LED brake light/blinker ideas - Page 3 - Electronic Circuits Projects Diagrams Free

Torben · 25th October 2007, 09:48 AM

Quote:

Originally Posted by ljcox

Torben,
That's interesting.

For the LED case, we could connect resistors in parallel with them in order to provide a path for the base currents.

And in the LED case, there won't be the cold current issue so the base resistors could be higher.

For example, say the resistors had to be 2k. Then we could connect 470 in parallel with the LEDs and the R1 ~ R4 resistors could be 1k5.

Oddly enough it works with LEDs now without the proposed parallel resistors, which means either A) I screwed up earlier, or B) I'm screwing up now.

Attached is an image of what I have for the LEDs on this right now. I'm only including one side here. In this version I have an incandescent lamp on the front flasher and an LED on the rear. It works just the same without the front flasher present, or with an LED in front instead of the lamp.

The sim seems to be doing exactly what I would want, and from what I (think I) know, what I would expect.

I'm confused about why the transistors wouldn't be getting current. Aren't the base resistors doing that?

Anyway, the brake (red line) is pressed 3 times slowly and the left flasher is on until 2/3 of the way through the simulation. Time is greatly compressed here. The rear flasher inverts when the brake changes state but flashes just fine.

Torben

[Edit: The lamp wattage and supply voltages in the image don't match my above explanation but I just changed them back to 12V and 6W and got my expected results. The lamp current line now evens out at 0.5A.]

ljcox · 25th October 2007, 10:59 AM

Torben,
I was a bit confused when I wrote the previous post. I was confusing the transistor on and off criteria.

If all the lamps are LEDs, then there may not be a problem.

If you look at the voltage on the brake line, you will see that it rises to about the LED voltage (if it's a red LED, it will be probably about 1.7 Volt) when Q1 is on.

So the base current is going through the brake LED.

There won't be a problem provided that the current is low enough so that LED does not glow AND the transistor has enough base current to saturate it.

But I'm not sure if both of these criteria can be met.

It is usual to make the base current about one tenth of the collector current in order to ensure that the transistor is saturated.

Can we meet this criterion and not cause the LED to glow?

EDIT
There will be more than one LED in the lamp. So it will also depend upon how many LEDs there are in series. With a 6 Volt system, you could probably have 3 LEDs in series.

If there are 3 in series, this will significantly reduce the base current.

So a resistor in parallel may be necessary.

Torben · 28th October 2007, 10:05 AM

Hi Len,

OK, I built the thing. I didn't really get what you were on about above until I did--not your fault; I just needed to see it in action. The sim is nice but doesn't really help explain it. Getting the build wrong a few times, thinking about it, etc., helped.

I couldn't get the base current to shut off the flasher LED when the brake and flasher were both HIGH until I added a resistor to ground from the Left brake line input. If I'm doing this right, that means this resistor is dissipating 120nA when the left flasher is on. Seems acceptable.

I haven't tried driving a front flasher from it yet though, and this circuit on the board is just a very simple exploration: 2 PNPs, 5 resistors, and some wire.

Anyway, thanks very much for the patient explanation. I think I'm getting what's going on in the circuit now. Perhaps I'll build one for a high brake/flasher on my truck canopy. Someday. I won't be using 3906s though.

Torben

[Edit: nb: This was an all-LED build, one LED per side. Just to check the basic idea before proceeding.]

[Edit 2: I added 2 more LEDs per side. These are all normal 20mA garden-variety red LEDs. It works fine as long as the voltage doesn't sag. At all.

]

ljcox · 28th October 2007, 09:56 PM

Quote:

Originally Posted by Torben

Hi Len,

OK, I built the thing. I didn't really get what you were on about above until I did--not your fault; I just needed to see it in action. The sim is nice but doesn't really help explain it. Getting the build wrong a few times, thinking about it, etc., helped.

I couldn't get the base current to shut off the flasher LED when the brake and flasher were both HIGH until I added a resistor to ground from the Left brake line input. Do you mean the left flasher input? If I'm doing this right, that means this resistor is dissipating 120nA Do you mean nA or mA or do you mean nW or mW? when the left flasher is on. Seems acceptable.

I haven't tried driving a front flasher from it yet though, and this circuit on the board is just a very simple exploration: 2 PNPs, 5 resistors, and some wire.

Anyway, thanks very much for the patient explanation. I think I'm getting what's going on in the circuit now. Perhaps I'll build one for a high brake/flasher on my truck canopy. Someday. I won't be using 3906s though.

Torben

[Edit: nb: This was an all-LED build, one LED per side. Just to check the basic idea before proceeding.]

