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Lamest question ever asked in this forum (by a consumer)

Michael Davis
Prodigy 40 points
Other Parts Discussed in Thread: OPA1611

Hi,

I'm a complete novice at this topic, so please forgive me the intrusion, but TI Technical Support directed me to this forum, because my question was "too technical."  Really?

Here goes...

I have purchased a Meier Audio Corda Stepdance headphone amplifier that uses the OPA1611.  See:  http://www.meier-audio.homepage.t-online.de/stepdance.htm

When I asked the amp's designer, Dr. Jan Meier, the questions below, he directed me to this datasheet for the OPA1611:  http://focus.ti.com/lit/ds/symlink/opa1611.pdf

I've attempted to interpret the datasheet for this opamp with the goal of figuring out how many Watts (or milliwatts) RMS (or even just peak Watts) the opamp can produce with a given supply voltage (Vs).

If I'm interpreting the specs correctly, specifically Figures 4 and 27, with a supply voltage (Vs) of +/-15V, at 25-degrees C, the maximum output voltage (Vppp) is about 28.5V across all audible frequencies (Figure 4) and the maximum output current will be about 47 mA.

Question #1:

Would it be correct to calculate the maximum (peak?) output Watts as follows?: 

Max Output Voltage (Vppp) * Max Output Current (mA)  = Max Output Watts

28.5V * 47mA = 1339.5 mW  (at 25-degrees C, with a +/-15V supply voltage)  

If not, can someone please correct my math (and/or my interpretations/assumptions)?

Question #2:

From reading elsewhere on the net, I understand that Watts RMS could be calculated as 0.707 * Peak Watts if only the signals were pure sine waves, which for audio amps is not the case, of course.

Can someone give a ballpark way of estimating Watts RMS (into 50-ohm headphones) when Peak Watts is known (if indeed I've calculated it correctly)?

Question #3:

Lastly, since the OPA1611's Output Voltage varies with the Supply Voltage, given that my headphone amplifier allows external supplies to range from 6VDC to 15VDC -AND- given that the amp designer has included a circuit that doubles the voltage coming from the external supply, as if a PSU that supplies 15VDC was actually supplying 30VDC, would I have to use a 7.5VDC external linear regulated power supply to mimic the conditions documented in Figures 4 and 27?   In other words, when the datasheet specifies a Supply Voltage (Vs) of +/-15V, would that be what I'm supplying with a 7.5VDC PSU (given that the amp's design doubles the supply voltage)? 

Along those lines, given that the amplifier's specs indicate users can connect PSU's that supply anywhere from 6VDC to 15VDC, how can there be a voltage-doubling circuit if the OPA1611's Supply Voltage must fit the range +/-2.25V to +/-18V?

Thank you in advance!

Mike

 

 

 

 

 

  • 0
    Guru 21975 points

    Michael,

    It is not possible to provide an accurate answer to your questions. The complete headphone amplifier has an internal circuit to "split" the battery voltage. The characteristics of this splitter circuit significantly affect any estimates of power output capability. I won't hazard a guess.

    It could be measured by driving a sine wave (maximum undistorted) into a 50 ohm resistor. Power output would be the square of the voltage (rms) divided by 50 ohms. You will probably get less when both channels are driven. (Not responsible for damage.)

    I refuse to calculate "peak power" as it goes against my vows as an electrical engineer. :)

    Regards, Bruce.

     

  • 0 in reply to Bruce Trump
    Prodigy 40 points

    Bruce,

    Thanks for the quick reply! 

    Learning of this circuit that splits the voltage explains how it is that the Stepdance headphone amplifier can double PSU voltages in the range of 6V to 15V, resulting in voltages as high as 30V, yet not exceed the OPA1611's maximum supply voltage of +/-18V.  

    It makes sense, too, that we couldn't hazard a guess, as you say, because only the amp's designer could know  what the actual supply voltage to the opamp is for any given PSU voltage.

    That pretty much kills my quest to "calculate" the Watts RMS into 50 ohms actually delivered to the headphones.    Doh!   At least I knew this was going to be a "lame" post.    :-)

    Regarding your last statement, however, I am bewildered.  Can you please elaborate?   I'd like to get in on the joke.  :-)

    Thanks again,

    Mike

     

     

  • 0 in reply to Michael Davis
    Guru 21975 points

    Mike,

    Power of an AC sine wave signal has a specific meaning--in this case it's voltage times current (both in rms and with a resistive load). So-called "peak power," the product of peak voltage and current in this same situation, is largely for the purpose of selling product and inflating egos. It might be likened to a car manufacturer promoting the peak horsepower produced at the instant a cylinder fires. Bigger number but the car does not go faster.

    In my opinion, there is a justifiable case for short-term power testing and rating that would involve short bursts of sine waves. This might be appropriate in the case of your headphone amp. Here, the power available from the battery is limited by its source impedance but power supply capacitors provide large currents for short periods.

    Regards, Bruce.