Halls sensor? Reed Switch?

MisterBill2

Joined Jan 23, 2018
8,716
Please explain how you feel this will result in self destruction. I see a high input turning on the lower transistor while holding the upper transistor off, causing a low at the output. The next scenario, a low at the input will turn the upper transistor on and turn the lower transistor off, causing a high at the output. Am I wrong? If so - I can take criticism. If I learn from my mistakes - I learn something. And I certainly don't purport to know "everything". In fact, there is more I don't know than what I DO know.

In a thread I posted recently I used the opposite setup with an emitter follower that worked. Unfortunately in my circuit, the device the buffer was powering has an internal short, causing excessive amounts of current to be drawn, which resulted in the destruction of a PNP transistor. Not being the most "highly skilled" person on this website, I make mistakes. Sometimes I give bad advice. When I learn it I amend my comments - or remove them altogether. So if you can teach me something then please - by all means, teach away.

If one is concerned about the circuit self destructing - then just use a CD4049 buffer. They can give either a high or low output, which can toggle the LED's as the thread starter is seeking to do. Only, with a 4049, there are six in a package and only one would be needed - provided (and I don't know this) they can handle the current for the LED's.
That circuit in post #13 has the bases tied to each other and also the collectors. So if the transistors have any leakage current at all they will quickly both be biased into full conduction. That will indeed cause problems.
And we do not yet have any information about the power source. But we do now understand the application, which is for a quite bright set of lights that are easy to see while backing up a truck to a loading dock. The motivation for a hidden magnetic triger is not clear, though.
So we need more information from the TS to avoid random guessing
 

Thread Starter

Mumeral

Joined Jan 28, 2020
12
The wiring for the switch less light is the same with the leads where the switch is. The flasher flashes the red light. The switch is a manual switch, so maybe a Reed toggle switch might be better? I don't have a power source selected besides probably the 110v light, but I am probably better off with the DC 12v version. Suggestions for power source and parts are welcome, I don't even have a magnet either.
 

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MisterBill2

Joined Jan 23, 2018
8,716
I will make a guess that the building is wired with 120 volts AC for lighting and not wired wit 12 or 24 volts DC. So here are what I see as the alternatives: If the 120 volt version with incandescent bulbs is selected the power can be taken directly from the lighting supply, both simple and direct, but the bulbs will have a definit life span and thus need replacement, probably on a yearly basis if they are always on. If the 12volt DC version is selected then a mains powered DC power supply will need to be installed, with wiring both from the mains to the supply and from the supply to the light assembly. Probably the LEDs will last for quite a few years. The only downside to both approaches is that the switch is accessible to anybody passing by, and that may, or not, be a problem.
 
Does Fairchild own Honeywell now? I've had good luck with the Honeywell SS495 and a few of the other analog out varieties.

I can't keep up with all the mergers and acquisitions - seems like every time I try to recommend a product it's under a different name than the last time I checked!
I have the same feeling...
I have used successfully components from Allegro, they have a wide portfolio of products. Here are som proximity switches:

https://www.allegromicro.com/en/Products/Sense/Switches-and-Latches/Three-Wire-Hall-Effect-Switches
 

Thread Starter

Mumeral

Joined Jan 28, 2020
12
So if I get this right I need an IC chip that's a logic gate that inverts the input signal Red light, Invert signal Green Light. Runs on 12v and a reed switch that runs parallel to the power and inputs the first inverter pin. I also will need a power supply (brick) that out puts 12v +/-5a? I just don't know which parts to buy, there's a huge variety of stuff out there.
 

MaxHeadRoom

Joined Jul 18, 2013
23,084
Unless i am missing something, why not a simple SPDT relay?
The only issue, unless the relay coil is very low current I would tend to use a SS400 product instead of reed switch to pick up the relay, relay energised to light the green, de-energised Red?
Why do you need logic semi conductors?
Max.
 
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Reloadron

Joined Jan 15, 2015
5,934
I am trying to setup a PCB to use a magnet for the indicator. I'm going to use a 12v 0.3a red/green LED stop light and MAYBE a flipflop relay. I am not sure which to use, and I'm just a hobbyist at best.View attachment 197774
Something like this.
I can share this much with you. The circuit, as drawn for your application won't work. This is the data sheet for the 74ALS109AN dual JK Flip Flop. Two things of importance are looking at the data sheet for the Stop/Go lamps the DC versions they come in 12 volt and 24 volt DC versions. The chip is a 5 Volt chip and if we look at the data sheet the IOH High-level output current and IOL Low-level output current are not adequate to drive the LED versions which require 0.030 amp (30 mA).