[Edit 2: I added 2 more LEDs per side. These are all normal 20mA garden-variety red LEDs. It works fine as long as the voltage doesn't sag. At all.

]

As you can see by the questions, I'm a bit confused. (not unusual)

It would help if you could post a diagram.

Torben · 28th October 2007, 10:29 PM

Quote:

Originally Posted by ljcox

As you can see by the questions, I'm a bit confused. (not unusual)

It would help if you could post a diagram.

Oh dear. I can see why you're confused. I posted complete gibberish.

I meant "left flasher line", not "left brake line".

And I meant that the new resistor to ground would pass 120uA (see below for why 120 is wrong too), not dissipate it. The nA thing was just me counting zeroes wrong, I guess. The resistor to ground is 100k. I got a bit confused because I did the calculation on paper first for 6V, got 60uA, and then simmed it and got 120uA, which confused me until I remembered that the sim was running 12V, not 6V like my breadboard. Then when I posted I just read the value off the sim instead of my paper. Argh. So that value should really be 60uA for a 6V supply, not 120nA.

I must try to remember to post *before* falling asleep. Apologies.

The schematic is the same as the one I posted earlier (the LTSpice image with the graph) except the front incandescent lamp model has been replaced with a 100k resistor.

Torben

ljcox · 29th October 2007, 01:23 AM

I've revised your previous post (partially quoted below) in the light of your latest.

Quote:

Originally Posted by Torben

Hi Len,

I couldn't get the base current What I think you mean is that you could not switch Q1 off. This needs the base - emitter voltage to be < 0.5 Volt. The base current is virtually zero when the transistor is off. to shut off the flasher LED when the brake and flasher were both HIGH until I added a resistor to ground from the Left flasher line input. If I'm doing this right, that means that the current through this resistor is 120uA when the left flasher is on. Seems acceptable. Yes. See comments below.

Torben

I expect that the 100k resistor is necessary for the simulation, ie. the software can't analyse the circuit without it.

Torben · 29th October 2007, 01:31 AM

Quote:

Originally Posted by ljcox

I've revised your previous post (partially quoted below) in the light of your latest.

Quote:

Originally Posted by torben

Hi Len,

I couldn't get the base current What I think you mean is that you could not switch Q1 off. This needs the base - emitter voltage to be < 0.5 Volt. The base current is virtually zero when the transistor is off. to shut off the flasher LED when the brake and flasher were both HIGH until I added a resistor to ground from the Left flasher line input. If I'm doing this right, that means that the current through this resistor is 120uA when the left flasher is on. Seems acceptable. Yes. See comments below.

Torben

Actually, it works fine in the sim without the resistor, but in real life it needed the resistor there to cause the transistor to block the current to the flasher LED when the flasher line went high. Otherwise, when the brake line was high both LEDs would light up, but when the flasher line would then go high, the flasher LED would not turn all the way off. With the 100k grounding resistor there, it worked perfectly.

Quote:

Originally Posted by ljcox

I expect that the 100k resistor is necessary for the simulation, ie. the software can't analyse the circuit without it.

Not sure why, but the sim doesn't seem to care whether it's there or not.

Torben

ljcox · 29th October 2007, 02:08 AM

Quote:

Originally Posted by Torben

Actually, it works fine in the sim without the resistor, but in real life it needed the resistor there to cause the transistor to block the current to the flasher LED when the flasher line went high. I assume you mean the case where both the brake and flasher lines are high. If the F is high and the B is low, the flasher LED should glow. Otherwise, when the brake line was high both LEDs would light up (as they should), but when the flasher line would then go high, the flasher LED would not turn all the way off. This does not make sense to me. When the F line is high, the resistor should have no effect. It is only has an effect when the F line is low - since it tends to pull the line low if the F line is (without the resistor) going from open to High. With the 100k grounding resistor there, it worked perfectly.

Not sure why, but the sim doesn't seem to care whether it's there or not. See below.

Torben

In the sim case, the voltage source driving the F line goes from low to high. When it is low, the F line is at 0 Volt. But in the real case, I assume that the F line is going from open to High (as I said above) hence there is no path for the current when it is open.

However, I still can't explain what you're observing. I'll need to think some more about it.

ljcox · 29th October 2007, 02:26 AM

I still can't make any sense of it. So I've drawn the attachement.

Please measure the voltages (without the 100 k resistor) at the C, B & E of both transistors for each case.

A point that just occurred to me is the that EB junctions will go into Zener breakdown if the supply is 12 Volt. And it may occur at 6 Volt.

Torben · 29th October 2007, 03:31 AM

Quote:

Originally Posted by ljcox

I still can't make any sense of it. So I've drawn the attachement.

