SparkFun Forums

[kolban]

The pins for the device listed here:

https://www.sparkfun.com/products/11125

have two apparent supplies. One is listed as VCC and the other is 3.3V. I haven't found any other descriptions for these. What connections do I need for these? I am assuming this is a 3.3V device ... do both or just one of the pins need to be connected to 3.3V?

[Valen]

Have you looked at the schematic? Vcc comes from the battery connector and goes into the MIC5219 voltage regulator. Where as the 3.3.v is coming out of the MIC5219 voltage regulator. So, either you supply it with a battery on the connector or other powersupply into the Vcc pin, or you provide it with external 3.3 volt on the 3.3 pin (bypassing the regulator).

[kolban]

Thank you Valen. What I hadn't appreciate was the apparent fact that the micro SD card needs to be powered at 3.3V while the sound module can be powered at 5V or 3.3V. So ... if I understand it correctly, if I am powering from external sources bypassing the input regulator, if I supply 3.3V I can connect that to both Vcc and 3.3V ... however, if I want to power at 5V bypassing the regulator, I still need to supply only 3.3V at the 3.3V connection.

[Ross Robotics]

No. If you power it with 5v+, connect it to Vcc. If you have a WELL regulated 3.3V supply, then use the 3.3V pin. My recommendation is you power it with 5V+ and leave the 3.3V pin alone.

[Valen]

Thank you Valen. What I hadn't appreciate was the apparent fact that the micro SD card needs to be powered at 3.3V while the sound module can be powered at 5V or 3.3V.

I don't know where you could have gotten that distinction from. The WTV020-SD datasheet says it's working voltage is up to 3.5 volt. So, 5 volt will likely damage it. In fact, on page 19 of the datasheet is suggests to use two diodes to reduce 5 volt to 3.3 volt. So, apparently 5 volt directly could kill the chip. But if you meant this Sparkfun board could be powered with 5 volt then you are correct, but only on the condition that this is connected to the battery connector/Vcc pin. As this would then be lowered by the onboard regulator.

So ... if I understand it correctly, if I am powering from external sources bypassing the input regulator, if I supply 3.3V I can connect that to both Vcc and 3.3V ... however, if I want to power at 5V bypassing the regulator, I still need to supply only 3.3V at the 3.3V connection.

No, you may have gotten my meaning of "bypassing" wrong. I did not mean to connect both the input and output together. I meant not using the voltage regulator if you have an external source of 3.3 volt. By leaving the Vcc/battery connector floating, and rather supply the external 3.3v source into the 3.3 pin directly. Eventhough this pin is also the output of the regulator, it remains non-operational if there isn't a higher supply going into it.

BUT!, since you have not said anything about what other controller or shield/board you are going to connect this to, it is impossible to judge from here if you have what it takes to supply this sound breakout board with 3.3 volt. Some 3.3 volt regulators on Arduino board may not be able to supply as much current as needed. So, I agree with Codlink. Apply 5 volt to Vcc if you have it , or at least .3 volt higher than 3.3v to Vcc for the regulator to operate correctly and provide a nice steady 3.3 volt. It will probably do the sound quality a lot of good as well.

[kolban]

Again, thank you guys ... I think the light bulb just went on for me. I'm looking at the schematic now. If I understand correctly ... I can power the Sparkfun module (and that's what I'm talking about ... not the WTV020SD chip generically) in 3 distinct ways:

1. I can plug in a JST plug with a voltage of 3.3V or 5V (and maybe higher) ... in these cases, I need to wire nothing to the Vcc or 3.3V pins of the board. Power has already been supplied.

2. I can omit the use of a JST plug source and apply a 3.3V or 5V source to Vcc. In this case, I would leave the 3.3V pin alone as the Vcc has provided an input voltage which has been regulated down to 3.3V by the MIC5219 device.

3. I can omit the use of a JST plug source and apply a 3.3V source to the 3.3V pin. In this case, I would leave the Vcc pin unconnected as that was the input to the voltage regulator and served no further purpose.

Is that the correct way of thinking? There seems to be a muttering on the Web by reviewers of this Sparkfun module that there is relatively little docs (for the beginner like myself) on how to get it "up and running" (simple tutorial or Instructable type thing). The ones that I have both use JST plugs and, unfortunately, I don't have those ... but was thinking that I could use the supply voltages from my Arduino ... and that was where I fell of the rails trying to understand the relationship between "the socket", Vcc and 3.3V. It may be apparent to those who can read schematics ... but to me, the value of Sparkfun is that it is aimed at the hobbyist and newbie who is new to electronics technologies and needs help getting things going.

[Valen]

Correct, the light bulb is working. You just got brighter.

I don't know how the WTV020SD chip works internally, or how it interfaces (aside from that it can do I2C). I ignored those pages in the datasheet. But now after reading it a bit further I can understand the complaints about documentation. Just short of Chinglish and cutting corners in terms of depth about adressing with I2C. However, this isn't Sparkfun's fault. That is usually what you get with Chinese or Asian products. The language barrier is big to overcome and this shows in documentation/datasheets. Sparkfun often makes a hookup guide, but not for this one I guess. At least they have example code available for Arduino in Github.

[Valen]

Ahh, I have to come back on point 2 with a detail: it must be a higher than 3.3 volt going into the regulator. Linear voltage regulators need excess voltage to burn to regulate. If the board only draws low amounts of current then this MIC5219 regulator only needs a few 10milivolts extra. But if the current draw is several 100ma then the voltage drop is also several hundred milivolt extra. Exactly 3.3 volt on the input is not enough for it to operate correctly. I don't know how much current this board draws (it can power a speaker so probably not a little) so I cannot suggest applying 3.3v to it.