0
\$\begingroup\$

I am trying to design a digital timer that receives a time, t, as input and after t seconds it triggers a buzzer. It is actually similar to a kitchen timer but we can adjust the time ourself.

A way to do this is using a "555 IC" but the time will be a function of the resistor and the capacitor we choose and we don't have control over the time after we build the circuit.

Can anyone please help me with this problem?

\$\endgroup\$
3
  • 2
    \$\begingroup\$ Use a micro.... \$\endgroup\$
    – Trevor_G
    Dec 13, 2017 at 13:11
  • 1
    \$\begingroup\$ How does it receive "a time, t, as input" exactly? \$\endgroup\$
    – Finbarr
    Dec 13, 2017 at 13:14
  • 2
    \$\begingroup\$ we don't have control over the time after we build the circuit Not true if you replace the resistor with a variable resistor or make it switchable to choose between different resistor and/or capacitor combinations. You must specify how much count down time you want, a 555 gets "troublesome" if you want more than a minute or so. \$\endgroup\$ Dec 13, 2017 at 13:44

1 Answer 1

2
\$\begingroup\$

Use a (simple) microcontroller. You can add electronics or whatever component to receive the time it is even easy to add a WIFI/RF type to make it wireless or you can use SPI, I2C, serial/uart/USB etc when using a microcontroller).

Store the time in a variable, and control the buzzer directly via the microcontroller after a simple comparison of elapsed time and current time.

Probably receiving the time will be most hard to do without a microcontroller, assuming a (decent) protocol is needed.

(btw, it is possible to make a variable resistor, but it is a bit of a 'dirty solution' if not using a programmable resistor: use a LED, let it glow for some percentage. Attach a photo diode to it and put tape around the LED and photo diode. The amount of light of the LED will result in a difference resistance of the photo diode. I don't know if there is also some trick for capacitors).

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Ha! Never saw that kind of variable resistor! Notice that it's still a diode, so it must be biased to some forward voltage to exhibit a (somewhat) linear current/voltage behaviour. You could of course just use the voltage (or current) you use to control the LED brightness to set the bias point of a non-photo diode, and get a similar behaviour. If you, instead of having a positive bias voltage, have a negative one (i.e. you flip the diode around), the diode is in reverse direction, and will act like a capacitor! There's diodes that are optimized for this usage, called Varicaps (PIN diodes). \$\endgroup\$ Dec 13, 2017 at 14:02
  • 1
    \$\begingroup\$ (cool idea, by the way, using brightness of an LED to control the effective resistance of a photodiode; I like that idea. Basically, you get isolation for free – maybe one could just abuse an optocoupler for that) \$\endgroup\$ Dec 13, 2017 at 14:03
  • \$\begingroup\$ Thanks ... and yes, but with a microcontroller it must be not so difficult to make a formula from the amout of brightness (taking the forward voltage into account) and the requested amount of resistance. Actually it can more than a default variable resistor and let the work be done by the microcontroller. \$\endgroup\$ Dec 13, 2017 at 14:13
  • 1
    \$\begingroup\$ The varicap diode is very commonly used in radio electronics, when, for some reasons, you need to have an adjustable filter, or a widely adjustable oscillator. \$\endgroup\$ Dec 13, 2017 at 14:16
  • 1
    \$\begingroup\$ You can probably actually do the same with a very "boring" standard rectifier diode! The higher the reverse voltage, the wider the isolating zone in between the P- and the N-region in the diode; that translates to a larger capacitor plate distance, which means sinking capacitance with increasing \$\left\lvert V_\text{reverse}\right\rvert\$. But: normal diodes have a capacitance of a few pF, so not very much and not very easy to make use of. See for example this 1N4001 datasheet, page 2, figure bottom right. \$\endgroup\$ Dec 13, 2017 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.