Yep, that's the truth table I've been working from too. This truth table matches the exhibited behaviour when I have the 100k resistor in place.

Quote:

Please measure the voltages (without the 100 k resistor) at the C, B & E of both transistors for each case.

A point that just occurred to me is the that EB junctions will go into Zener breakdown if the supply is 12 Volt. And it may occur at 6 Volt.

I'll have to measure them later tonight but I'll do it.

Torben

Torben · 29th October 2007, 08:46 AM

Quote:

Originally Posted by ljcox

I still can't make any sense of it. So I've drawn the attachement.

Please measure the voltages (without the 100 k resistor) at the C, B & E of both transistors for each case.

A point that just occurred to me is the that EB junctions will go into Zener breakdown if the supply is 12 Volt. And it may occur at 6 Volt.

I did my measurements and got what's in my attachment.

From the fact that I need 2K2 base resistors instead of 2.2K, which I also get from basic calculations, I think my Radio Shack "similar to 3906" transistors are not that similar to 3906s after all. The 2k2s work in the sim but not on the breadboard.

As an aside, I tried simplifying and going back to my books, and found that in the sim, the 100k resistor to ground was not needed if I put 22K pulldowns on the bases of both transistors. I don't think the pulldown is needed on the base of the PNP which has its emitter to the brake line.

This bombed on the breadboard. But the version with 2K2 base resistors and the 100k resistor to ground from the left flasher input line works great.

I'm stumped for tonight. I'll have another look tomorrow.

Torben

[Edit: Oops. Uploaded smaller image. Also, I just tried this with 2K2 base resistors, a 220K pulldown on the base of Q2 (the transistor bridging the brake and flasher lines; I call the one interrupting the flasher line Q1), and no 100k resistor--and voila: it works fine. i.e. it follows the truth table you posted.]

ljcox · 29th October 2007, 11:30 PM

Torben,
What do you mean by "pulldowns on the bases of both transistors"?

Are the resistors connected from base to emitter?

I'll look at your measurements later.

ljcox · 30th October 2007, 02:21 AM

Torben,
I think I know what you mean by "pull down resistors". I assume you mean resistors from the bases to gnd. This is not necessary or sensible.

I've looked at your measurements.

They show that, without the 100k resistor, there is insufficient base current to turn Q2 on. (4.63 - 4.62)/2.2 = 5 uA approx.

This is not a surprise to me.

As I stated originally, you need either incadescent lamps for the front flashers and the centre lamp in order to provide a low resistance path for the base currents.

If you use LEDs, then you need a resistor of 470 Ohm in parallel with the LED lamps. Your figures show that 100k is not low enough to saturate the transistors. For example, if Q2 was saturated, then the collector voltage would be about 5.0 Volt (given that you measured 5.21V on its emitter), not 4.07 Volt.

So I suggest that you install 470 resistors across the LED lamps and use 2.2k resistors for the bases as in my diagram.

Torben · 30th October 2007, 05:13 AM

Hi Len,

OK. I wasn't prepared to trust the thing, since neither intuition nor my calculator liked the 22K base resistors with 220k pulldowns (you're right about what I meant by pulldown, by the way: base resistors directly to ground). I was using the pulldowns to try to ensure that the bases weren't floating when not connected to anything (i.e. when the switch/relay was open).

I'm going to go back to the 2k2s and try the 470 resistors in parallel with the LEDs later tonight. My solution did work for me, but without understanding why, I wouldn't trust it. So I'll also run the numbers for both ideas (mine which "worked" but probably for the wrong reasons, and yours, which is far more likely to be the "right" solution) and see if I can't figure that much out.

So basically this whole thing came down to me not getting the point that the LEDs are not providing the base current path to ground? Seems that way to me.

Torben

Torben · 30th October 2007, 08:40 AM

Len,

OK, I tried it without the 100k resistor and pulldowns, but with the 470 ohm resistors parallel to the LEDs. With that configuration, connecting only the brake input to 6V would light the brake LEDs but not the flasher LEDs. However, replacing the 100k resistor did the trick. I changed the 100k resistor to 22k and it worked perfectly.

I think I'm going to cram for a while with my books and calculator to try to figure out exactly what's going on here, what the current paths are, and maybe order some real 3906s to work with instead of these so-called "similar" PNPs I've got here.

