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Unread 23-02-2015, 17:30
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Re: Using the TI Launchpad

It appears from the schematic:

The top port is the 3.3VDC power from the TI Launchpad.
The bottom port is the digital ground reference to the TI Launchpad.
The middle port goes to the I/O pin on the TI Launchpad.

When the normally open button is depressed the 3.3VDC power is put across the resistor (current limit) and LED (light emitting diode).
With the button depressed the I/O will see a 'high' of 3.3VDC.
With the button not depressed the I/O will be pulled 'low' because the digital ground is still connected.

Assuming that's a 20mA forward current in series with a voltage drop of 1.8VDC across the LED (check the LED specifications) then R1 would be around:
3.3VDC - 1.8VDC = 1.5VDC across R1
E = I * R (Ohm's law)
1.5VDC = 0.02A * R
R = 1.5VDC / 0.02A
R1 = 75 Ohms

Alan posted before I could finish.

Normally to get that 3.3VDC regulated power for simple projects I use something like this:
LM2937-3.3

In the case of the TI Launchpad the 5V rail that powers it can come from the USB port 5V power.
It has an onboard regulator circuit that makes 3.3VDC based:
On the MSP-EXP430G2 the 3.3V regulator is the TPS77301DGKR.
On the MSP-EXP430F5529LP the 3.3V regulator is the TPS62237DRY.

That 3.3VDC is available at all the edge pins marked:
VCC on the MSP-EXP430G (TP1 on this PCB will give access to 5VDC)
3V3 pins on the MSP-EXP430F5529LP

It appears the board you have is this: (based on the image in the first post of this topic)
MSP-EXP430F5529LP

The other board (which I have laying around) is this:
MSP-EXP430G2

Sorry I revised this post a few times to capture accurate information about both 'TI Launchpads' I have seen.

Last edited by techhelpbb : 23-02-2015 at 18:30.