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So I'm pretty new at this, so I'm sorry for any dumb questions/statements.

I'm trying to build a puzzle box where you activate a circuit with a reed switch using a hidden magnet. Activating this circuit needs to make a sound and make 2 LED's light up. I'm using a 9V battery, 2 green LEDs (no datasheet available, no multimeter of potentiometer either. But the operating voltage range is 2.9-3.6 V at 20 mA) and a mechanical buzzer rated at 2-4 V at 20 mA.

So here is my thought process: 2 LEDs with a voltage drop of about 3V, will leave about 3V for my buzzer (well in the operating range if the LEDs happen to drop more/less). Since all of the components are using 20 mA, I can safely put them in series. The only thing I'm not sure about, is the current that will flow through. I know you normally would place a resistor in front of the LEDs to limit the current, because the LEDs are basically a dead short, but will the buzzer do the same job?

So:

  1. Is this possible: all of them in series without a resistor? Will this have any effect on the brightness/sound level?
  2. if not, can I just put the 2 LEDs in series with a resistor (needs about 140-150 ohm resistor in front of it) and those LEDs in parallel with the buzzer (with a 300 ohm resistor?
  3. If both of them are wrong, what's the solution?

Thanks!

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    \$\begingroup\$ Why don't you experiment? It's easy enough to do. You have 2-4 V for the buzzer and about 3-3.6 for the LED, so that's anywhere from 5-7.6 V total. This may provide enough overhead for a simple BJT current-limiting circuit using two BJTs and requiring perhaps about \$1\:\text{V}\$ of overhead (maybe a little more.) Worth an attempt. If you want an example, I'd be happy to suggest one. It's not hard. But of course, no guarantees that it does everything you need. (I prefer active control of the current rather than passive using a resistor.) \$\endgroup\$
    – jonk
    Oct 16 '18 at 10:26
  • \$\begingroup\$ The correct way to do this is put all of the elements in parallel, and give each LED its own resistor. \$\endgroup\$
    – DerStrom8
    Oct 16 '18 at 10:35
  • \$\begingroup\$ @jonk I'm using 2 green LEDs (each with a Vf of ±3V) in series with a 2-4 V buzzer. The problem is, that I don't have the parts here to experiment. I need to order them, so I want to be sure to order the right parts because I don't want to pay for shipping twice. What I thought was: the LEDs will both drop their forward voltages and whatever voltage is left (around 3V) will be used by the buzzer. Since the 2 LEDs in series wil guarantee a voltage drop of 6V, they will always leave 3V for the buzzer, which draws 20 mA from the battery. But I'm worried it will draw more and fry the LEDs. \$\endgroup\$
    – Sam
    Oct 16 '18 at 10:47
  • \$\begingroup\$ @Sam All you've done is to propose your own internal state of mind -- your hypothesis. You might get the right parts and get a result you want. But it's not a design. A design is about putting vagaries of devices under management. If you were to ask someone to engineer a house design, for example, you would expect them to perform a design that takes into account variations in the strength of materials available. (Not all lumber is exactly the same.) But we can't do that here. All we can say is to "try it." Your idea doesn't have the overhead for a managed design, as stated. \$\endgroup\$
    – jonk
    Oct 16 '18 at 16:14
  • \$\begingroup\$ Hi there! I know that we can't test it here, but I just wanted to know whether it may be possible to begin with. Because everyone says you need a resistor to limit the current, but I can't find whether or not I can use another part. Like I said, I'm new at this. I also don't understand things like "overhead" and "BJT current-limiting circuit". I just want to know if it is at all theoretically possible to replace that resistor with another part like a buzzer, or does it always have to be a resistor. It's not at all for professional use by the way. \$\endgroup\$
    – Sam
    Oct 16 '18 at 20:30
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Putting them all in series should work okay.

If the buzzer is 3V and 20 mA then it has a resistance of 150 Ω.
The buzzer would act as your 150 Ω current liming resistor.

I would first power the LEDs with a 150 Ω resistor to see the brightness then use the buzzer in place of the resistor and see if the brightness is about the same.

The bigger problem I see is the 9V battery. It will provide 9V for only a short period. Notice below how quickly the battery voltage drops to 8V. This may not be a problem if the buzzer is not activated for long periods of time and you use a fresh alkaline battery.

enter image description here

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  • \$\begingroup\$ Thanks! I was planning to test first with a 150 ohm resistor. It will only be used in very short periods of time (seconds) with a fresh battery. That's why the buzzer is there, to stop people from activating the circuit with a magnet (accidentally) for too long, because the light will only be visible in a dark room (it's inside a box). The place where the light comes out is a clue, the buzzer is to annoy. But as stated the operating range for the buzzer is 2-4 volt, so if the battery voltage drops to 8, it will still draw 13 mA, which is enough to power the buzzer and the LEDs. \$\endgroup\$
    – Sam
    Oct 16 '18 at 20:59

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