Thanks again,

Torben

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25th October 2007, 10:59 AM	(permalink)
ljcox	Torben, I was a bit confused when I wrote the previous post. I was confusing the transistor on and off criteria. If all the lamps are LEDs, then there may not be a problem. If you look at the voltage on the brake line, you will see that it rises to about the LED voltage (if it's a red LED, it will be probably about 1.7 Volt) when Q1 is on. So the base current is going through the brake LED. There won't be a problem provided that the current is low enough so that LED does not glow AND the transistor has enough base current to saturate it. But I'm not sure if both of these criteria can be met. It is usual to make the base current about one tenth of the collector current in order to ensure that the transistor is saturated. Can we meet this criterion and not cause the LED to glow? EDIT There will be more than one LED in the lamp. So it will also depend upon how many LEDs there are in series. With a 6 Volt system, you could probably have 3 LEDs in series. If there are 3 in series, this will significantly reduce the base current. So a resistor in parallel may be necessary. __________________ Len Last edited by ljcox; 25th October 2007 at 11:03 AM.

28th October 2007, 10:05 AM	(permalink)
Torben	Hi Len, OK, I built the thing. I didn't really get what you were on about above until I did--not your fault; I just needed to see it in action. The sim is nice but doesn't really help explain it. Getting the build wrong a few times, thinking about it, etc., helped. I couldn't get the base current to shut off the flasher LED when the brake and flasher were both HIGH until I added a resistor to ground from the Left brake line input. If I'm doing this right, that means this resistor is dissipating 120nA when the left flasher is on. Seems acceptable. I haven't tried driving a front flasher from it yet though, and this circuit on the board is just a very simple exploration: 2 PNPs, 5 resistors, and some wire. Anyway, thanks very much for the patient explanation. I think I'm getting what's going on in the circuit now. Perhaps I'll build one for a high brake/flasher on my truck canopy. Someday. I won't be using 3906s though. Torben [Edit: nb: This was an all-LED build, one LED per side. Just to check the basic idea before proceeding.] [Edit 2: I added 2 more LEDs per side. These are all normal 20mA garden-variety red LEDs. It works fine as long as the voltage doesn't sag. At all. ] Last edited by Torben; 28th October 2007 at 10:18 AM.

29th October 2007, 11:30 PM	(permalink)
ljcox	Torben, What do you mean by "pulldowns on the bases of both transistors"? Are the resistors connected from base to emitter? I'll look at your measurements later. __________________ Len

30th October 2007, 02:21 AM	(permalink)
ljcox	Torben, I think I know what you mean by "pull down resistors". I assume you mean resistors from the bases to gnd. This is not necessary or sensible. I've looked at your measurements. They show that, without the 100k resistor, there is insufficient base current to turn Q2 on. (4.63 - 4.62)/2.2 = 5 uA approx. This is not a surprise to me. As I stated originally, you need either incadescent lamps for the front flashers and the centre lamp in order to provide a low resistance path for the base currents. If you use LEDs, then you need a resistor of 470 Ohm in parallel with the LED lamps. Your figures show that 100k is not low enough to saturate the transistors. For example, if Q2 was saturated, then the collector voltage would be about 5.0 Volt (given that you measured 5.21V on its emitter), not 4.07 Volt. So I suggest that you install 470 resistors across the LED lamps and use 2.2k resistors for the bases as in my diagram. __________________ Len Last edited by ljcox; 30th October 2007 at 02:24 AM.

30th October 2007, 05:13 AM	(permalink)
Torben	Hi Len, OK. I wasn't prepared to trust the thing, since neither intuition nor my calculator liked the 22K base resistors with 220k pulldowns (you're right about what I meant by pulldown, by the way: base resistors directly to ground). I was using the pulldowns to try to ensure that the bases weren't floating when not connected to anything (i.e. when the switch/relay was open). I'm going to go back to the 2k2s and try the 470 resistors in parallel with the LEDs later tonight. My solution did work for me, but without understanding why, I wouldn't trust it. So I'll also run the numbers for both ideas (mine which "worked" but probably for the wrong reasons, and yours, which is far more likely to be the "right" solution) and see if I can't figure that much out. So basically this whole thing came down to me not getting the point that the LEDs are not providing the base current path to ground? Seems that way to me. Torben

30th October 2007, 08:40 AM	(permalink)
Torben	Len, OK, I tried it without the 100k resistor and pulldowns, but with the 470 ohm resistors parallel to the LEDs. With that configuration, connecting only the brake input to 6V would light the brake LEDs but not the flasher LEDs. However, replacing the 100k resistor did the trick. I changed the 100k resistor to 22k and it worked perfectly. I think I'm going to cram for a while with my books and calculator to try to figure out exactly what's going on here, what the current paths are, and maybe order some real 3906s to work with instead of these so-called "similar" PNPs I've got here. Thanks again, Torben