(88.9k and 110k)
Yes 0603.Are you sure you're approaching this correctly?
From the SD8942 datasheet:
View attachment 372164
Vout = 5.4V
Vfb = 0.6V
It follows that R2 = (R1)/8 in your application.
If you take R2 = 15k and R1 = 120k you'll be right on the money and both values are E12 series, so common as muck.
The footprint on that board looks like something like 0603 or so. I bet you can get 0402, 0603 or even 0805 soldered onto those pads manually.
If you get an SMD resistor kit from e.g. AliExpress, you'll have values to match your application. Costs only a few bucks. 1% accuracy will be plenty good enough in all likelihood.
However, as mentioned above, I can't seem to delete the existing resistors.
Excellent, I did find the schematic ( on this site ) but it opens in a separate window. I CAN delete the resistors in the schematic.I don't think you're supposed to.
The logic of most EDA suites I've used is that you use the schematic capture to select and connect components. Then in the PCB layout you do the physical routing. This means that any changes to components or nets should be done in the schematic capture, and then cascaded to the PCB layout. I've not used EasyEDA much, but I suspect this is how it's supposed to work there as well.
As Koraks pointed out - you have to modify the schematic to do what you want to do.
There should be a project file - the project file links the schematic and the PCB - you probably won't be able to do the linkage on your own. Often you won't be able to open a PCB file without the project and schematic files, or at least shouldn't be able to.
...
I'm working on modifying a PC board ( link to the regulator ) to output 5.4V for mercury battery replacement.
The issue at hand is the PC board is using SD8942 with two resistors to set the voltage. All fine, but my specific problem is as such:
I'd like JLPCB to make the board and install the resistors. To do this I need to locate the two resistors they have in stock in the correct size and put them in my EasyEDA project so they can install the resistors when they make the board.
The equation is: R1 and R2: V = 0.6 * (R1 / R2 + 1)
I need to make it output 5.4V
I'll take a flying leap of faith and postulate that your application will do just fine with a bog-standard 5 volt regulator.
I suspect so, too.
"Worst" case scenario you connect a small Schottky diode between the regulator's GND lead and actual circuit GND (cathode to circuit GND, anode to regulator GND pin), effectively lifting it by about 300mV. This will make 5.3V or thereabouts.
I'm stuck on removing the knob. The set-screw is out and the trim piece on the end is off. It appears the knob is frozen or glued on the shaft. I'm pretty sure a second screw holding the halves togeter is under the knob.
I might try wrapping the knob with a lot of tape and use a gear puller.
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