12V LED light timer circuit

djsfantasi

Joined Apr 11, 2010
9,237
And back to the shift register solution, the 74HC595 is commonly used in the Arduino world.

Here is a tutorial on its use.

My suggestion is to tie the Ser input (pin 14). ) to Vcc.

Then send a clock signal with a period equal to the delay between lighting the individual LED segments. This signal will be input to the Clk (pin 12).

Finally, use the unused outputs to create the desired pause. The pause can be calculated as n * clock period. The nth unused pin would connect to the Clr
Input (pin 11).

One chip solution! And you could use @crutschow ’s clock circuit.
 

AnalogKid

Joined Aug 1, 2013
12,133
And back to the shift register solution, ...
One chip solution!
Post #20.

There are several 74xx 8-bit shifters that would work, including some with 24 mA output stages to drive single LEDs directly, but given that the TS has 12 V for the LEDs, I went with all 12 V logic.

For even fewer parts, look into a 1-transistor oscillator that uses the Q in avalanche breakdown mode. Unijunction - ish.

ak
 
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AnalogKid

Joined Aug 1, 2013
12,133
You guys are awesome. I would have hammered it out with the decade counter.
That would have been messy. Binary output to a 3-bit to 8-line decoder, each of the first 4 outputs clocking latches to hold the LEDs on, and the 5th output resetting the latches and the counter.

Or a 4017 with each of the first 4 output setting flipflops to drive the LEDs and the 5th output resetting everything.

You can save a chip in both approaches by using a quad S-R flipflop, a CMOS part that (I think) does not have a 74xx equivalent. Still, I think the shift register approach is cleaner; plus, fewer parts.

ak
 
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PeteHL

Joined Dec 17, 2014
580
My idea is a sawtooth at input to 4 paralleled comparators. Given that only 12 V is available, I'm not sure that a suitable sawtooth oscillator could be devised. If unijunction transistors are still available, the unijunction transistor configured as an oscillator has a sawtooth at one point in the circuit. In one of my texts, there is a schematic diagram of a sawtooth oscillator that is a configuration of a 741 and 301 op amps, one of each. This might be a quite simple solution but lacking in being able to independently control time on/ time off.

Regards,
Pete
 

Wolframore

Joined Jan 21, 2019
2,619
Since we’re using an Arduino I would use a 4017, one digital pin for clock out and have a reset pin connection. LED would be on 0-3. The loop would just advance the state (count pulses then reset based on count stage). You can stall on final state as long as you want. I’ll draw it out later if I’m not busy. You can add mosfet array ADL810027 if you need more than 5-10mA per LED and need to run them from 12v. I’m still amazed with you guys with your CMOS solutions. In some ways I like the CMOS solutions better. It’s instantaneous and doesn’t take any resources away from the uP depending on what else you’re doing.

The circuit is really simple... you can add the mosfets if required.

upload_2019-4-23_10-4-37.png
 
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AnalogKid

Joined Aug 1, 2013
12,133
The circuit is really simple...
Again, this circuit does not do what is spelled out in post #1.

ak

ps. The ENA input must be grounded for the counter to count.
Never leave CMOS inputs floating.
There is no current limiting for the LEDs.

AND - Once you have an Arduino in the project, why bother with an external counter/shifter/whatever? There is nothing in post #1 about the Arduino doing any other tasks.
 

AnalogKid

Joined Aug 1, 2013
12,133
Here is another version using 5 V logic, with a single 8-bit shifter instead of a dual 4-bit. The 164 has a negative-true Reset input, so Q1 is needed to invert the reset signal. The circuit function and chip count are identical, but the schematic looks prettier, less busy.

ak
LED-Stepper-3-c.gif
 

djsfantasi

Joined Apr 11, 2010
9,237
Here is another version using 5 V logic, with a single 8-bit shifter instead of a dual 4-bit. The 164 has a negative-true Reset input, so Q1 is needed to invert the reset signal. The circuit function and chip count are identical, but the schematic looks prettier, less busy.

ak
View attachment 175644
That’s the kind of circuit I’ve been suggesting! Haven’t been able to get to my installation of LTSpice. But, I’m working on a schematic using an 74HC595 that’s very similar.
 

Wolframore

Joined Jan 21, 2019
2,619
So the correct termination for unused Q4-Q9 and CO is pull down resistor? Is 10K ok? They will never activate due to counter won't be allowed to get that high. But if floating pins create capacitance issues then is that the right termination or just ground directly?


upload_2019-4-23_11-42-17.png
 

Wolframore

Joined Jan 21, 2019
2,619
@bertus could you please explain the stress you're talking about? I thought that the outputs were buffered with clamping circuits and would limit the current itself within the recommended voltage ratings of the chip... especially the CD variety... Maybe I've been abusing them for a while but they seems to robust and tolerant of this... HC version is another story.
 

bertus

Joined Apr 5, 2008
22,925
Hello,

When you look at the datasheet, at a powersupply of 15 Volts, an output voltage of 13.5 Volts will appear when 6.8 mA is drawn.
How much current will be drawn when a led of say 3 Volts is connected on the output?

Bertus
 

Attachments

Wolframore

Joined Jan 21, 2019
2,619
I just found something interesting... there's a difference between the B and the UB ... the voltage doesn't matter because it's a current mirror type circuitry in the Buffered versions. http://www.ti.com/lit/an/scha004/scha004.pdf

And except for the HC chips all the CMOS I've used have been TI "B" versions...

I believe you have to do a power calculation and you will see that it's under the power limit of the LED in the buffered cases.
 
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