LED sequencing circuit

[daffy]

Hi daffy,

how many groups of LEDs are you planning to use over a distance of 10m?

If a group of LEDs is used with the LEDs arranged in horizontal order a group of 5 LEDs will require a total of 2,500 LEDs, if arranged vertically the total number of LEDs will be 25,000.

A single counter with 5 usable outputs won't suffice for your application. So you require multiple counters cascaded for the proper sequence.

Also, do all groups have to be lit in sequence, or do you want to light group 1 and 6, then 2 and 7 and so on?

Boncuk

Yes...on a horizontal line. I bought 10,000 so qty is not an issue.

 

[daffy]

i think its 500 LEDs, and if its grouped then 5 LEDsx100 groups.

so use a 4017 as primary counter X axis, and another 4017 to count its resets Y axis . so totally a matrix of 10x10 = 100.
by taking out from 1st counter and another out from 2nd counter and making them by and AND out driver transistor (ie when both are high the transistor will drive the group of 5) this task can be completed.

i think he may need
555 - 1
4017-2
ressitors and transistors - 100 x2 each (its like switch on two transsitors to get one group on.

Edit: one transistor can be used as a emitter follower and other as a switch, use current limiting resistor and the group of LEDs to connect both transistors emitter and collector respectively.

Do you have a diagram with what you propose? That would be great.

 

[daffy]

Sorry I didn't think it was vague. Think of a line that is 10 meters long with one row of LED's spaced at 2cm apart. If I group every 5 LED's it should look like a 'bar' moving from one side to the other. Get it? There will be a total of (10x100)/2 = 500 LED's with a total group (or so called LED 'bars') of 100. I have 10,000 LED's available.

Application? A fool proof 'follow me this way' path.

 

[Boncuk]

I guess vne's design is not what the OP wanted.

I understood to light up the groups in sequence, one at a time. Johnson counters must be cascaded, or otherwise each stage will count (and light up the connected LEDs) individually. Cascading multiple counters means eight usable outputs per counter.

Here is a design cascading 6 counters for a total of 48 groups of LEDs. If one group contains three LEDs 144 of them can be used, if it contains five LEDs the number increases to 240.

The timer circuit is also a CMOS type, but it has a higher fanout than logic devices. To make sure the clock signal reaches the last stage each "module" of six uses one AND-gate to refresh the clock signal, which supplies each counter.

I have simulated the circuit using four stages and it works well. The PCB layout won't be easy to do if it has to be single sided and will contain a lot of wire jumps.

Since only one group of LEDs is lit a time one single LED current limiting resistor for all LEDs will suffice. For easier combining the different groups each group could contain its own limiting resistor.

Boncuk

 

[mbarazeen]

I thought of this last night. Is there any reason this idea wouldn't work? It uses 1 555, 3 4017s, 65 NPNs, and 10 current limiting resistors to sequence 500 LEDs each staying on for ~ .5 seconds. I think given a high enough supply voltage, the idea could be expanded indefinitely. For 2 4017s 100 LEDs, 3 4017s 1000, 4 4017s 10000, etc. What does everyone think?

View attachment 38018

it may not work, you can have 10x10 (using two 4017) or 10x20 (using 3 ), 20x20 (using 4)...so on.
example when pin 3 of V1, V2 & V3 are high, then LED 1 will be ON, also 101, 201, 301, 401 may get ON may be with less brightness via base to emitter forword current of particualr transistors connected to V1. its like another diode in series with it.

so you can not make more matrix to use more than two axis.

Daffy,

the circuit is more similar to vne147 posted, but i only sugest to use 10x10 matrix and substitute a single LED by a group of 5 LEDs with propper resistor.

 

[vne147]

Sorry I didn't think it was vague. Think of a line that is 10 meters long with one row of LED's spaced at 2cm apart. If I group every 5 LED's it should look like a 'bar' moving from one side to the other. Get it? There will be a total of (10x100)/2 = 500 LED's with a total group (or so called LED 'bars') of 100. I have 10,000 LED's available.

