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I want to light up a couple of LEDs using a relay to trigger this event. I have a very low DC voltage, about 0.15mV to 0.20mV and I want to somehow take this voltage to a relay, that will be also connected to a battery to light up the LEDs. So far, I know how to calculate the resistors value not to blow up the leds, but don't know how to drive this low current to the relay to make it work?

Is this possible? If so, Can anyone point me in the right direction?

EDIT: Per one suggestion, let me explain my scenario. I have a regular alarm clock that I want to use to turn those LEDs on. My grandpa can't hear good anymore and we think this would help him to remind him about certain situations. The DC voltage I'm talking about is coming from the little buzzer this thing have, and I said 0.15mV because that's what I could get from the voltimeter hehe... So the general idea is to use that small voltage to turn those LEDs on.

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    \$\begingroup\$ Millivots is not a measure of current. It appears you meant to say you have a low DC voltage. Also, in engineering we try to keep 1-3 significant digits left of the decimal point and adjust the units multiplier accordingly. In other words, 0.15 mV is better written as 150 uV. \$\endgroup\$ Commented Aug 30, 2013 at 16:53
  • \$\begingroup\$ Where exactly did you measure this 150 uV with your voltmeter? There should be a much stronger signal in there if a buzzer is going off. Trying to tap into the alarm clock electrically is a bad idea because it could be danergous. A small piezo microphone mounted on the alarm clock where the noise comes out should produce more than 150 uV. Also, what makes you think this is a DC signal? It probably is AC with a 0 average level, which is why is looks so small to a DC voltmeter. Set the meter to AC volts. Again, what exactly are you measuring? \$\endgroup\$ Commented Aug 30, 2013 at 20:54
  • \$\begingroup\$ It's a small alarm clock running on one 1.5 volts battery... \$\endgroup\$
    – Sergio A.
    Commented Aug 30, 2013 at 21:44

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It appears you want to make some LEDs go on a signal reaches 150 µV. Let's forget about your imagined implemetation of a relay. A low side transistor switch can do this.

However, the tricky part is how to detect this 150 µV signal so that a solid on/off signal of appropriate voltage can be generated. 150 µV is a very small signal, smaller than what many microphones would produce when you talk into them normally. This won't be easy if this signal really needs to be DC coupled.

If this signal can be AC coupled, like a audio signal can, then this isn't too bad. Such signals can be amplified, then AC coupled again to eliminate the inevitable DC offset caused by the amplification. Otherwise 150 µV will be tricky to detect. Most comparators will have more offset than that. For a one-off do it yourself project, you can add a trim pot to adjust the threshold by experimentation, but be aware that even modest temperature changes can require re-calibration.

To give more details on the solution, you need to provide more details on the problem. Pop up two levels and explain what you are really trying to accomplish, but leave out any imagined solution on your part, like whether to use a relay or not. In other words, the question should be when some well described thingy does this, a string of LEDs should light.

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  • \$\begingroup\$ Thanks Olin. I just explained the scenario in the original question. \$\endgroup\$
    – Sergio A.
    Commented Aug 30, 2013 at 18:37
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I'm guessing that the voltage on the buzzer is AC, that's how buzzers are typically driven. Measuring with a DC voltmeter, you would read close to 0, ie 150uV.

That signal most likely swings close to rail-to-rail and should be able to drive your relay. If the buzz frequency is high enough, all you need is to drive it through a diode, to clip the negative cycle. If the relay is still chattering and the LED blinking, you may need a small RC to smoothen it out.

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