Audio Sound Transmitter

Discussion in 'General Electronics Chat' started by alannia, Oct 8, 2010.

  1. alannia

    Thread Starter New Member

    Nov 8, 2009
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    Hi, I've designed an audio sound transmitter which continuously transmits an audio output of about 4.3kHz, such that a matching receiver can pick up the signal about 50 cms away.

    "Circuit 1 & 2": View attachment transmitter_circuits.pdf

    The 555 output is amplified by a gain of 10 by the LM741 which runs off a dual voltage supply. However, I want the circuit to run off just one 9V battery (as I want the the transmitter to be as light as possible since it is meant to be portable).

    1. What is simplest way to change "Circuit 1" to run from a single voltage? I can hook up two 9V batteries for plug and minus operation, but like I mentioned earlier, weight is an issue so would prefer not to resort to this method.

    2. I'm thinking of replacing the inverting amplifier with BD139 View attachment BD139.pdf (NPN transistor) so that the circuit runs from a single voltage (see "Circuit 2"). Can someone kindly explain how I can calculate R3 and R4 values so that the voltage gain would still be close to 10 like in "Circuit 1"?

    3. Since the transmitter currently pulses every 0.233ms, it'll drain the battery very quickly. What would be the simplest method, in terms of circuit additions/changes, so that the transmitter is more power efficient but still pulses enough such that the receiver can pick up the transmitted signal 50cm away and interpret it?

    Thank you for taking the time to read my post.
     
  2. tom66

    Senior Member

    May 9, 2009
    2,613
    214
    The simple solution is to connect the 555 timer between 9V and 0V, and the noninverting input of the op-amp to 4.5V, created by a voltage divider. A decoupling capacitor should be added on the op-amp output to create a virtual AC signal.

    Also, you shouldn't use the LM741 for this kind of application if you want to use it from a single supply. It won't work, due to the 1.5V drop on either rail. Use something like an LM358. That has two op-amps, which would allow you to create a little buffer for the voltage divder.
     
  3. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    Just use a transistor but I don't agree with that circuit either. I'd run it as an emitter follower with the speaker going to ground, sure would cut the power requirements.
     
  4. alannia

    Thread Starter New Member

    Nov 8, 2009
    7
    0
    Thank you for the help so far.

    If I was to not worry about how much power the circuit uses, and just fix the problem of just using one battery by going along with "Circuit 2" which uses the BD139 as the amplifier stead of LM741. Using Circuit 2, I've tested R3=21k ohm, R4=2.2k ohm, but the sound was too soft. So I went with the LM741 inverting amplifier to get a gain of 10 which made the sound alot louder because my transistor understanding is very poor. Would really appreciate it if someone can explain how would I calculate R3 and R4 values to drive a speaker that has 8ohms resistive load so that it's the equivalent of using LM741 to get a voltage gain of 10? Sorry if it's just something very basic.
     
  5. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    If your 555 is powered from a 9v battery, the output will swing from ~0 to ~7 volts. If your 741 or transistor has a gain of 10, your output is now supposed to be ~0 to ~70 volts. You can't get there from here.
    The 7 volt/200ma output from the 555 should be plenty to drive a speaker by itself.

    Circuit #2's transistor is not a voltage amplifier, it is a switch. The collector resistor value is determined by the desired current flow through the speaker and the voltage drop across the resistor. Once that current is established, 1/10 of it will be the base current. That along with the voltage drop across the base resistor (Vin-Vbe) will determine its value.
     
    Last edited: Oct 9, 2010
    alannia likes this.
  6. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Sorry, this was a double post.
     
    Last edited: Oct 9, 2010
  7. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    [NOTE] This is all just rough rambling, not meant to be accurate but something to try.

    A BD139 can take up to 0.5A into the base and output as much as 1.5A so it's fairly low gain. (or very forgiving)

    A 555 shouldn't really be called upon to source or sink any more than 100 mA and has a 2V drop at that level.

    To feed the transistor base 100 mA you'd take 9V - 2V = 7V - 1V for the transistor's stated Vbe junction drop = 6V

    R = E/I so you've got about 6V you're trying to get 100 mA out of = 60 ohms, 62 or 68 would be close enough for R3 and since the transistor probably has a lot more gain than you think that value could probably be much higher.

    Since that output circuit is an oddball in that it just shorts the speaker out you're pretty much limited on R4 to what you can get away with getting from your 9V supply without exceeding the 1.5A collector limitation of the transistor which works out to be around 6 ohms.

    These are maximum values, that transistor can't handle more than 12.5W dissipation with a proper heatsink and you're already putting about 1W into the BE junction. If we limit it to a wattage of 10 all you need to know is the Vce(sat) of the transistor to figure out a value of R4 which will limit it to this but I wouldn't use anything less than 10 ohms.

    In all honesty I'd use a different single transistor output circuit specifically designed to directly drive a speaker.

    Probably 100 errors above, I'm busy building a PC at the moment so my mind is elsewhere.
     
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