Using 555 timer to drive LEDs with high current draw

Thread Starter

mrobbo15

Joined Jan 2, 2017
4
Hi all,

I'm planning to recreate this sequential turn signal circuit:

... and actually use it in a custom LED turn signal assembly. Problem is, apparently the 555 is only rated to 200mA, whereas my proposed assembly will draw 1.33A @14V with all LEDs powered (19 arrays of 4 LEDs, each with a forward current of 70mA - 76 LEDs total).

Is there any way I can get around this and provide enough current for the LEDs? Also, is it possible to just add on strings of LEDs to this diagram so that there are 19 steps instead of 5?

I'm a novice when it comes to circuit design, just hoping to make sense of it enough to create this circuit.
 

AnalogKid

Joined Aug 1, 2013
10,986
Many "Instructables" circuits have serious design flaws, so beware of anything they post. Also, those are not standard schematic symbols for resistors, capacitors, transistors, or a battery. The circuit as designed is to cute for its own good. Each successive LED is dimmer than the one before it, and the base current for the bottom transistor increases to over 50 mA by the end of a cycle. Think transistor death. Also, the timing between the LED steps is not constant. Separate from that, none of the LED current comes from or goes through the 555, so your main concern does not apply.

In this circuit each LED stays on when the next one comes on, so the visual is one LED, then two at the same time, then three, etc. This is a Mustang or Cougar display, originally done with a timing motor, machined cam shaft, and three micro switches, similar to a music box mechanism; as opposed to a Knight Rider display where only one LED is on at any given moment. Which display type are you after, and how many LEDs?

In your circuit, the 555 drives an R-C network that produces an exponential ramp. The transistors act as a series of voltage comparators, each one turning on at a higher and higher voltage. If you want to stay with this technique, better to replace the transistors with quad opamps acting as comparators, driving power MOSFETs. You can adjust the comparison resistors to "straighten out" the timing.

A better way to go, especially if you want a 19-step circuit, is to keep the 555 as the timing clock and drive shift registers, with the first 19 stages driving power MOSFETs. How much of an all-dark pause do you want between all on and one on?

ak
 
Last edited:

Thread Starter

mrobbo15

Joined Jan 2, 2017
4
Many "Instructables" circuits have serious design flaws, so beware of anything they post. Also, those are not standard schematic symbols for resistors, capacitors, transistors, or a battery. The circuit as designed is to cute for its own good. Each successive LED is dimmer than the one before it, and the base current for the bottom transistor increases to over 50 mA at the end of a cycle. Think transistor death. Also, the timing between the LED steps is not constant. Separate from that, none of the LED current comes from or goes through the 555, so your main concern does not apply.

In this circuit each LED stays on when the next one comes on, so the visual is one LED, then two at the same time, then three, etc. This is a Mustang or Cougar display, originally done with a timing motor, machined cam shaft, the three micro switches, similar to a music box mechanism; as opposed to a Knight Rider display where only one LED is on at any given moment. Which display type are you after, and how many LEDs.

In your circuit, the 555 drives an R-C network that produces an exponential ramp. The transistors act as a series of voltage comparators, each one turning on at a higher and higher voltage. If you want to stay with this technique, better to replace the transistors with quad opamps acting as comparators, driving power MOSFETs. You can adjust the comparison resistors to "straighten out" the timing.

A better way to go, especially if you want a 19-step circuit, is to keep the 555 as the timing clock and drive shift registers, with the first 19 stages driving power MOSFETs. How much of an all-dark pause do you want between all on and one on?

ak
Thanks for the info! It's a shame the design is so flawed.

I'm after the Mustang-style sequential pattern, not the Knight Rider style. Interesting that you say they're run by motors and switches, I wouldn't have guessed that. The plan is for 76 LEDs total, but that may end up being a bit less, like 60-odd. They'll be in strings of 4, hence the 19 steps.

Ok MOSFETs, sounds good. I'd like it to just be in time with the original turn signals. I had a thought - if I just wire them up to the turn signal power as is, the on/off timing can be retained (controlled by the in-car relay). This means all I'd need from the 555 circuit is for it to run the sequential effect and stay on, as it will be turned off automatically by the relay. It will then come on again when power is supplied again via the relay. So I shouldn't need the 'dark time' programmed into the circuit, it just needs to instantly start as soon as it receives power.

