Electrical – Wiring on-off switch with resistor to SoC

bad-wiringraspberry pisocswitcheswiring

Please note: although this question involves a Raspberry Pi (hereafter RPi), it is really a pure electronics question at heart!

I am trying to connect my RPi 1 Model A to a breadboard with a single, simple on-off-on-off switch on it. The kicker here is that although I'm using an on-off-on-off switch, I really just want it to function like a normal (on-off) switch. That is: push it once, the circuit is closed and sending an input signal on to my RPi. Push it again, and the circuit open. Rinse and repeat.

I was given this diagram to follow for wiring things up:

I then subsequently inferred my own rendition of this, which more clearly shows the joining of the switch's left and right pins together (giving it the desired on-off behavior) as well as how to wire the 3.3V power on the RPi to the GPIO input pin:

So to behin with if anything above looks incorrect or awry to you, please begin by correcting me!

Assuming I'm on track, I'm now trying to actually wire this up in real life, between my RPi and my breadboard. Here's my best attempt:

Nevermind the LED and resistors in the bottom right corner of the breadboard, they're leftover from another experiment and aren't connected to anything else.

So at the top-left we have the 3.3V power source from the RPi connected to the top-most rail on the breadboard via a red jumper; then
A smaller red jumper wire forwards the power onto a column that is then connected to a 10-kOhm resistor; more on this in a second

On the left-side of the above photo we have the switch, here's a better look at the wiring/setup:

Notice the small orange jumper connecting the left and right switch pins; I believe this is what accomplishes the on-off-on-off -> on-off behavior I'm looking for
The same column that joins left and right pins (via orange jumper) is also connected to a red jumper which is also connected to the same 10-kOhm resistor we talked about up above
Finally, the switch's middle pin is connected to the GND rail via a black jumper

This brings us to the center of the beadboard, where that big fat 10-kOhm resistor is hanging out:

The resistor is connecting the RPi's power to the switch (both pins at the same time)
The resistor is also connecting back to the GPIO's input pin via gray jumper wire

Finally, my question!

Remember, at the end of the day, all I want to do is:

Convert this on-off-on-off switch to a on-off switch
When the user presses the switch (closing it), an appropriate signal is sent on to the GPIO input pin, which is then handled at the software layer

So I ask: will my circuit accomplish the following behavior? Is it wired correctly (correct junctions of wires, correct usage of resistor, etc.)? Or will it "fry my pi"?! If anything is incorrect, what's the fix/solution?

Update

Several users have pointed out to me that my wiring around the switch is incorrect, here's a Fritzing diagram of what I think the solution is:

Final update

Wiring when I set the internal pin resistor at the software layer, and omit the breadboard resistor:

Best Answer

Figure 1. Incorrect pull-up.

Your pull-up resistor is connected to the wrong rail. It is pulling down. Your button just pulls it down better.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Figure 2. (a) What you intended. (b) The way you wired it. (c) With internal pull-up enabled.

The wire link on the right end of your resistor is wired to GND instead of V+.

If you can program the GPIO with internal pull-up then omit the resistor. It is not required. In the present configuration it is competing with the pull-up and the input voltage can never rise above that set by the potential divider created by the pull-up and pull-down.

Extreme clarification:

Figure 2. The pull-up resistor jumper wire needs to be moved to terminate in position 2 - not position 1.

From the comments:

OK, so I will initialize my pin at the software layer via gpioController.provisionDigitalOutputPin(RaspiPin.GPIO_23, 'RunningLed', PinState.HIGH), which will cause the GPIO pin #23's internal resistor to "pull up" (PinState.HIGH), so that when someone pushes the switch (which in turn closes the circuit) the internal resistor will kick in, yes?

That will work correctly but your description of the action is incorrect.

schematic

^{simulate this circuit}

Figure 3. Internal pull-up with external pull-down switch.

Once enabled the pull-up resistor will be in-circuit as though it is wired in place. It doesn't "kick-in" at any point but it will be effective under certain conditions.
Fig. 3a: With SW1 open the pull-up, R1, connects the internal buffer to V+. The input is now "pulled-up" to Vcc. It will be read as "high" by your program.
Fig. 3b: With SW2 closed the pull-up is still trying to pull but is overcome by the much, much lower resistance of the switch. The input voltage will drop to 0 V. The input will be read as "low" by your program.

Note that earlier in the discussion you were going to have an external pull-down and an internal pull-up. This would not work as the voltage (with the switch open) would be neither fully high nor fully low and would probably be in an undefined state alternating between the two when read by your program.

Related Solutions

Electronic – Reed switch wiring

My first test simply lights up a LED when the switch lets the current flow. As my LED is 2V 20mA, a resistor of (3.3-2)/0.02 = 65 Ohm should be enough, but I only have 100 Ohm, it should be ok right?

That will be fine.

The next step is to capture the switch event in the GPIO 5 listening for 3.3v input. I want a max current of 1 mA to flow through my pin, so I pick a resistor of around 3 k-ohm. Say 5k for safe purposes.

You are getting mixed up at this point.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A typical micro-controller input with optional internal pull-up resistor.

The input impedance (resistance) of a GPIO pin is very high. If you connect it to the V+ it might draw a couple of micro-amps even with no resistor present. This makes it very sensitive to stray voltages or electro-magnetic fields and, if not addressed, can result in random switching of the input on and off. The way to solve it is to add a pull-up or pull-down resistor of sufficiently low resistance to do the job. 10k is usually a good balance between being sufficiently low and not so low that we're wasting power by increased current through it.

The need for pull-ups is so common that most manufacturers include them internally with a configuration option to enable them.

In your third diagram you have the switch on the "high side" which means that you need a pull-down resistor. If the chip doesn't have an internal pull-down you'll have to add an external one. 10k should be fine.

Electronic – How to i drive this 5v Relay board with RPi

Here is a likely schematics of your relay module:

Relay to Raspberry connection:

VCC to +5V
IN+ to GPIO
IN- to GND
GND to GND

GPIO is configured as push/pull output, no pull-up, no pull-down. Relay is turned on by writing 1 to the port.

Alternate connection if the first one can't activate relay:

VCC to +5V
IN+ to +5V
IN- to GPIO
GND to GND

GPIO is configured as open-drain output, no pull-up, no pull-down. Relay is turned on by writing 0 to the port.

As voltage between IN+ and IN- is more then 2V, GPIO is never forced above 3.3V, so it should be pretty safe.

Finally, my question!

Update

Final update

Best Answer

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Electronic – Reed switch wiring

Electronic – How to i drive this 5v Relay board with RPi

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