Trying to understand a rotary encoder's pinout

[Acorn Eye]

I'm not entirely clear on this pinout.

 

https://www.mouser.com/datasheet/2/15/EC18A-1370731.pdf (EC18AGB20401 [no 3])

 

Pins 2,4,3,1 are all data outs. Pin 5 is a COM terminal. 

 

There's no VIN and GND.

 

I'm not clear on what a COM terminal is but I believe it acts as the VIN.

 

I'm going to guess the ground is part of the chassis?

 

Reading the encoder would involve comparing outputs of 1,2,3,4 to each other. In the case where none of them output 1 its 0, and when all of them do, it's 225deg.

[AbydosOne]

30 minutes ago, Acorn Eye said:

I'm not entirely clear on this pinout.

Without firsthand knowledge, here are my thoughts: pin 5 should always be connected to the voltage source (up to 10V, per the spec, but probably 3.3 or 5V depending on your microcontroller), and then you sample the inputs from pins 1-4.

 

The encoder isn't a smart device (i.e. no silicon), so it doesn't have VIN or GND. It's basically a fancy switch changing connections between the pins and COM as you rotate it.

 

On your microcontroller, if you can, set an interrupt on all four of your input pins for rising and falling edge, and run the same code (to handle encoder state change) for any of those four interrupts. Otherwise, sample all four pins periodically (every tenth of a second?) and look for changes.

[mariushm]

You have the codes on page 2 of the datasheet

Here's EC18AGB20401

 

It can work both ways, depending on how you configure the microcontroller.

You can connect the COM to ground and send power through the four pins ... when you rotate, the device sinks that pin to ground and the microcontroller detects the drop.

Alternatively, you can connect the COM to voltage and when you rotate, some pins get voltage and your microcontroller detects a digital 1 on the pin.

Traditionally a COM indicates going to ground, and if i remember correctly, from a hardware point of view, it's easier for a microcontroller to "feel" the current being sunk to ground, then to receive voltage in the pin... but you as a microcontroller user shouldn't care about it, it won't affect you in any way, you can use it either way.

 

 

Now if you pay attention to the codes, you will see that you have groups of signals that mirror .. first 4 mirror the next 4 and so on

 

 

So you can do something like this :

 

value = bit1  [ 0..1]

if bit2 = 1 then value = 3-value  [ 0, 1, 2, 3 ]

 

so far you have :

0,1,2,3 | 3,2,1,0 | 0,1,2,3 | 3,2,1,0

now you add bit 3 in the mix , when bit 3 is 1, do 7 - value so now you have
 

0,1,2,3 | 3,2,1,0 | 0,1,2,3 | 3,2,1,0

0,0,0,0 | 1,1,1,1 | 1,1,1,1 | 0,0,0,0  =

0,1,2,3 | 4,5,6,7 | 7,6,5,4 | 3,2,1,0

Now add bit 4 in the mix : when bit 4 is 1, do 15-value

 

 0, 1, 2, 3 |  4, 5, 6, 7 |  7, 6, 5, 4 |  3, 2, 1, 0

 0, 0, 0, 0 |  0, 0, 0, 0 |  1, 1, 1, 1 |  1, 1, 1, 1

 0, 1, 2, 3 |  4, 5, 6, 7 |  8, 9,10,11 | 12,13,14,15

 

And that's your sequence in order

 

Oh... and use some resistors in series (between micro pin and knob pin) on each pin, something like 1-100 ohm,  maybe even a tiny capacitor (1-100nF) from pin to ground, as a sort of debouncing and to reduce the amount of current that will instantly jump between contacts in the knob thing (reduce wear on the metal bits inside)

[Acorn Eye]

On 10/12/2019 at 6:52 AM, mariushm said:

You have the codes on page 2 of the datasheet

Here's EC18AGB20401

 

It can work both ways, depending on how you configure the microcontroller.

You can connect the COM to ground and send power through the four pins ... when you rotate, the device sinks that pin to ground and the microcontroller detects the drop.

Alternatively, you can connect the COM to voltage and when you rotate, some pins get voltage and your microcontroller detects a digital 1 on the pin.

Traditionally a COM indicates going to ground, and if i remember correctly, from a hardware point of view, it's easier for a microcontroller to "feel" the current being sunk to ground, then to receive voltage in the pin... but you as a microcontroller user shouldn't care about it, it won't affect you in any way, you can use it either way.

 

 

Now if you pay attention to the codes, you will see that you have groups of signals that mirror .. first 4 mirror the next 4 and so on

 

 

So you can do something like this :

 

value = bit1  [ 0..1]

if bit2 = 1 then value = 3-value  [ 0, 1, 2, 3 ]

 

so far you have :

0,1,2,3 | 3,2,1,0 | 0,1,2,3 | 3,2,1,0

now you add bit 3 in the mix , when bit 3 is 1, do 7 - value so now you have
 

0,1,2,3 | 3,2,1,0 | 0,1,2,3 | 3,2,1,0

0,0,0,0 | 1,1,1,1 | 1,1,1,1 | 0,0,0,0  =

0,1,2,3 | 4,5,6,7 | 7,6,5,4 | 3,2,1,0

Now add bit 4 in the mix : when bit 4 is 1, do 15-value

 

 0, 1, 2, 3 |  4, 5, 6, 7 |  7, 6, 5, 4 |  3, 2, 1, 0

 0, 0, 0, 0 |  0, 0, 0, 0 |  1, 1, 1, 1 |  1, 1, 1, 1

 0, 1, 2, 3 |  4, 5, 6, 7 |  8, 9,10,11 | 12,13,14,15

 

And that's your sequence in order

 

Oh... and use some resistors in series (between micro pin and knob pin) on each pin, something like 1-100 ohm,  maybe even a tiny capacitor (1-100nF) from pin to ground, as a sort of debouncing and to reduce the amount of current that will instantly jump between contacts in the knob thing (reduce wear on the metal bits inside)

So I only had 4 100Ω resistors lying around, the only capacitor I have is 330uF, so I tied pin 5 to ground with a 10.2Ω resistor.

Do I need to tie each pin to 5v as well?

[mariushm]

The resistor placement is confusing in your picture.

No, you'd connect the 4 pins to the microcontroller and the 5th pin (COM) to voltage or ground, depending on how you want to read the pin states with your microcontroller. Probably easiest would be to connect 5v to voltage (5v).

You don't need a resistor on the 5th power to limit amount of energy going into the device.