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I'm using a Maxim DAC (the MAX541) with a TI microcontroller. I output a digital audio signal from the microcontroller to the DAC. I have the DAC's output hooked up to a typical 3.5mm audio jack. When I plug the 3.5mm jack into typical computer speakers the volume is quite low. I have to crank up the speakers to hear the audio.

Hooking up the 3.5mm audio jack to my oscilloscope I see the max peak-to-peak voltage output is only about 0.12 volts:

enter image description here

When I play the same digital audio using my phone at a reasonably loud level and analyze it with my oscilloscope I see the peak-to-peak max voltage is around 1 volt:

enter image description here

When purchasing the MAX541 DAC I suspected it would need some amplification so I also purchased a MAX492 op amp. I have the analog output of the DAC going to the input exactly as shown in the MAX492 datasheet's "typical operating circuit" example:

MAX492 typical operating circuit

Here's a link to the MAX492 datasheet.

However, this is only amplifying the signal to a peak-to-peak max of about 0.23 volts. I of course need the MAX492 to amplify the signal to around 1 volt (like my phone's output level), however, I've read through the datasheet and it's not clear to me how to do this or if it's even possible.

Any help would be most appreciated. Thank you.

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    \$\begingroup\$ You should be able to get more than 150mV out of the DAC - what are you supplying to Vref? \$\endgroup\$
    – Frog
    Dec 31, 2020 at 1:55
  • \$\begingroup\$ I'm supplying 2.5V to Vref. The TI microcontroller is a TM4C1294G dev board. It has a 5V power pin. I use a voltage divider to get 2.5V. \$\endgroup\$
    – Terence D
    Dec 31, 2020 at 3:57
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    \$\begingroup\$ Have you got decoupling on the divider? Perhaps check the actual divider voltage in case the DAC is loading it. \$\endgroup\$
    – Frog
    Dec 31, 2020 at 5:32
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    \$\begingroup\$ Sure, I’m happy to help. Decoupling is where you place a capacitor in parallel with one of the resistors in a divider, typically between ground and, in this case, the 2.5V. This has the effect of reducing the effective impedance of the divider at high frequencies. In other words, if the DAC suddenly applies a load to the divider, the voltage that it presents won’t change immediately. However, if the DAC loads the divider constantly it won’t help, and you’ll need a better way to produce Vref, such as a regulator or a divider followed by a buffer. \$\endgroup\$
    – Frog
    Dec 31, 2020 at 22:59
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    \$\begingroup\$ Looks good - 330R is a relatively low resistance so should hold up against a DC load too. That aside, do you know whether the digital source is using the full range? If not then clearly you’re not going to get full range in analogue \$\endgroup\$
    – Frog
    Jan 2, 2021 at 20:55

1 Answer 1

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You have the voltage gain of the opamp at \$Gain = 1 + \frac{10k}{10k} = 2\$ times. If you want the voltage gain to be 8.3 times then use 75k for the negative feedback resistor (R1 in the image below).

opamp

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  • \$\begingroup\$ Hi, thank you for this information. I'll give it a try. I take it this is a standard equation for a non-inverting amp? I'm a software engineer learning about electronics (with no formal training), so if so, I was not aware of this. \$\endgroup\$
    – Terence D
    Dec 31, 2020 at 4:02
  • \$\begingroup\$ Thank you! This seems to work brilliantly! I also googled some and found this: circuitdigest.com/calculators/op-amp-gain-calculator with the same equation, so I assume it is indeed a standard equation for a non-inverting amp. \$\endgroup\$
    – Terence D
    Dec 31, 2020 at 5:51

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