Missing pins

Published: 2022-01-03
Last update: 2023-06-30

In this tutorial you will learn what to do if the number of available pins you have at your disposal is not enough to connect everything you want your system to interact with.

Table of contents

• Buttons, buttons... to many buttons
• The resistor ladder
  • The voltage divider
  • The resistor ladder
    • Step 1 -- real test
    • Step 2 -- fixing resolution
  • Pressing more buttons
• Multiplexing
  • Creating a digital 2-to-1 multiplexer
  • Creating a digital 4-to-1 multiplexer
  • Digital demultiplexer
  • Ready to use (de)multiplexer
• Expanders
  • PCF8574: communication
  • PCF8574: pinouts
    • PCF8574: addresses
  • PCF8574: usage examples
    • LED blinking
    • Detect button press
    • Interrupts

Although most of microcontrollers have many input/output pins, there are no problems to spend all of them and find yourselves in a situation when there would be no option to attach anything else. In this part I will show what you can do in case of such a situation. Other words, I will show how to "multiply" your pins.

Consider the following simple system in which you can use four buttons to enter and accept code/password (defined as CODE in source code). If correct, you will have an access to protected resources. System signals the fact of obtaining the appropriate authorizations by changing the status of the LEDs: the red one will go OFF and the green one will go ON for 2 seconds, after which the system will return to the default state, indicated by the red LED ON and the green one OFF.

The role of buttons is as follow (from left to the right):

• First, 0: allows to enter character (digit) 0.
• Second, 1: allows to enter character (digit) 1.
• Third, ENTER: allows to accept the entered sequence.
• Fourth, Delete: allows to delete the last entered character.

Both system and code is rather simple but I want to focus your attention on pins. At this moment you use 5 of total 19 pins of which 13 are digital and 6 are analog (but may be also used as digital). 14 pins are at your disposal, but will that be enough to handle typical alpha numeric characters? Typical, very simple membrane keyboard:

with only digits and some "function" keys require 8 pins. It is highly probable that you will lack of pins. This is how missing pins problem arise.

Recall voltage divider idea discussed in Voltage divider tutorial. General schema looks like it is showed on the following image

where the formula used to calculate $V_{out}$ takes a form
$$V_{out} = \frac{V_{in}R_{2}}{R_{1}+R_{2}}$$

	$$V_{out} = \frac{V_{in}\cdot 0}{R_{1} + 0} = 0$$
	$$V_{out} = \frac{V_{in}R_{2}}{R_{1}+R_{2}} = \frac{V_{in}R_{2_{1}}}{R_{1}+R_{2_{1}}} = \frac{V_{in}}{\frac{R_{1}}{R_{2_{1}}}+1}$$
	$$V_{out} = \frac{V_{in}R_{2}}{R_{1}+R_{2}} = \frac{V_{in}(R_{2_{1}} + R_{2_{2}})}{R_{1}+R_{2_{1}}+R_{2_{2}}}= \frac{V_{in}}{\frac{R_{1}}{R_{2_{1}}+R_{2_{2}}}+1}$$
	$$V_{out} = \frac{V_{in}R_{2}}{R_{1}+R_{2}} = \frac{V_{in}(R_{2_{1}} + R_{2_{2}} + R_{2_{3}})}{R_{1}+R_{2_{1}}+R_{2_{2}}+R_{2_{3}}}= \frac{V_{in}}{\frac{R_{1}}{R_{2_{1}}+R_{2_{2}}+R_{2_{3}}}+1}$$

Examples given in previous subsection show few similar but different dividers. You can combine all of them into one schema:

How it works is explained in a set of images below.

When you press the first button, current bypass all resistors so you obtain voltage divider with only resistance $R_{1}$ without resistance $R_2$ ($R_2=0 Ohms$).

When you press the second button, current flows through only one resistor ($R_{2_{1}}$) so resistance $R_2$ is equal to $R_{2_{1}}$.

When you press the third button, current flows through two resistors ($R_{2_{2}}$ and $R_{2_{1}}$) so resistance $R_2$ is equal to sum of $R_{2_{2}}$ and $R_{2_{1}}$.

When you press the fourth button, current flows through three resistors ($R_{2_{3}}$, $R_{2_{2}}$ and $R_{2_{1}}$) so resistance $R_2$ is equal to sum of $R_{2_{3}}$, $R_{2_{2}}$ and $R_{2_{1}}$.

This way you have had created a variable voltage divider called a multiple voltage divider or more often a resistor ladder.

Now it's time to make a real test.

