L.E.D. Marquee - Electronic Circuits Projects Diagrams Free

ParkingLotLust · 2nd August 2006, 12:42 AM

As seen here, I would like to do the exact same thing, except with a different name. Ive been in contact with him (and he's a very nice guy!) however he doesnt have the schematic. Through trial and error, Ive drawn up a schematic in eagle from looking at how his breadboard is connected. I was wondering two things: what (if any) are the errors in my schematic; and will someone design a pcb for me if I give you a paint drawing of where I want the LEDs to go? (single sided please, something I can etch myself). The .sch file is zipped and attached, as is the gif schematic.

on1aag · 2nd August 2006, 02:40 AM

You should turn around the led's and connect them to Vcc.

on1aag.

ParkingLotLust · 2nd August 2006, 02:57 AM

Why? It works perfectly fine with them sourcing current from the chips.

on1aag · 2nd August 2006, 07:24 AM

All TTL outputs are designed to sink current rather than to source current.
Which doesn't mean that your leds won't turn on if you use them the way
you do. You'll get a lot more light from them if you use them the other way
round. I just did the test, I connected four green leds to a 74LS164 via 120
ohm resistor's. The average sink current was a little over 20 mA while the
average source current was just under 10 mA. The effect on the leds was
clearly visible. If you want the leds to be visible during the daytime...
You should also consider using high efficiency leds.
But there is a but, there always is. You probably gonna have to insert a
buffer for each last output of each shiftregister to drive the led. A buffer,
an OR gate or even an AND gate will do the trick. The logical zero will not
be "zero" enough because the logical "0" level will rise to ca. 0,5 volt.
The unused B input and the clear should be connected to Vcc.

on1aag.

ParkingLotLust · 2nd August 2006, 02:36 PM

But if I source current, then I wont need the buffer? Could you give me a rough schematic of what Id have to add, so I know what you mean? I dont mind if theyre a bit dimmer, its just a little project for myself, its not a big deal. Also, thanks for the tip about connecting the stuff to vcc, Ill change the schematic accordingly

on1aag · 3rd August 2006, 06:37 AM

The problem might arise at the Qh outputs that are connected to the next
shiftregister's A input, so you need eight buffers to drive the leds.
A 74LS245 is a suitable candidate for the job.
Don't forget that your power supply must deliver an extra 750 mA if you
decide to turn around the leds.

on1aag.

ParkingLotLust · 3rd August 2006, 10:26 PM

Ill have to pick up some of those chips then. Why do they need to be buffered when theyre turned around? Also, will it be possible to spell out a name using a single sided pcb, or will there be too many crossed wires (jumpers are okay)?

on1aag · 3rd August 2006, 11:29 PM

You only need one 74LS245 and only the last led of each shiftregister should be buffered. If you use source mode the output voltage at that output will drop, but it still will be recognized as a logical "1". If you use sink mode, you will get
the double output current, but the logical "0" level will rise to
about 0,5 volt, and there you enter the gray zone between logical "0" and logical "1" level. The next shiftregister might not recognize that 0,5 volt as a logical "0" level.

on1aag.

ParkingLotLust · 4th August 2006, 12:45 AM

Ohhh, I see. Hmm. Does it matter if I get a 74HC245 or even just any 74XX245? Or does it specifically have to be the LS version?

EDIT This work?

ParkingLotLust · 5th August 2006, 06:51 PM

Hmm, just realized I might need one more 74164 due to the length of the name. Guess Ill have to add another 74245 to the mix. Anyone up for designing the pcb for me?

on1aag · 6th August 2006, 03:06 AM

After taking another look at your last schematic I noticed that you moved
the data inputs to the outputs of the buffer. That is wrong, they should
stay where they were in the first schematic. If you take a close look at
the internal household of the shiftregister, you will find that all outputs
are buffered. No matter how much you load the outputs, it will not affect
the working of the shiftregister. But when data is passed from one shift-register to another, the last output must have clearly defined logic levels
that have to be recognizable by the next shiftregister! The buffers don't
load the Qh outputs as much as the leds would do, the logic level will not
be affected by the buffers. The logic level at the output of the buffers
will be affected by the load of the leds, but that's no problem since there
is nothing connected to it.
About the use of another logic family for the 74LS245, you could use a
74HC245. But you could also use a 74LS- or 74HC244 or 74LS- or 74HC541
or similar.
If you decide to build the entire circuit with 74HCXXX chips, the extra
buffers might not be needed. You can even choose if you want them to
source or sink because they have symmetrical outputs.
The best thing is to use the same logic family.

on1aag.

