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Help with a simple amplifier design for an audio signal.

  1. Oct 19, 2007 #1
    Hi, I'm designing an amplifier and need help with some basic ideas (Filtering is already taken care of). My main question is bolded.

    The output is an audio signal, and it drives an 8 ohm speaker. The power is 1 watt through this speaker.

    The input signal is coming from another amplifier that says its full-scale output is 1Vrms.

    So the output specification is 1 watt through 8 ohm speaker, so (Vo^2)/R = 1W, Vo = sqrt(8) = +or- 2.828427V. The power rail of the amp will be 5V and it can source over 2 watts, so this will be ok.

    So now I know my gain is to be Vo/Vi = 2.828427V/1Vrms. . .
    Do I need to convert my output audio voltage to RMS, or do I need to convert my 1Vrms input to normal V? Also, I would imagine my Vi of 1Vrms is going to actually be greater since the input impedance of my audio amplifer is going to be more than a 10k ohm load as specced for the input signal.
  2. jcsd
  3. Oct 19, 2007 #2
    I've decided to convert RMS to peak to peak voltage and get 1V * sqrt(2) = 1.41421 Vpp

    which is funny that it works out to an exact gain of 2V/V. I'm going to give a little extra gain to 2.1V.

    Is this reasonable?
  4. Oct 19, 2007 #3


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    One thing. You computed the 2.82v as RMS (or DC).
    You might want to reconsider what this means when converted to a peak to peak sine wave in terms of your power supply voltage and output amp headroom requirements (unless its a rail to rail amp in which case headroom does not apply).

    Also your gain will be wrong unless input and output are specified the same. You don't want mixed RMS and Peak.
  5. Oct 22, 2007 #4
    Okei well my dilemma has been that the gain units did not match up (V and Vrms) so I did then just change my Vrms to Vpp instead, which should be my magnitude of highest signal, which will be best to use as the DC gain. Is this right to you?

    Also, the power supply is 5V to 0V, and I should never go over 5V according to my power calculation giving me a voltage max below 3V. I look through the datasheet, but see nothing about thresholds or clipping, so I am optimistically assuming it is rail to rail.

    What you mentioned has just put another thing of consideration into my mind though, is what the audio signal is biased at going into the amp. . it better not be biased at 0V or else I'm in trouble.

    The power supply to the chip that is sourcing the audio signal into my amp is GND to 3.3V. So at best, its bias should be 1.65V so that there can be equal negative and positive swing in the signal. If this is the case, I have to worry about clipping at the negative end if my gain is too high.
  6. Oct 22, 2007 #5
    This question is important. On the inputs of the amplifier, I was told to put a cap between the inputs, to "keep the noise potential the same" between them. I always thought to put noise through the cap between the signal and ground, not between the two signals. I looked through my electronics book and cannot find a single example where a cap is AC coupling the two inputs to the amp.. does anyone know about this?
  7. Oct 22, 2007 #6


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    RMS and DC are equivalent representations.
    That is the definition of RMS.
    You had it right in the first post :smile:

    This is related to the confusion you are having in the prior question.
    Perhaps it would help you to graph a 5v Peak to Peak sine wave,
    the max for your 5v power supply, then compute the RMS voltage from the graph.
    Tell us what you get for the RMS voltage.

    You did say you had the datasheet for your power amp.
    Most of these are just a variation on the op-amp.
    The manufacturer usually includes a generalized circuit for their chip, which can be quite useful as a starting point for a design.

    You could use AC coupling in which case this is not a problem.
    If you really need DC coupling then you will need to do some level shifting.
  8. Oct 22, 2007 #7


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    Perhaps with a resistor as a low pass filter to reduce the high frequency response.
    One of amp inputs would be at a ground equivilant.
    It seems unlikely that you would want to do this.
    Where did you get the "keep the noise potential the same".
    That just sounds wrong.
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