What does need to be controlled is:
a) The speed of the sequence and
b) To keep the on/off rate the same as normal, usually done by wiring a load resistor in parallel.
 

AnalogKid

Joined Aug 1, 2013
10,986
While there is no standard, a normal turn signal is approx. 0.5 s on and 0.5 s off, or about 25 ms for each of your 19 stages. So the perception will be one of a continuously moving bar rather than something stepping across. What is the current through each of the 4-LED strings? If it is more than a gate or comparator can handle, consider a ULN2004 or 2804 darlington transistor array. 7 or 8 channels in one package, low cost, rugged.

Another way to go is the LM3914 dot/bar graph driver. Two of them gets you 20 steps, with adjustable current limiting. This would be driven by a linear ramp rather than a 555 clock. If you look at the original 60's Mustang signal, it steps 2 times at equal periods per step, then holds briefly at the last step (all on) before going dark and restarting. Adjusting a ramp into 3914's to get the same effect gives you a small "rubber band" period at the end of a cycle so minor timing variations caused by temperature changes and component aging don't cause the display to turn off before it reaches the last step. Plus, displays that reach an end and then instantly reset rather than hold a bit on the last step are irritating/agitating/disturbing (I worked with research psychologists for 6 years). Depending on your LED requirements, this could be a more simple solution; a linear ramp is easy to achieve, and the car delivers the timing. The ultra-techy-cutsie approach is to use a logarithmic ramp straight out of a simple R-C network into a pair of 3915's, which have a logarithmic response curve, for a pseudo linear motion effect. Looks good on paper...

ak
 

dannyf

Joined Sep 13, 2015
2,197
Also, is it possible to just add on strings of LEDs to this diagram so that there are 19 steps instead of 5?
the design is a little bit weird in that it doesn't actually have "steps". looks like the person designed it wanted to have progressive "kick-in" points for the leds but that particular circuit isn't going to do it.

I think you will get more help if you outline what you are trying to do, what you have (parts and capabilities) and people may be able to better help you.
 

Thread Starter

mrobbo15

Joined Jan 2, 2017
4
While there is no standard, a normal turn signal is approx. 0.5 s on and 0.5 s off, or about 25 ms for each of your 19 stages. So the perception will be one of a continuously moving bar rather than something stepping across. What is the current through each of the 4-LED strings? If it is more than a gate or comparator can handle, consider a ULN2004 or 2804 darlington transistor array. 7 or 8 channels in one package, low cost, rugged.

Another way to go is the LM3914 dot/bar graph driver. Two of them gets you 20 steps, with adjustable current limiting. This would be driven by a linear ramp rather than a 555 clock. If you look at the original 60's Mustang signal, it steps 2 times at equal periods per step, then holds briefly at the last step (all on) before going dark and restarting. Adjusting a ramp into 3914's to get the same effect gives you a small "rubber band" period at the end of a cycle so minor timing variations caused by temperature changes and component aging don't cause the display to turn off before it reaches the last step. Plus, displays that reach an end and then instantly reset rather than hold a bit on the last step are irritating/agitating/disturbing (I worked with research psychologists for 6 years). Depending on your LED requirements, this could be a more simple solution; a linear ramp is easy to achieve, and the car delivers the timing. The ultra-techy-cutsie approach is to use a logarithmic ramp straight out of a simple R-C network into a pair of 3915's, which have a logarithmic response curve, for a pseudo linear motion effect. Looks good on paper...

ak
Yeah, I've figured it will look like a smooth sweep rather than 'steps', which I think is cool! Each string will need to see 70mA.

I feel like I'm starting to get out of my depth! I like the idea of the car delivering the timing and the 3914 seems well-suited to my requirements. Holding the lights on momentarily at the end of the sequence seems ideal too.

the design is a little bit weird in that it doesn't actually have "steps". looks like the person designed it wanted to have progressive "kick-in" points for the leds but that particular circuit isn't going to do it.

