimly lit. Some might not even light at all!

I operate a fairly complicated machine that accelerates charged particles. The particles impinge on targets loaded into a specially-built carousel with 25 positions. When we contracted to have the carousel built, I insisted that a positive means be incorporated to read-out each carousel position on an LED seven-segment numeric display. So a few months and several tens of thousands of dollars later it was delivered and appeared to work as specified. Within a year it failed to produce a reliable position read-out. Troubleshooting was pretty easy. There was a disk with radially positioned, binary-coded-decimal, slots attached to the carousel shaft, behind which disk six infra-red LEDs shined through the slots to illuminate a corresponding photo-transistor. There was also a small hole associated with each position that had it's own LED/photo-transistor sensing arrangement to allow precise positioning of the carousel. It was the BCD-coded slots that were causing the problem. Some of the slots acted as if the photo-transistor wasn't turning on when exposed to the LED through the slot. The problem would appear at random. I finally decided to remove the LED array and try to measure the output of each LED individually. That is when I discovered that the original "designers" had wired all the LEDs in parallel and were driving them through a single current-limiting resistor. So they saved a couple dollars on resistors with their faulty design. Didn't help their bottom line though. That a company went out of business. Meanwhile, I replaced the one resistor with six individual resistors, one for each infra-red LED. It has worked, as we originally specified, ever since.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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That is when I discovered that the original "designers" had wired all the LEDs in parallel and were driving them through a single current-limiting resistor.

It's hard to agree with you more.

Whenever you see this (especially in your case in some expensive -- and probably custom designed -- piece of equipment) all that comes to mind is "what were they thinking?" (perhaps without the "what").
 

solder followed by anger

Oct 8, 2014
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@hevans1944

thank you for you detailed explanation of why i should not use a single current limiting resistor. i am very very new to all of this electronics

stuff. I was building this lamp off of plans by charles platt from make magazine this is my first project that wasn't a through hole kit like to

solder not so good at the other stuff. was wondering if you could tell me what value of resistor to use from ic 2 to the leds and also if i

should remove the resistor r3 from the circuit as well.
 

hevans1944

Hop - AC8NS
Jun 21, 2012
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Yes, remove resistor R3 and follow the instructions below.

The resistor you use in each string of three LEDs depends on the forward voltage drop across the string at the current you intend to operate each string. The value will be the voltage from pin 3 of the 555 IC minus the forward voltage drop of the LED string divided by the current in the string. For example, if you measure 10 V from the 555 when its output is high, and you measure 6 V across the LED string with 20 mA current, then (10 - 6)/0.02 = 200 Ω.

You should start with 200 Ω and adjust the resistor value up or down slightly while measuring the current in the LED string until the desired forward current is obtained. A simple way to do this is with a 9V "transistor radio" battery connected to the resistor in series with one LED string. That will allow you to measure both the current and the forward voltage drop across the string until you find the resistor value that gives you the current you want, typically about 20 mA, but more if you want a brighter LED. Make a note of the forward voltage drop across the LED string at whatever current you decide is necessary for adequate brightness. Subtract this voltage from whatever voltage you measure from pin 3 on the 555, then divide by the current you measured when using the 9V battery. The value will not be too far from the resistor you used in the 9 V battery test, but will be a little higher if the 555 output is substantially greater than 9 V. Just remember this resistor is "dropping" the "excess" voltage from the 555 output to the LED string. Whatever that "excess" is, the value of the resistor will be the "excess" voltage divided by the LED current.

That's why you must perform the 9 V battery test to determine the LED string forward voltage at the desired operating current. The forward voltage across an LED is a function of the current through the LED: higher currents will produce more forward voltage until you reach a level that "lets the smoke out" of the LED. All electronic devices depend on keeping the smoke inside for proper operation.

Note also that the LEDs are flashing on and off, so they will appear dimmer than they do during the 9 V battery test. If they appear to be too dim, repeat the 9 V battery test at a higher current, not to exceed the maximum rating for your particular LED. That may be information that is difficult to find, so expect to "let the smoke out" of a few LEDs until you find the maximum current at maximum brightness your particular LEDs will accommodate. Then, having found that maximum current (hopefully after burning out only one or two LEDs), reduce it by at least ten percent for reliable operation.

A really nice article on how to connect multiple LEDs in series and determine the value of the dropping (current limiting) resistor can be found here.
 
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