2
\$\begingroup\$

I need to trigger a speaker if my input voltage is higher than 0.6 V, otherwise the speaker must not be triggered. That's the reason I'm using an AMPOP Comparator:

enter image description here

and, as far as I know, the voltage Vout should be:

$$ V_{\text{out}} = \begin{cases} V_{\text{S}+} & \text{if } V_1 > V_2, \\ V_{\text{S}-} & \text{if } V_1 < V_2, \\ 0 & \text{if } V_1 = V_2\end{cases}$$

But when I simulate my circuit, my output voltage is not Vs+ or Vs-. There is a loss in voltage and you can check all voltages on the following image.

enter image description here

V1 = 1 V and V2 = 0.6 V, Vout = 13.011, but Vs+ = 15 V;

enter image description here

V1 = 0.2 V and V2 = 0.6, Vout = 1.9889, but Vs- = 0 V.

I need Vout near 15 V (and 0 V if V1 < V2). It doesn't have to be exactly 15 volts, but near it. Am I doing something wrong? Is there any way to do it without another amplifier after the output? I'm beginner sorry if I misunderstood something.

BTW, if you know an easier way to trigger the speaker, tell me ;)

\$\endgroup\$
  • 2
    \$\begingroup\$ Not sure exactly what you mean by "triggering" a speaker, nor why you think a op-amp wired as a comparator will do that. But, the output voltages you see are easily explained by the fact that many op-amps have a compliance voltage in that range... that is, the outputs can not drive the whole supply range... compliance of ~2V is not unusual. Read your datasheet! You don't mention what op-amp you are using. A "rail-to-rail" op amp will drive the output closer to the rail voltages. Even better, use a dedicated comparator IC for this function. \$\endgroup\$ – B Pete Nov 26 '12 at 3:55
  • 1
    \$\begingroup\$ @BPete, good enough for an answer, I think. \$\endgroup\$ – The Photon Nov 26 '12 at 4:09
  • 1
    \$\begingroup\$ Why you need 15V or a DC signal on a speaker that would imply 2 Amps (but not deliver) in the speaker coil is beyond me. But you can't create a tertiary level output window detector with a simple comparator.. try again? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 26 '12 at 4:14
  • \$\begingroup\$ @Richman: Good points. We could use some detail about what the poster is trying to accomplish. \$\endgroup\$ – B Pete Nov 26 '12 at 4:43
  • \$\begingroup\$ I want an alarm system. If the voltage goes over 0.6 volts, the speaker is activated, otherwise no sound. \$\endgroup\$ – Doon Nov 27 '12 at 16:01
1
\$\begingroup\$

Do you even need +15V to -15V output???

Power on the speaker is ~Vpeak^2/R, so for an 8 ohm coil this system will blast out 15^2/8 ~= 28 Watts, which is pretty loud. Also placing DC current through a speaker is bad for it.

If you wanna stick with this design: you're going to need a Comparator with Rail-Rail operation, and even then that only gets you to Vdd-0.3V and Vss+0.3V . Check to see if this suits your needs. Since you are also driving a speaker, use Ohm's law on the speaker impedance to figure out how much current you're gonna need. An 8 ohm speaker will need at least a 2 Amp power supply for both +15 V and -15V. Best bet is to drive the system with +-18V 3A Rails. Find a comparator that can handle +-15V outputs, and get an output buffer that can handle 2 Amps both ways.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.