ParkingLotLust · 6th August 2006, 04:00 AM

There, that better?

Hmm, so 74HC164's wouldnt need buffers? I might have to get ahold of some of those then.

Still looking for someone to design a pcb for me. Id be willing to throw a bit of money your way if you can do it single-sided.

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2nd August 2006, 02:40 AM	(permalink)
on1aag	You should turn around the led's and connect them to Vcc. on1aag.

2nd August 2006, 02:57 AM	(permalink)
ParkingLotLust	Why? It works perfectly fine with them sourcing current from the chips.

2nd August 2006, 07:24 AM	(permalink)
on1aag	All TTL outputs are designed to sink current rather than to source current. Which doesn't mean that your leds won't turn on if you use them the way you do. You'll get a lot more light from them if you use them the other way round. I just did the test, I connected four green leds to a 74LS164 via 120 ohm resistor's. The average sink current was a little over 20 mA while the average source current was just under 10 mA. The effect on the leds was clearly visible. If you want the leds to be visible during the daytime... You should also consider using high efficiency leds. But there is a but, there always is. You probably gonna have to insert a buffer for each last output of each shiftregister to drive the led. A buffer, an OR gate or even an AND gate will do the trick. The logical zero will not be "zero" enough because the logical "0" level will rise to ca. 0,5 volt. The unused B input and the clear should be connected to Vcc. on1aag.

2nd August 2006, 02:36 PM	(permalink)
ParkingLotLust	But if I source current, then I wont need the buffer? Could you give me a rough schematic of what Id have to add, so I know what you mean? I dont mind if theyre a bit dimmer, its just a little project for myself, its not a big deal. Also, thanks for the tip about connecting the stuff to vcc, Ill change the schematic accordingly Last edited by ParkingLotLust; 2nd August 2006 at 05:35 PM.

3rd August 2006, 10:26 PM	(permalink)
ParkingLotLust	Ill have to pick up some of those chips then. Why do they need to be buffered when theyre turned around? Also, will it be possible to spell out a name using a single sided pcb, or will there be too many crossed wires (jumpers are okay)?

3rd August 2006, 11:29 PM	(permalink)
on1aag	You only need one 74LS245 and only the last led of each shiftregister should be buffered. If you use source mode the output voltage at that output will drop, but it still will be recognized as a logical "1". If you use sink mode, you will get the double output current, but the logical "0" level will rise to about 0,5 volt, and there you enter the gray zone between logical "0" and logical "1" level. The next shiftregister might not recognize that 0,5 volt as a logical "0" level. on1aag.

5th August 2006, 06:51 PM	(permalink)
ParkingLotLust	Hmm, just realized I might need one more 74164 due to the length of the name. Guess Ill have to add another 74245 to the mix. Anyone up for designing the pcb for me? Last edited by ParkingLotLust; 6th August 2006 at 03:56 AM.

6th August 2006, 03:06 AM	(permalink)
on1aag	After taking another look at your last schematic I noticed that you moved the data inputs to the outputs of the buffer. That is wrong, they should stay where they were in the first schematic. If you take a close look at the internal household of the shiftregister, you will find that all outputs are buffered. No matter how much you load the outputs, it will not affect the working of the shiftregister. But when data is passed from one shift-register to another, the last output must have clearly defined logic levels that have to be recognizable by the next shiftregister! The buffers don't load the Qh outputs as much as the leds would do, the logic level will not be affected by the buffers. The logic level at the output of the buffers will be affected by the load of the leds, but that's no problem since there is nothing connected to it. About the use of another logic family for the 74LS245, you could use a 74HC245. But you could also use a 74LS- or 74HC244 or 74LS- or 74HC541 or similar. If you decide to build the entire circuit with 74HCXXX chips, the extra buffers might not be needed. You can even choose if you want them to source or sink because they have symmetrical outputs. The best thing is to use the same logic family. on1aag.