How to make a sensor bypass a resistor?

Question

I have a very basic circuit question.

I bought these LED bracelets with the intent of modifying them. The idea is to make it so the LED brightness would be dimmer than normal when switched on (by adding a resistor in series.) Then I would have a vibration sensor that would make the LEDs go back to normal brightness when it senses motion.

I thought that if I added the sensor in parallel with the resistor, it would bypass the resistor, since the circuit would choose the path of least resistance.

I got the resistor to dim it the way I want, but the sensor in parallel doesn't seem to do anything. I tested the sensor separately so I know it works. What's the right way to do this?

Additionally, in testing the sensor separately, I observed that the LED only stays on for a fraction of a second. What can I do to make the lights stay on longer, like at least half a second?

The bracelet has another switch mode to make the lights slow fade on and off. If there was a way for it to toggle a single slow fade on/off on motion, that would be a desired behavior too, but I wouldn't know how to do that.

The board has a 6V battery, 2 LEDs, 2 resistors, 1 capacitor, and a 6pin chip, though I don't know in what order they are.

This is the vibration sensor I got. It has a spring in a cylinder with 2 pins. uxcell SW-58010P High Sensitivity Spring Electronic Vibration Sensor Switch Straight Pin 10Pcs

Answer 1

This appears to be what is inside your sensor.
The centre conductor makes contact with the spring when the sensor is touched.

On resistance is probably low - in the order of a few ohms - but occurs intermittently and for short periods. To alter an LEDs brightness you need to extend the period for which the vibration sensing lasts.

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A low tech solution would be to have the sensors turn on a relay with a capacitor across the relay coil to hold the relay on. This is functionally the same as the circuit below but uses fewer and simpler to use parts. The relay requires as low an operate current as practical. I can specify component values for an example circuit if this is of interest.

R1 is shown as a short circuit. Using sat 10 Ohms here will reecuce capacitor inrush current and possibly increase sensir lifetime.
The relay could be a miniature reed relay.

Example only This reed relay in stock at Digikey for $1.91 in 1's, operates at <= 3.75V and releases at >= 0.5V and draws 10 mA at 5V coil voltage. C1 of 1000 uF will probably give you in the 0.5 1s hold in time with a 5V battery. Size is ~~= 20 x 5 x 8 mm - nit utterly tiny but probably bearable.

You could wind your own using a reed switch - many turns of fine wire on the switch body - posiblywith manet biasing to get more sensitivity.

^{simulate this circuit – Schematic created using CircuitLab}

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Here is a simple circuit that would probably meet your need.
MOSFET NFET1 is turned on when its gate is pulled high by sensor closure.
C1 charges almost instantaneously to battery voltage.
When the switch opens C1 keeps the FET turned on and is discharged by R3.
R3 and C1 together control the "bright up" period.

LED current is limited by R1 and R2 in series.
When the FET is on it shorts R2 and allows increased LED current.

R4 is shown as 0 Ohms - a wire link - if desired a low resistor value would limit sensor current and possibly proling its lifetime.

FET NFET1 is almost any small N Channel MOSFET - we can discuss suitable FETs if you wish to proceed with this design.

Cost of components other than battery, LED, sensor and on/off switch is not much over $1. The switch can be obtained from scrap electronic equipment or toys. If bought new it might cost as much as the other small parts combined, depending on source.

^{simulate this circuit}

Answer 2

I love Russell McMahon's idea, it's definitely the right way to go, but if you want to try something simpler:

^{simulate this circuit – Schematic created using CircuitLab}

It needs a big capacitor to work properly, a super-capacitor, and R3 must be beefy too, at 1W.

When vibrations happen, the sensor switch closes, discharging the capacitor, bringing the voltage across it to 0V, maximising the current through the LED.

When vibrations stop, and the sensor switch is open, the capcitor slowly charges, reducing the total voltage available for the R1 and the LED, dimming the LED.

Discharging is about 50 times faster (with the values of R2 and R3 shown), so the effect will be that the capacitor takes a couple of seconds to discharge (for maximum brightness) but a lot longer to charge again, dimming the LED.

This circuit is horribly inefficient, and R3 will get hot, but it could work.