Application? A fool proof 'follow me this way' path.

I think I understand what you want now. Maybe I didn't read this thread close enough the first time but all I posted was a 500 LED chaser but you want a 100 group chaser with 5 LEDs each per group. That is actually much simpler than the circuit I posted. I'll post an updated schematic later today. BTW, what supply voltage will you be working with?

 

[vne147]

Daffy,

Here is the updated schematic. It sequences 100 banks of 5 LEDs each. VR1 is a potentiometer that you can use to vary the speed from between ~ .2 - 1.5 sec between banks. I made the supply voltage 18V so I could get away with placing all 5 LEDs in series. I don't expect the circuit to draw more than 50 mA or so, so you can power it with a relatively small 18V AC adapter. Make sure you get a regulated 18V AC adapter as the 555 and 4017 are already operating at their maximum allowed voltages. I hope you have some cheap laborers to put this thing together.

To everyone else, please check my circuit over for mistakes. One thing I wasn't sure about was whether or not R13 in between the 555 and the first 4017 was necessary. The data sheet of the 555 (which I think is TTL) said it could source up to 225 mA and the input of the 4017 (which is CMOS) said it could only withstand ± 10 mA at any input pin. Is this the proper way to interface TTL and CMOS? Is there a proper way? Can someone comment on that please? Thanks.

 

[BrownOut]

Pretty cool. I wonder if the column switches should be PNP's, and the resistors should be in the collector(s) line.

 

[vne147]

Pretty cool. I wonder if the column switches should be PNP's

I may be wrong but I think if the column transistors were PNPs, all but 1 column at a time would be on. Maybe I'm wrong or maybe you know that and just thought it would create a cooler looking output.

and the resistors should be in the collector(s) line.

I considered that but thought I would also have to add base resistors if I did it that way. Once again I may be wrong but I thought that by placing the resistor at the emitter, it would also limit the base current making a base resistor unneccesary.

Any comments, corrections, input, or constructive criticism is welcome from all. Thanks.

 

[dougy83]

Nicely drawn up.

A couple of notes:
* it's better to use a voltage below the 'absolute maximum' rating of the ICs. Not all 555s are the same; I have 2 datasheets here, one has 16V abs max and the other 18V abs max. I would suggest reducing the supply voltage.
* The emitter resistors on the column drivers are not a good way to do it. The output of the 4017 will be loaded down to the voltage across the emitter resistor + 0.7V. There should be a base resistor and a collector resistor.

 

[vne147]

Nicely drawn up.

A couple of notes:
* it's better to use a voltage below the 'absolute maximum' rating of the ICs. Not all 555s are the same; I have 2 datasheets here, one has 16V abs max and the other 18V abs max. I would suggest reducing the supply voltage.

I know. It didn't feel right using that high a voltage but I thought it was necessary to place the LEDs in series. I didn't want to place the LEDs of each individual gang in parallel. Maybe I'm remembering this wrong but I thought I read somewhere that if you place LEDs in parallel, because of slight variations from LED to LED almost all the current would go through just one of them either making it really bright compared to the others or burn it out. I didn't want this to happen and I also didn't want to have to use a resistor for each LED to prevent it. In hindsight I think I could decrease the supply to 16 V. From the BC548's data sheet the Vce at 20mA is ~ .09V so:

2 BC548 = 2 x .09V = .18V
5 LED with Vf of 3V = 15V

That leaves 16 - (15 + .18) = .82V left over which means I should change the 150Ω resistors to:

.82/.02 = 41Ω.

Does that seem right?

* The emitter resistors on the column drivers are not a good way to do it. The output of the 4017 will be loaded down to the voltage across the emitter resistor + 0.7V. There should be a base resistor and a collector resistor.