The 74ALS109AN is two independent J-K positive-edge-triggered flip-flops. A low level at the preset (PRE) or clear (CLR) inputs sets or resets the outputs regardless of the levels of the other inputs. What that means is that for each positive going clock transition the Q and /Q outputs will toggle state. Now you can clock the chip with a magnetic reed switch but if doing so you may want to include some form of switch debounce at your flip flop clock in. There are other chips which would work and you still need debounce and using a few transistors or MOSFETs to get the drive current for your lights. So as drawn for your application things simply won't work.

If you want to do this using a magnetic reed switch I would consider Mr. Bill's suggestion and just use a simple SPDT Magnetic reed switch. With magnet present you have one color and less the magnet, the other color.
I think that there are also 3-terminal reed switches, which would make the scheme much simpler.
You can also use a Hall Sensor to trigger the change depending on exactly what you want to do.

Ron
 

MaxHeadRoom

Joined Jul 18, 2013
23,084
The dual contact reed switches I have used were rated around 200ma, which I suspect is typical, from experience I much prefer solid state sensor over the reed switches, which seem to exhibit much shorter life spans than the reed version,
My two ¢. ;)
Max..
 

Tonyr1084

Joined Sep 24, 2015
5,964
if we look at the data sheet the IOH High-level output current and IOL Low-level output current are not adequate to drive the LED versions which require 0.030 amp (30 mA).
We keep getting stuck on the idea of a single LED, the type we solder to circuit boards. The lights the TS has linked are much bigger than a single LED and likely will draw much more current than 30 mA.
The dual contact reed switches I have used were rated around 200ma, which I suspect is typical, from experience I much prefer solid state sensor over the reed switches, which seem to exhibit much shorter life spans than the reed version,
My two ¢. ;)
Max..
I tried the dual contact reed switch a few years back. I found their functionality quite unsatisfying. That may have been due to their origin - China. Nevertheless, I have yet to find one that is robust. Not that I've been looking for such.
 

Tonyr1084

Joined Sep 24, 2015
5,964
I remember a thread like that. It seems someone wanted to build a PWM that could be bought for ~$5.
If we're going to Buy everything then why do we need to ask how to build anything? Some people enjoy building simply for the satisfaction of accomplishment. I used a spare DD flip flop to create a clock signal. Why? Because I didn't want to add another chip to my board. And I was only going to use one of the flip flops. Why not use the other to create a clock pulse and not have to build another chip onto the board? Of course I could have just Bought something. But where's the fun in that?!
 

Reloadron

Joined Jan 15, 2015
5,934
We keep getting stuck on the idea of a single LED, the type we solder to circuit boards. The lights the TS has linked are much bigger than a single LED and likely will draw much more current than 30 mA.
Tony, this is the data the thread starter linked to for the lights. That link spells out the current required for the Stop and Go lights. I am not stuck on a single LED?
Stop And Go Light Technical Specifications
Housing:
Safety Yellow Polypropylene
Power Source: 12 Or 24 volt DC operation or 115 volt AC operation
Lens diameter: 4 1/4″
Light Set Dimensions: 11 3/8″ H x 6 3/8″ W x 3 3/4″ D
Retrofit Kit Dimensions: 4.125″ H x 4.12″ W x 1.125″ D

Both the 12 volt and 24 volt versions spell out 0.030 Amp (30 mA) which is what I based my post on. Granted it seems like a low number but I needed to base things on something so used their number. Possibly as a matter of perspective and in keeping with 12 volts. Typical motorcycle tail light about 600 mA (low element of an 1157 lamp) while a brighter LED replacement about 200 mA. I have no idea where they got the 30 mA but it is their data for their product. I merely pointed out the original circuit as drawn would not work and they why or a few why.

I tried the dual contact reed switch a few years back. I found their functionality quite unsatisfying. That may have been due to their origin - China. Nevertheless, I have yet to find one that is robust. Not that I've been looking for such.
If you ever are then look towards the guys who make them like Hamlin. Littlefuse is another reliable manufacturer. That or suffer the consequences.

Ron
 

Sensacell

Joined Jun 19, 2012
2,820
Please explain how you feel this will result in self destruction. I see a high input turning on the lower transistor while holding the upper transistor off, causing a low at the output. The next scenario, a low at the input will turn the upper transistor on and turn the lower transistor off, causing a high at the output. Am I wrong? If so - I can take criticism. If I learn from my mistakes - I learn something. And I certainly don't purport to know "everything". In fact, there is more I don't know than what I DO know.

In a thread I posted recently I used the opposite setup with an emitter follower that worked. Unfortunately in my circuit, the device the buffer was powering has an internal short, causing excessive amounts of current to be drawn, which resulted in the destruction of a PNP transistor. Not being the most "highly skilled" person on this website, I make mistakes. Sometimes I give bad advice. When I learn it I amend my comments - or remove them altogether. So if you can teach me something then please - by all means, teach away.