I think you will get more help if you outline what you are trying to do, what you have (parts and capabilities) and people may be able to better help you.
Ok, good idea. The outline:
- A circuit that runs 19 strings of LEDs in a sequential 'bar graph' style
- Timing controlled by the car's turn signal... so it will receive power for roughly half a second, then be turned off for half a second and repeat.
- LEDs need to perform the full sweep in the half second they are powered on, then stay on until the turn signal relay turns them off.
- 70mA per string
- I'm capable of getting any common parts easily attainable from an electronics shop
- Capable of soldering
- Not capable of designing the actual circuit (only studied basic electrical engineering)
 

hp1729

Joined Nov 23, 2015
2,304
Base current problem? I think not. TIP31, max base current of 1 Amp. He is not talking about a small transistor here. Even the circuit as shown base current at the bottom transistor is still going to be okay. 2N4401 has a max base current of about 30 mA. Assuming white LEDs, 20 mA, baser current will be less than 1 mA. Less than 10 mA at the bottom transistor. It looks like it will work. I didn't bother to build it, but it looks reasonable.
 

AnalogKid

Joined Aug 1, 2013
10,986
Assuming white LEDs, 20 mA, baser current will be less than 1 mA. Less than 10 mA at the bottom transistor.
Each base is driven by the emitter of the transistor above it, so it has all of the LED current from that transistor and every transistor above it. The current path from the top LED is through each lower transistor's base-emitter junction, down to the last transistor's emitter which is grounded through a standoff diode, and each transistor's emitter current includes its own LED's current. It all adds up.

Also, a BC547 is not equivalent to a TIP31. It is a small signal TO-92 transistor with an absolute max collector current of 200 mA.

ak
 

Thread Starter

mrobbo15

Joined Jan 2, 2017
4
Why 19 strings. That's fine for two 3914's, but they cannot sink 70 mA. 14-16 strings works out better for most logic circuits.

ak
19 Was just based on how many LEDs I wanted to try fit in, but I can definitely do 14-16 instead. In fact, it might fit better that way.
 

dannyf

Joined Sep 13, 2015
2,197
The outline:
Fairly easy to handle:

1. Build 19 timer units, each delays on 500ms/19 or 25ms.
2. Cascade the timers so one would turn on 25ms after the turn signal on, thee 2nd will turn on 25ms later, and so on and so forth.
3. The turn off signal rears then all.
4. Since each string is only 70ma, they can be powered by the 555 easily.
 

AnalogKid

Joined Aug 1, 2013
10,986
Fairly easy to handle:
1. Build 19 timer units, each delays on 500ms/19 or 25ms.
That is at least four times as many component pins to deal with than any other method suggested so far. Please show a schematic of the first two or three stages as an example.

ak
 

AnalogKid

Joined Aug 1, 2013
10,986
While a counter and decoder will work, the decoding into a bar graph display is overly complex and quickly drowns itself in gates. That leaves three basic methods:
1. Analog (!) ramp and comparators using two LM3914's.
2. Analog (!!) ramp and comparators using 4-5 quad comparators (LM339, etc.).
3. Clock oscillator and digital shift registers.

Each of these methods is followed by current amplifiers to drive the LED strings, such as ULN2004/2804. The most simple is the shifter for 16 stages; 5 chips, only two timing components, 72 total pins plus decoupling, LED current limiting, etc.
1 - 555 clock, 1 resistor, 1 capacitor
2 - SIPO (serial-in, parallel-out) shift register chips
2 - ULN2804 octal darlington transistor arrays
6 - decoupling capacitors; 5 small, 1 large

Or something like that.

ak
 

dannyf

Joined Sep 13, 2015
2,197
The beauty of a 555 here is its current capability. As it can drive directly too 200ma for the BJT version.

If you don't care about that, a better route to go for part count is a MCU. A 28pin MCU can easily handle the number of output you want. You have to use a driver here, however.
 

dannyf

Joined Sep 13, 2015
2,197
here is a quick demo of 3 leds being driven sequentially:

1) the turn signal comes in on PD0 (active high);
2) three leds on PD1..3 (all active high);
3) each led turns on 50ms after the previous one.
4) all leds are reset when PD0 goes low.

ledflasher-sim.PNG

the programming is done graphically:

ledflasher-programming.PNG

you can easily add more output pins.

hard to beat it in terms of part count.
 
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