That makes total sense. I can see why that's a problem now. I'll swap them to collector resistors per your and BrownOut's recommendation and add some appropriately sized base resistors.

Thanks for the comments. Keep em' coming!

 

[Boncuk]

5 LED with Vf of 3V = 15V

That leaves 16 - (15 + .18) = .82V left over which means I should change the 150Ω resistors to:

.82/.02 = 41Ω.

Does that seem right?

Thanks for the comments. Keep em' coming!

Hi vne147,

white LEDs normally have a Vf of 3.5 to 3.8V which means a minimum LED supply voltage of 17.5V for an LED string of 5 pcs. Supplying the timer (and the counters) with that voltage will fry them all.

Low power transistors can stand a VCEO of at least 32V. Consequently two power sources should be used: 12V for the digital circuit and 19 to 21V for the LEDs. This is no problem if both voltages use common ground.

Regards

Boncuk

 

[vne147]

Here is an updated shematic incorporating all the suggested changes and improvements I've recieved so far. Let me know if I've missed anything and what everyone thinks. Daffy, are you MIA or just lurking?

 

[BrownOut]

The problem is still going to be the column drivers. Although you have them connected to a higher voltage, the voltage delivered at the emitters is going to only be the level output by the IC's minus a diode drop. You're either gonna need a pair of transistors at each comumn (NPN/PNP), or else a row of inverters and PNP pullup transistors. I can't think of which is better right now, cause I'm in the middle of somethings.... but those drivers will need to be fixed.

 

[mbarazeen]

as brownout told, those transistors wil fuction as emitter followers and only give the output little less than IC high out.
you better shift all LEDs directly from high voltage side in series with the current limiting resistor, then use both transistors for switching with base resistors.

otherwise you have to connect the group LEDs in parellel by connecting a single resistor for each LED. the power consumption of the circuit will increase by this.

 

[Ubergeek63]

If you want 1 driver per group of LEDs. It's going to work out to be very expensive and not simple to wire up. I must say I prefer vne147's matrixed version which uses minimal parts.

actually i thought he was talking about a rope....it would be simple to wire three wires to be exact: power, ground, and the trigger for the next stage.

 

[Ubergeek63]

If you want 1 driver per group of LEDs. It's going to work out to be very expensive and not simple to wire up. I must say I prefer vne147's matrixed version which uses minimal parts.

vne147's circuit does not work - there is no way for the 12V logic source to forward bias a BE junction into 16V of LEDs. easily corrected with an extra set of transistors and resistors but kind of kills the "low componewnts count" bragging point

 

[dougy83]

oh, they're white leds. vne147, as you've 2 supply rails anyway, you can run the top counter between the 21V & 12V rail, and the lower counter between 12v and gnd (see attached). A resistor from 12V to gnd might help regulation in this configuration.

 

[vne147]

I think I now understand the issue with the column transistors that mbarazeen and BrownOut have been talking about. When the output of the 4017 is high, the voltage at the base of the NPN will be 12V but the voltage at the emmiter will be almost the full 21V from the supply. That makes sense I think and I see why that wouldn't work. But BrownOut said they should have been PNPs way back when I still had the circuit with only one supply voltage. I don't understand why the column transistors wouldn't have worked if there was only one supply rail. Maybe someone can explain it to me?

Dougy,

I like the idea with your mod. It's pretty simple and seems like it would work. That might be the simplest fix.

 

[dougy83]

When the output of the 4017 is high, the voltage at the base of the NPN will be 12V but the voltage at the emmiter will be almost the full 21V from the supply.

The base voltage would be 12V, the emitter voltage would be ~11.3V & the collector voltage would be 21V. The transistor would drop ~9.7V (VCE).

I don't understand why the column transistors wouldn't have worked if there was only one supply rail. Maybe someone can explain it to me?

The column (those with emitter connected to ground) transistors will work (they'll sink current). I might reduce the 33k resistor but it's not very important.