If one is concerned about the circuit self destructing - then just use a CD4049 buffer. They can give either a high or low output, which can toggle the LED's as the thread starter is seeking to do. Only, with a 4049, there are six in a package and only one would be needed - provided (and I don't know this) they can handle the current for the LED's.
In this configuration, the base-emitter junctions of both transistors are connected in series, across the power supply.
The base-emitter junction 'looks' like a diode, with a drop of around 0.65V - essentially you then have two of these diodes in series - shorting the power supply.
An unlimited amount of current flows until something melts.

By exchanging the transistors (see below) they become complementary emitter followers, only one of the B-E junctions can be forward-biased at any instant.
This configuration works as advertised. It's a subtle but critical difference.


good_buffer.png
 

Tonyr1084

Joined Sep 24, 2015
5,964
@Sensacell in your LT Spice circuit - it's the one I used on a project. I received some flack saying my configuration - same as yours - would not work, that I was going about it all wrong. It was then recommended I swap the NPN and the PNP for the configuration I drew. To be honest I never built it that way - the way I drew. But I was convinced that it would work and that it was the right way to go about it. I'm finding a lot of conflicting information here, which makes learning more difficult.

As you have (in LT Spice) is the way I built my circuit (with a 555 driving it). My PNP burned out after about 15 minutes of continuous testing due to excessive heat. At a loss to understand why, I finally - at the behest of others here - tested my IGBT, which is what my buffer was driving. Turned out my IGBT had gone semi-short circuited and it was drawing excessive current through the PNP transistor. After changing my IGBT for another one (the last one I had) I powered it directly from the 555 and it works. It's been tested for a number of continuous hours of operation running an automotive light bulb. VERY little heat was measured using an IR thermometer. So I never got back to using the transistor buffer inverter I drew. So I have no experience that shows it working. However, like I said, I was sort of directed to believe it would work. So now I've learned even more.

Thank you for explaining it to me. But this is not my thread. Since the TS has a possible solution waiting to be ordered, the only thing I can add in the way of commentary is that I've gotten plenty of 12 V 5 A power bricks from places like DirecTV and Xfinity. Their service centers take back old units and upgrade their customers with newer units. Almost always they get power bricks back but they don't re-use them. I've asked if I could have one or two every now and then when I'm passing by and they just GIVE them to me. So a fairly reliable PS of 12V 5A is easily obtained from the likes of Xfinity or DirecTV. I'm sure other places may also be willing to hand them out.

Post #6 has been edited. I've removed my drawings because they were in error.
 
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Tonyr1084

Joined Sep 24, 2015
5,964
Since it's a quick edit - here's the circuit I've built and worked.

1580356211302.png

And here's the LED's in the edited version:

1580356250189.png
When signal is High, the upper transistor conducts, bypassing the green LED and the red LED is lit.
When signal is Low, the lower transistor conducts, bypassing the red LED and the green LED is lit.

With that - I'm done.
 

atferrari

Joined Jan 6, 2004
4,321
In this configuration, the base-emitter junctions of both transistors are connected in series, across the power supply.
The base-emitter junction 'looks' like a diode, with a drop of around 0.65V - essentially you then have two of these diodes in series - shorting the power supply.
An unlimited amount of current flows until something melts.

By exchanging the transistors (see below) they become complementary emitter followers, only one of the B-E junctions can be forward-biased at any instant.
This configuration works as advertised. It's a subtle but critical difference.


View attachment 197865
Sorry @Sensacell , has this been actually built?
No chances to do it right now myself, wondering if we can rule out even marginal shoot through.
 

Tonyr1084

Joined Sep 24, 2015
5,964
As I mentioned, in another thread (here) I was advised that I had it backwards. The first 10 posts pretty much cover how I came to the inverter I initially drew (but have since edited and removed the bad drawing). Some times we don't always get good advice. If you look at that post, follow it through if you wish, you'll see how I was steered the wrong direction. Nevertheless, in my post I have solved my issue, as is apparent on page 4 of that post.
 

Sensacell

Joined Jun 19, 2012
2,820
Sorry @Sensacell , has this been actually built?
No chances to do it right now myself, wondering if we can rule out even marginal shoot through.
No shoot through possible- to turn both transistors on simultaneously, the input would need to be 1 Vbe above the output and 1 Vbe below at the same time!

A class AB amplifier achieves simultaneous current flow in both transistors with a bias scheme that places the bases of the transistors at two different voltages- not the case here as they are tied together.
 

atferrari

Joined Jan 6, 2004
4,321
No shoot through possible- to turn both transistors on simultaneously, the input would need to be 1 Vbe above the output and 1 Vbe below at the same time!

A class AB amplifier achieves simultaneous current flow in both transistors with a bias scheme that places the bases of the transistors at two different voltages- not the case here as they are tied together.
Ok thanks.
I will keep it in my "mental file".
 
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