# Racing start lights ( like F1 tracks)

M

#### [email protected]

Jan 1, 1970
0
Hello, i would like to know if someone has a link to a circuits to
create starting lights for races ? I would like it to turn on 5 red
LEDs one at a time without turning off the other till the 5 are on.
Then turn them off altogether. and to be nicer. would love it to light
5 green LED's as soon as the reds turn off,

Thanx for the help

This is how i would like it to look like a . is for lights off a R is
for red light on ang G is for green lights on:

at start

......
......

1 sec later

R....
......

1 sec later

RR...
......

1 sec later

RRR..
......

1 sec later

RRRR.
......

1 sec later

RRRRR
......

1 sec later

......
GGGGG

C

#### Chris

Jan 1, 1970
0
Hello, i would like to know if someone has a link to a circuits to
create starting lights for races ? I would like it to turn on 5 red
LEDs one at a time without turning off the other till the 5 are on.
Then turn them off altogether. and to be nicer. would love it to light
5 green LED's as soon as the reds turn off,

Thanx for the help

This is how i would like it to look like a . is for lights off a R is
for red light on ang G is for green lights on:

at start

.....
.....

1 sec later

R....
.....

1 sec later

RR...
.....

1 sec later

RRR..
.....

1 sec later

RRRR.
.....

1 sec later

RRRRR
.....

1 sec later

.....
GGGGG

Good morning. This is one of those jobs that pretty much cries out for
a $2 18-pin PIC. This is one of them. But I'd guess you'll want to be doing this with discrete logic. It's not that difficult (view in fixed font or M$ Notepad):

 Lightbar Clock
VCC VCC
 + +
| | .------------<R
 .-. | |
470K| | | |
 | | .---o----o----.
'-' | 8 4 |
 | | |
| | |
 o-----o7 |
| | |
 | | 3o------->CLK
.-. | LM555 |
 470K| | | |
| | .--o6 | VCC
 '-' | | | +
| | | | |
 o--o--o2 | --- .01uF
+| | 1 5 | ---
 --- '---o----o----' | RST
1uF--- | N.C. o--->
 | | |
| | .-.
 === === | |47K
GND GND | |
 '-'
|
 ===
GND

 Lightbar Logic 2 VCC
VCC
VCC __ + +
 + .------. .------------------| \ ___ ~~ | |
| | | | |00 )o-|___|-|<-o ___ ~~ |
 '----oIN Q1o---' .--|__/ 470 R | .--|___|-|<--o
| | | | | 470 G |
 | 1/2Q2o------------------ | __ | | ___ ~~ |
|4015 | ')--| \ ___ ~~ | o--|___|-|<--o
 CLK>---oCLK Q3o---------------. | |00 )o-|___|-|<-o | 470 G |
| | | o--|__/ 470 R | | ___ ~~ |
 .-----oRST Q4o--o----------. | | | o--|___|-|<--o
| | | | | | | __ | | 470 G |
 | '------' | | '---)--| \ ___ ~~ | | ___ ~~ |
| | | | |00 )o-|___|-|<-o o--|___|-|<--o
 | .-------------' | o--|__/ 470 R | | 470 G |
| | | | | | ___ ~~ |
 | | .------. | | __ | o--|___|-|<--'
| | | | '-----)--| \ ___ ~~ | | 470 G
 | '---oIN Q1o----------------. | |00 )o-|___|-|<-o |
| | | | o--|__/ 470 R | |
 | | 1/2Q2o---. | | | | VCC
| |4015 | | __ | | __ | | +
 CLK>---oCLK Q3o '---| \ '--)--| \ ___ ~~ | | | __
| | | |00 )o-. | |00 )o-|___|-|<-' | o--| \
 o-----oRST Q4o VCC+-|__/ o----o--|__/ 470 R | | |00 )oNC
| | | | | '--|__/
 | '------' | |
^ ___ | __ |
 RST VCC+-|___|-o---o---| \ ___ |/
4.7K | | |00 )o-|___|-o--| 2N3904
 R<--------------' '---|__/ 1K | |>
___ | |
 GND-|___|-' ===
10K GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Above you've got an LM555, a CD4015, and two 74LS00s, along with a
transistor to drive the 5 green LEDs.

Here's how it works: The 555 gives a 1 second clock. It has a control
pin (4) that inhibits the clock when it's low. That control signal is
called R above. The 4015 is a shift register, which starts out at all
logic 0s, due to the RST signal. Since the 6th output (Q2 on the
second half of the 4015) starts out at 0, the inverted output is 1,
which lets the 555 clock go. Now, as the 555 clocks, 1s are shifted
out one at a time, causing the outputs going down the diagram to turn
to "1" one at a time. These are NANDed with the inverted 6th output to
turn on the red LEDs one at a time. The LS00 has enough current
sinking capability to drive an LED directly.

The red LEDs light up until all 5 are lit. On the next clock, the 6th
output goes high. That does several things. First, the inverted
output goes low, which turns off the 5 red LEDs. That output is again
inverted, and is used to drive an NPN transistor. That will drive the
5 green LEDs. The third thing the 6th output does is turn off the 555.
Thus, the green LEDs will remain ON until you cycle power, which also
resets the circuit. When you turn it on, it starts over.

I should note here that this circuit is made for high intensity LEDs,
as you only have 7 mA or so going through each LED. If you use
standard LEDs, you might be a little disappointed by the results.
Also, it's made for a 5VDC regulated power supply. If you don't have
one of those, you might want to get a regulated switching +5V wall wart
rated for 150mA or more. Remember that you turn off the power supply
to reset the circuit. And please remeber that the logic ICs have power
and GND pins. On the LS00s, Vcc is pin 14 and GND is pin 7. On the
4015, Vcc is pin 16, and GND is pin 8.

