LED Chaser -- IN REVERSE?

Shamrock_Don · Post by **Shamrock_Don** » Mon Jan 19, 2009 1:52 pm

Does anyone know of modification to this circuit --> http://www.aaroncake.net/circuits/chaser.asp <--- that will have all the LED's lit except one. The unlit LED is the one that "chases".

I sure this could be done with microcontrollers but after a little research into the subject my head feels like it will explode -- to complicated...

Thanks for your time.

Pat Amaral · Post by **Pat Amaral** » Mon Jan 19, 2009 2:03 pm

I'm not sure of the proper connections but you should be able to put a hex inverter (4049 chip) on the output of the counter (U2 in the circuit you linked to). The inverter will tell the LEDs to turn off then the counter says "on" and vice versa. I believe that will accomplish what you're looking for but there may be additional components involved. It's been a while since I tried to do this sort of thing but I'm sure the experts here can tell you how to hook it up. It should be simple enough.

jwrjr · Post by **jwrjr** » Mon Jan 19, 2009 4:27 pm

I mostly agree. You will need 2 of the 4049s, though. Remember that the 4049 outputs are only rated for 5ma max.

tetsujin · Post by **tetsujin** » Mon Jan 19, 2009 5:07 pm

There's a fair chance that you could do this by reversing the polarity of each LED, and then connecting each through a series resistor to +5V rather than ground.

The reasoning is that if the 4017 outputs alternate between +5V (on) and 0V (off) (as opposed to +5V and open-circuit or something) then,

When an output is "on", feeding positive voltage to the cathode end of a LED whose anode end is connected to +5V, the drop across the resistor will be zero, and hence it'll emit no light.
When an output is "off", i.e. tied to ground, connected to the cathode end of a LED whose anode end is connected (through a resistor) to +5V, the drop across the resistor will be some number of volts, in the proper direction for the LED to operate.

Note that, since you're going to be operating multiple LEDs at once, each should have its own resistor. Also there may be limits to how much current each terminal on the 4017 can sink when it's in its low state - you should avoid exceeding that amount.

There is, as I previously suggested, a chance that the outputs of the 4017 will alternate between +5V and an "open-circuit" - that would mean that when the 4017's output is in its "off" state it's like it's not connected to anything at all. That would be pretty unusual I think, but I don't know offhand whether that is the case. If that is the case then you could make it work by adding "pull-down" resistors connecting each 4017 output to ground.

(EDIT): It's been a long, long time since I built 4017 circuits. But I seem to remember actually doing this at one point. Memory's fuzzy on this, I could be wrong, but I think the active LEDs were dimmer than when I had the circuit set up for one active LED at a time. I'm not sure why - it could just be a matter of tweaking the circuit (using smaller resistance values) or it could be a limitation of the 4017.

Pat Amaral · Post by **Pat Amaral** » Mon Jan 19, 2009 5:13 pm

jwrjr wrote:I mostly agree. You will need 2 of the 4049s, though. Remember that the 4049 outputs are only rated for 5ma max.

Yeah, I forgot to mention that. You need one inverter (not gate) per output from the counter. Since the 4049 only contains six inverters (that's why it's called a hex inverter) and your circuit is driving 10 LEDs, you would need to use two of the inverter chips. I also didn't realize the 4049 outputs are only rated at 5ma. That might be a problem since the LEDs probably need around 20ma to run. I'm not sure how to overcome that limitation without getting complicated.

Sorry about that (and here I was thinking I was being helpful

Guess I should leave thess things up to the experts).

tetsujin's solution sounds like an easier one to me.

jwrjr · Post by **jwrjr** » Mon Jan 19, 2009 6:37 pm

Yes. Tetsujin has a point.

macfrank · Post by **macfrank** » Mon Jan 19, 2009 10:03 pm

tetsujin wrote:There's a fair chance that you could do this by reversing the polarity of each LED, and then connecting each through a series resistor to +5V rather than ground.

The question is can the 4017 sink the current of 9 LEDs on at the same time?

The 74HC4017 (PDF data sheet) can sink/source up to 25ma per output... but it'll get pretty warm with 9 outputs on.

jwrjr · Post by **jwrjr** » Mon Jan 19, 2009 11:39 pm

I also thought of that. But I don't believe that a 74hc IC will be happy with a 9 volt supply. (Of course if the data sheet says otherwise, that is the final authority.)

macfrank · Post by **macfrank** » Tue Jan 20, 2009 12:26 am

jwrjr wrote:I also thought of that. But I don't believe that a 74hc IC will be happy with a 9 volt supply. (Of course if the data sheet says otherwise, that is the final authority.)

Only goes up to 6 volts. So drop the voltage and change the resistors.

tetsujin · Post by **tetsujin** » Tue Jan 20, 2009 11:17 am

macfrank wrote:
tetsujin wrote:There's a fair chance that you could do this by reversing the polarity of each LED, and then connecting each through a series resistor to +5V rather than ground.
The question is can the 4017 sink the current of 9 LEDs on at the same time?

