Hall Effect - Trigger and Reset.

Thread Starter

Andrew216

Joined Jun 24, 2015
29
I am using a monostable multivibrator (TLC555) to turn on a LED for 30 secs. I am running this off a 3v coin battery which goes into a voltage doubler(MAX1682). I want to use a hall effect to trigger the LED timer. (A car door [with a magnet attached] opening will trigger the hall effect which will trigger the timer.) I have a Texas Instruments DRV5013ADEDBZRQ1 Hall Effect that I will be using. I have pins 1(Vcc) and 3(Ground) tied to together with a .01uf cap. and Pins 1(Vcc) and 2(output) tied together with a 10k resistor. I have the output of the Hall Effect going to the trigger pin. When I use a magnet and pass it by the Hall Effect everything works perfect. So my question is... How do I make it so if the car door were to close before the 30 secs is up it will turn the timer and LED off until the door is reopened which will start the timer again. I know this will rely on the reset pin but how?!

I have attached a copy of my schematic but it does not include the hall effect. Only the timer circuit and voltage doubling circuit are pictured. Any help would be greatly appreciated.
 

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crutschow

Joined Mar 14, 2008
34,201
Below is my take on a circuit to do what you want.
I added a CD4049 inverter to get the proper polarity for the Reset signal and some capacitor coupling to differentiate the signal so that the Trigger and Reset signals are a short pulse during the Hall output pulse transitions.
The LTspice simulation shows the 555 output being reset when the door closes after 5s (0V is door open) but gives a 30s pulse when the door stays open.

555 one-shot.gif
 

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Thread Starter

Andrew216

Joined Jun 24, 2015
29
Thank you! I am going to get an inverter and try this ASAP. Right now I'm only using a momentary switch to reset it. I'm really pushing for no external parts minus the magnetic strip to activate the circuit so this sounds great. I will post my results as soon as I have something put together!
 

Alec_t

Joined Sep 17, 2013
14,263
Bear in mind that the CD4049 IC is static-sensitive and that the inputs of the five unused gates in the IC should be connected to either supply rail so that they are not left floating.
 

AnalogKid

Joined Aug 1, 2013
10,971
A bouncy hall effect sensor seems like an undue complication. Here is another approach. If you want to stay with a magnetic actuator, the reed switch is any common alarm door/window sensor. Or it can be a normally open pushbutton switch that is pushed to the closed condition by the car door. For a normally closed switch, swap SW1 and R1. Note that the components are what is in my design library. C1 can be an aluminum electrolytic, and all of the resistors can be 5% tolerance.

ak
 

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ebeowulf17

Joined Aug 12, 2014
3,307
A bouncy hall effect sensor seems like an undue complication.

I'm curious what you mean by bouncy here. A Hall effect switch with appropriate hysteresis (which this one appears to have built in) should have very clean output switching. Reed switches, being mechanical switches, naturally have some amount of contact bounce when switched. I don't know if it will matter either way for this project, but it seems to me that a Hall effect switch would be the less bouncy option if bounce was a concern.
 

Thread Starter

Andrew216

Joined Jun 24, 2015
29
I am trying to rid as much bounce as possible. Checking on my O scope there is still a tiny spike or two so I will need to fix that as well but that can come later after the proof of concept is there.
 

AnalogKid

Joined Aug 1, 2013
10,971
I'm curious what you mean by bouncy here. A Hall effect switch with appropriate hysteresis (which this one appears to have built in) should have very clean output switching. Reed switches, being mechanical switches, naturally have some amount of contact bounce when switched. I don't know if it will matter either way for this project, but it seems to me that a Hall effect switch would be the less bouncy option if bounce was a concern.
Page 10 of the data sheet shows 4 output state changes shortly after power up.

With a little protection, the circuit I put up will run directly off of vehicle power. It can be adapted to use the TI sensor.

ak
 

Thread Starter

Andrew216

Joined Jun 24, 2015
29
I'm running this off the 3v coin battery not the cars battery. And to save on that battery life I want the light to only stay on for 30 secs when the car door is left open so its not constantly on and draining battery life. I'm waiting on an Inverter and Reed Switches from Digi-Key so I can try all 3 ways. 1) Hall Effect and a Rocker Switch for reset 2) Hall Effect with Inverter for reset 3) Reed Switch with no timing circuit.
 

AnalogKid

Joined Aug 1, 2013
10,971
Static current through the Hall sensor is 3.5 mA. In my circuit, static current through R1 (10 K) is 10% of that. Increase R1 to 100k, total static current is less than 40 uA. A CR2460 is one of the biggest, fattest coin cells, and is rated at 620 mAh. At 3.5 mA off-time current, that's 177 hours, or just over 7 days. At 40 uA, the total possible off time is 1.75 years. If the car door is opened 10 times per day for the full 30 seconds each time, that's a total LED time of only 5 minutes out of a 24 hour day, so the off-time energy is 1435 minutes x 3.5 mA = 83 mAh. Unless the LEDs are drawing over 1 A, the original circuit is burning more power when it is off than when it is on. In your schematic it looks like the max LED current is 2 mA. a) that's not much; will the LEDs be bright enough? b) that's less than the sensor draws.

Both your circuit and mine can be reworked to greatly increase the battery life if the sensor is changed to a switch that is open when the door is closed, and closes when the door is opened. With this switch configuration, the switch can act as a power switch to the circuit, so the static current is zero when the door is closed. I know this is a significant change from what you described in your early posts, but the battery life extension is huge.

ak
 

Thread Starter

Andrew216

Joined Jun 24, 2015
29
I just realized that I have no clock in this circuit. Need an 80% duty cycle. I know this can be done with an Astable circuit but is there an easier way that wont involve another 555 timer?
 
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