And yes, the transistor is a bit underdriven for its load, and the LS00
output is being overdriven. This is a compromise between the current
drive capability of the LS output and the LED requirements. That LS00
output will definitely not be at logic level, but that doesn't matter.
The transistor should be pretty well saturated, and the extra drive
current won't hurt the LS00. The purists among us might take the extra
gate, invert the signal again, and drive a PNP transistor properly,
with LED load being ground-referenced. That's actually a better
solution.

I hope this has been of help.

Good luck
Chris

M

#### [email protected]

Jan 1, 1970
0
Great thanx alot chris.
Good morning. This is one of those jobs that pretty much cries out for
a $2 18-pin PIC. This is one of them. But I'd guess you'll want to be doing this with discrete logic. It's not that difficult (view in fixed font or M$ Notepad):

 Lightbar Clock
VCC VCC
 + +
| | .------------<R
 .-. | |
470K| | | |
 | | .---o----o----.
'-' | 8 4 |
 | | |
| | |
 o-----o7 |
| | |
 | | 3o------->CLK
.-. | LM555 |
 470K| | | |
| | .--o6 | VCC
 '-' | | | +
| | | | |
 o--o--o2 | --- .01uF
+| | 1 5 | ---
 --- '---o----o----' | RST
1uF--- | N.C. o--->
 | | |
| | .-.
 === === | |47K
GND GND | |
 '-'
|
 ===
GND

 Lightbar Logic 2 VCC
VCC
VCC __ + +
 + .------. .------------------| \ ___ ~~ | |
| | | | |00 )o-|___|-|<-o ___ ~~ |
 '----oIN Q1o---' .--|__/ 470 R | .--|___|-|<--o
| | | | | 470 G |
 | 1/2Q2o------------------ | __ | | ___ ~~ |
|4015 | ')--| \ ___ ~~ | o--|___|-|<--o
 CLK>---oCLK Q3o---------------. | |00 )o-|___|-|<-o | 470 G |
| | | o--|__/ 470 R | | ___ ~~ |
 .-----oRST Q4o--o----------. | | | o--|___|-|<--o
| | | | | | | __ | | 470 G |
 | '------' | | '---)--| \ ___ ~~ | | ___ ~~ |
| | | | |00 )o-|___|-|<-o o--|___|-|<--o
 | .-------------' | o--|__/ 470 R | | 470 G |
| | | | | | ___ ~~ |
 | | .------. | | __ | o--|___|-|<--'
| | | | '-----)--| \ ___ ~~ | | 470 G
 | '---oIN Q1o----------------. | |00 )o-|___|-|<-o |
| | | | o--|__/ 470 R | |
 | | 1/2Q2o---. | | | | VCC
| |4015 | | __ | | __ | | +
 CLK>---oCLK Q3o '---| \ '--)--| \ ___ ~~ | | | __
| | | |00 )o-. | |00 )o-|___|-|<-' | o--| \
 o-----oRST Q4o VCC+-|__/ o----o--|__/ 470 R | | |00 )oNC
| | | | | '--|__/
 | '------' | |
^ ___ | __ |
 RST VCC+-|___|-o---o---| \ ___ |/
4.7K | | |00 )o-|___|-o--| 2N3904
 R<--------------' '---|__/ 1K | |>
___ | |
 GND-|___|-' ===
10K GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

Above you've got an LM555, a CD4015, and two 74LS00s, along with a
transistor to drive the 5 green LEDs.

Here's how it works: The 555 gives a 1 second clock. It has a control
pin (4) that inhibits the clock when it's low. That control signal is
called R above. The 4015 is a shift register, which starts out at all
logic 0s, due to the RST signal. Since the 6th output (Q2 on the
second half of the 4015) starts out at 0, the inverted output is 1,
which lets the 555 clock go. Now, as the 555 clocks, 1s are shifted
out one at a time, causing the outputs going down the diagram to turn
to "1" one at a time. These are NANDed with the inverted 6th output to
turn on the red LEDs one at a time. The LS00 has enough current
sinking capability to drive an LED directly.

The red LEDs light up until all 5 are lit. On the next clock, the 6th
output goes high. That does several things. First, the inverted
output goes low, which turns off the 5 red LEDs. That output is again
inverted, and is used to drive an NPN transistor. That will drive the
5 green LEDs. The third thing the 6th output does is turn off the 555.
Thus, the green LEDs will remain ON until you cycle power, which also
resets the circuit. When you turn it on, it starts over.

I should note here that this circuit is made for high intensity LEDs,
as you only have 7 mA or so going through each LED. If you use
standard LEDs, you might be a little disappointed by the results.
Also, it's made for a 5VDC regulated power supply. If you don't have
one of those, you might want to get a regulated switching +5V wall wart
rated for 150mA or more. Remember that you turn off the power supply
to reset the circuit. And please remeber that the logic ICs have power
and GND pins. On the LS00s, Vcc is pin 14 and GND is pin 7. On the
4015, Vcc is pin 16, and GND is pin 8.

And yes, the transistor is a bit underdriven for its load, and the LS00
output is being overdriven. This is a compromise between the current
drive capability of the LS output and the LED requirements. That LS00
output will definitely not be at logic level, but that doesn't matter.
The transistor should be pretty well saturated, and the extra drive
current won't hurt the LS00. The purists among us might take the extra
gate, invert the signal again, and drive a PNP transistor properly,
with LED load being ground-referenced. That's actually a better
solution.

I hope this has been of help.

Good luck
Chris

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