Well not high power, low resistance. I have an unusual situation where I am directly driving speakers from an oscillating signal ( its not audio its more like a buzz )

I only have 2 inputs

  • 12v
  • square wave signal that oscillates at audio frequencies between 0v and 12v ( not open circuit and ground )

so the speaker basically sees a negative signal which is no problem in this situation and I block too much current with a capacitor.

Ideally though I would like to reduce the voltage to make it quieter, so I have gone through the following ideas.

  • Cant use resistors as they would have to be low resistance = heat and power!
  • diodes in series work but to get down to a good voltage I need about 10+ diodes
  • I tried a LM317 but it has ripple protection? It was too quiet

I know ideally if I had a steady 0v it would be easy, but does anyone have any other ideas I could use? or perhaps I am doing something futile?

Ok edit:

I am not making sense , the title makes no sense sorry so I corrected it.

clear info

I am driving an 8 ohm speaker off a circuit that drives LEDs.

It supplies 12v and another connection between 0v and 12v that pwm fades the led at audio frequencies.

I can't touch the circuit driving it and I want to passively cut down the voltage to reduce the volume on the speaker as it is fairly loud.

Ideally with not too much concentrated heat.

What is the best way to do it?

  • \$\begingroup\$ Did you actually do the calculations for a resistor? \$\endgroup\$ – PlasmaHH Mar 31 '15 at 12:00
  • \$\begingroup\$ yep speaker is around 8ohm ( not really right? ) so resistors would have to be near 1-5ohm so 30+ watts? Is this wrong? \$\endgroup\$ – Shorthair Mar 31 '15 at 12:08
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    \$\begingroup\$ 12v square wave peak-peak = 6V RMS more or less. For n 8 Ohm impedance speaker P= V^2/R = 6^2/8 = 36/8 ~~= 4.5 Watts. | A 10 Watt resistor of suitable value would easily suffice to dissipate as much of the power desired (and a 5W one would be OK in many cases) and cost is small compared to alternatives. \$\endgroup\$ – Russell McMahon Mar 31 '15 at 12:12
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    \$\begingroup\$ Anything passive will inevitably need to dissipate the excess voltage as heat. \$\endgroup\$ – PlasmaHH Mar 31 '15 at 12:30
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    \$\begingroup\$ @Matt: Not quite. A transformer, in theory, will dissipate no power. It presents the speaker as a different impedance to the voltage source such that less power is drawn from that source. All that power then goes to the speaker. Real transformers have real losses, but it doesn't seem that's what we're talking about here. Even with a series resistance, the overall power drawn from the voltage source goes down, although only a fraction of that will go into the speaker with the rest dissipated in the resistor. \$\endgroup\$ – Olin Lathrop Mar 31 '15 at 17:34

The power level you are talking about isn't a big deal for the right resistors to dissipate. You have a 0-12 V square wave, which after AC coupling thru a capacitor results in a ±6 V square wave. The power into 8 Ω is

(6 V)²/(8 Ω) = 4.5 W.

First, dissipating 4.5 W is as simple as using a 10 W power resistor. Second, when you put a resistor in series the total power drawn from the ±6 V square wave will go down too. Your fixation on the resistor getting too hot doesn't make sense.

To find the right resistor value, do the math. Let's say you want only 500 mW into the speaker instead of the full 4.5 W. Assuming the speaker is 8 Ω, the voltage across the speaker you want is

sqrt((500 mW)(8 Ω)) = 2 V

This means the series resistor needs to drop 4 V when the speaker drops 2 V, so is twice the resistance of the speaker, or 16 Ω. The power this resistor will dissipate is

(4 V)²/(16 Ω) = 1 W

so a "2 W" power resistor should be fine.

Note that the total power is also significantly less than with just the speaker connected to the 6 V source. The resistor is dissipating 1 W and the speaker 500 mW, for a total of 1.5 W. You can see this another way too. The total resistance of the speaker plus series resistor is 24 Ω.

(6 V)²/(24 Ω) = 1.5 W.

  • \$\begingroup\$ Thanks! I tried out the equations and I understand now. Great . \$\endgroup\$ – Shorthair Mar 31 '15 at 14:34

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