LED brake light/blinker ideas

[Torben]

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]

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]

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]

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]

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

 

[ljcox]

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.

This does not make any sense. How can 100k resistors work and 470 Ohm resistors not work?

What LEDs are you referring to?

What I mean is the LEDs that are in place of the lamp I called "Front Lamp" in my drawing. This and the "Centre Lamp" need to have a 470 Ohm resistors in parallel with them.
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.
The transistor type should make no difference. Virtually any PNP should do.
Thanks again,

Torben

Torben,
I'm confused by your descriptions. I feel we may be talking at cross purposes. So I'll try to clear the air.

If you look at my diagram; if the Brake line is high and the flasher line is low, current should flow from the Brake line, though the emitter of Q2 to the base, then via R2 (the 2.2k) through the "Front Lamp" to gnd.

If the "Front Lamp" is a LED, there will be insufficient base current since the LED represents a very high resistance. That is why you need a 470 Ohm resistor in parallel with it.

The lamp I marked as "Left" on the right hand side of my drawing (ie. the rear brake/flasher) does not need a 470 in parallel.

 

[Torben]

Len,

Oh damn. OK, here's the thing which is probably getting us talking about different things: I'm doing only the back end. No front lamps at all in my sims or on my breadboard. Just one 6V connection for the brake input, and another 6V line for the flasher input.

I guess I never mentioned that in my mind I was never considering replacing the front lamps with LEDs--I would just tap into the existing brake/flasher line. My apologies.

This would explain why I'm talking about a 100k (it is where the parallel resistor on the *front* LEDs would be) and you're talking about 470 ohms: I thought the 470s were supposed to parallel the *rear* LEDs.


Sorry about the confusion,

Torben

 

[ljcox]

Torben,

Yes, I eventually came to the conclusion that there was a "crossing of wires" as your resposes were not making much sense.

The attachment shows the approximate voltages and currents for 2 cases.

1. with the Brake line high.

2. with the Flasher line high.

If the front flashers are incadescent, then the 470 resistors are not required.

I have not indicated the LED currents as I don't know what you used. But with a base current of about 1.95 mA, the collector will be able to source at least 20 mA with the transistor still in saturation.

EDIT. The exception to the above is that if the centre lamp is a LED one, then it will need a 470 in parallel.

 

[Torben]

Len,

Yes, I'm glad that is was simply a misunderstanding. I can deal with learning that I may need to be more clear in my questions. I was not happy thinking that everything I thought I'd learned about transistors up to this point was wrong.

It is now clear. Once again, thanks for the persistence! This must have been agonizing.


Torben

 

[ljcox]

,
No problem. As someone once said, a picture is worth a thousand words.

Verbal descriptions are often not precise enough.