The 74HC4017 (PDF data sheet) can sink/source up to 25ma per output... but it'll get pretty warm with 9 outputs on.

Good question. I don't really know the answer.

I figure each output is probably driven by a pull-up resistor connected to an NPN (on the Fairchild 4017 data sheet I found, each output is an inverted NAND - but probably it ends with an NPN and pull-up) - in which case the only source of heat from sinking current on an output is that voltage drop across the transistor, plus the heat generated by driving its output low... (but 9 out of 10 outputs are always driven low anyway...)

So if you're driving 9 LEDs at once, at 20mA, assuming a voltage drop of about 1V across each of those transistors when it's sinking current - it's an extra 180mW of heat.

According to that datasheet the device can dissipate up to 700mW - though I don't know how much of that would be accounted for by the normal operation of the device itself. But I think it'd be safe, since the SOIC version can only dissipate 500mW - meaning the device within either of those two packages must itself release no more than that as heat. That leaves at least a 200mW margin, which is enough to sink current for 10 LEDs.

Of course... in practice? I'd just try it. Run the thing off a wall wart for a couple weeks. If I come back and it's not working any more, then I'd rethink my design. :D (Though if you have inverters handy, and you can spare the space to use 'em, there's no reason not to go that route, either!)

Shamrock_Don · Post by **Shamrock_Don** » Wed Jan 21, 2009 11:00 pm

Thank you everyone for the replies so far..

I KNEW I had seen this "reverse chaser" before...

http://www.kc6sye.com/techmages_3_11_04_b.html
http://www.kc6sye.com/techmages_3_11_04_c.html

Shamrock_Don · Post by **Shamrock_Don** » Tue Jan 27, 2009 8:49 am

Anyone want to venture a guess how this was done ?

Joseph C. Brown · Post by **Joseph C. Brown** » Tue Jan 27, 2009 11:01 am

Have you tried sending a PM to Sparky? He is the mod for this area, and that's his webpage you put up the links to.

Shamrock_Don · Post by **Shamrock_Don** » Wed Jan 28, 2009 9:36 am

Oh Shoot! I did not know that -- thanks for the heads up. I'll message him now...

Sparky · Post by **Sparky** » Thu Jan 29, 2009 12:23 pm

The way to change the circuit to reverse chase, appears to be as mentioned. I'll reiterate though:

The counter turns on an output (raises it to Vcc) while all other outputs are pulled to (GND). [Off outputs are not left floating, aka tri-stated).

To generate the walking Off LED effect you flip the LEDs around (form perspective of the diagram in the circuits sticky thread). The negative of each LED goes to each counter output. All LEDs' positives go to the Vcc.

If you need a dropping resistor, please use one per LED. Since during switching the resistor could experience a different load than calculated resulting in spikes of extra power being delivered to the other LEDs.

Also if you are putting this in a larger model (those rings are very small and the LEDs use very little power)
Consider these 2 things:

Remember that some counter chips have problems with delivering power consistently to all outputs:
http://www.kc6sye.com/techmages_2_5_06_c.html
If you will have a large number of bright LEDs you may need to change the circuit to have some [power] transistors driving the banks of LEDs.

You can reach a greater number of simulated fan blade 'vanes' by increasing the number of quadrants the circle is divided into. However there is a limit for any given diameter circle. Too many 'vanes' and your eye will no longer see a rotating effect. This is why my little ring only had 2 Off LEDs. Any more and it was hard to 'see' a rotating effect.

I think a little diagram is needed to help explain the physical layout here. . .
Change the 4017 circuit diagram:
So that no LED is attached to output 9,
R (reset pint) is not connected to - (GND).
Then hook output 9 to the R (reset pin).

http://www.kc6sye.com/images/circuits/Q ... Layout.jpg
I will make another version of this that shows what I ended up doing with those led rings seen in the clips you found. . .

Sparky · Post by **Sparky** » Wed Feb 04, 2009 6:57 pm

Here's 2 sample circuits. They are not strictly tested though, and there is no Clock circuit in here (use 1 clock circuit wired to all your chasers that way your nacelles are running at the same pace).

http://www.kc6sye.com/images/circuits/R ... series.jpg

http://www.kc6sye.com/images/circuits/R ... d_para.jpg

Once you go to layout you will see a huge disadvantage to the series setup. Lots of jumper wires are needed to lace the LEDs of each quadrant.

Id say 90% of the work with the bussard lighting is building and wiring the dang ring of LEDs.

Shamrock_Don · Post by **Shamrock_Don** » Thu Feb 05, 2009 7:21 am

Oh goodie -- more reasons to break in my new breakboard.

THANKS AGAIN!!!!!!!!!!

D.