1. Create circuit given in the schema below:
   

   Left: Schema of resistor ladder. Right: Close to real view of resistor ladder

   
   
2. Attach multimeter to our circuit:
   

   Left: Schema of ready to test resistor ladder circuit. Right: Close to real view of ready to test resistor ladder circuit

3. Use the following code to program Arduino
   
   resistor_ladder_01.ino

   C

   #define VOLTAGE A0 int sensorValue = 0; int i = 0; void printVolts(int val) { float voltage = val * (5.0 / 1023.0); Serial.print(" ADC value: "); Serial.print(val); Serial.print(", volts: "); Serial.println(voltage); } void setup() { Serial.begin(9600); } void loop() { sensorValue = analogRead(VOLTAGE); Serial.print(i); i++; printVolts(sensorValue); delay(1000); }

   1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

   #define VOLTAGE A0    int sensorValue = 0; int i = 0;   void printVolts(int val) {     float voltage = val * (5.0 / 1023.0);   Serial.print(" ADC value: ");     Serial.print(val);     Serial.print(", volts: ");   Serial.println(voltage);   }   void setup() {   Serial.begin(9600); }   void loop() {   sensorValue = analogRead(VOLTAGE);   Serial.print(i);   i++;     printVolts(sensorValue);       delay(1000); }

4. Write down results: voltage displayed on multimeter and analog value read by Arduino and changed into numerical values by analog-digital converter (ADC in short). In this case I got:

   Switch pressed	Resistance value of variable divider part $R_{2}$ (for constant $R_{1}=10kO$)		Voltage
   	used	measured	as integer from ADC	converted to volts from ADC value	measured by multimeter	theoretical value
   1	0kO	0	0	0.0	0.00	0.0
   2	$R_{2}$ = 1kO	0.98kO	95	0.46	0.47	0.45
   3	$R_{2}+R_{2}$ = 1kO+1kO=2kO	1.97kO	180	0.89	0.83	0.83
   4	$R_{2}+R_{2}+R_{2}$ = 1kO+1kO+1kO=3kO	2.95kO	238	1.16	1.16	1.15

This allows you to interpret each reading as a different button being pressed with a few lines of code -- see listing below. Obviously rather than telling the microcontroller to look for the exact values you expect the analog pin to read, you set a range of values that can be interpreted as belonging to a specific button. I do this by calculating the points halfway between each expected ADC reading (I called them midpoints) and set these as as the boundaries of my ranges (see also image):

Switch	Range
	from	to	midpoint
1	0	95	47
2	95	180	137
3	180	238	209
4	238	1023	630

If you have all midpoints defining boundaries for each range, you can use them in software to detect which button you press:

The downside is that output from voltage divider varies when different voltages are used so it require an update to your software.

So far, so good but this solution has two drawbacks which makes it working only for a project with a small number of buttons.

1. Every time you change the number of buttons, you have to measure what have changed -- you have to measure ADC values for newly added button(s). Next you have to update your source code.
2. What is worse, voltage resolution is not constant. The more buttons you have, the shortest voltage ranges you will have.
   Adding another switches, each with a 1kO resistor, you obtain results presented in the table below. To make it more visible, also results for 21, 31 and 41 buttons obtained with 10kO resistors instead 10 times 1kO are included.

   Switch	Range
   	from	to	midpoint	Midpoint distance
   1	0	95	47	---
   2	95	180	137	90
   3	180	238	209	66
   4	238	294	266	57
   5	294	342	318	52
   6	342	384	363	45
   7	384	424	404	41
   8	424	458	441	37
   9	458	491	474	33
   10	491	518	504	30
   11	518
   21	682
   31	766
   41	816	1023	919

   
   and depicted in the image:
   

   
   The situation could be even worse if you use 1kO pull-up resistor:
   

   Switch	Range
   	from	to	midpoint	Midpoint distance
   1	15	518	266	---
   2	518	687	602	336
   3	687	769	728	126
   4	769	820	794	66
   5	820	853	836	42
   6	853	877	865	29
   7	877	895	886	21
   8	895	909	902	16
   9	909	920	914	12
   10	920	930	925	11
   11	930
   21	974
   31	990
   41	998	1023	1010

   
   Variable and short voltage ranges may lead to errors both in code and in real devices as a result of noise and components tolerance. Anyway, magic numbers are always bad solution and should be avoided whenever possible. If you should somehow predict ADC value based on ADC resolution and number of buttons then you could implement better solution.

So, lets try find out resistors values to have equidistance measured in voltage, based on the following specification:

• $V_{in}=5V$,
• four buttons,
• pull-up resistor $R_{1}=1kO$,
• voltage distance $V_{d}=0.5$, e.g. $V_{S1}=0$ (voltage output for switch $S1$), $V_{S2}=V_{d}$ (voltage output for switch $S2$), $V_{S3}=2V_{d}$, $V_{S3}=3V_{d}$.

You know the input voltage, output voltage, and resistor $R_{1}$, but not $R_{2}$ which should be calculated. Starting from previously given formula:
$$V_{out} = \frac{V_{in}R_{2}}{R_{1}+R_{2}}$$
rearranging its components:
$$V_{out}(R_{1}+R_{2}) = V_{in}R_{2}$$
$$V_{out}R_{1}+V_{out}R_{2} = V_{in}R_{2}$$
$$V_{out}R_{1} = V_{in}R_{2}-V_{out}R_{2}$$
$$V_{out}R_{1} = R_{2}(V_{in}-V_{out})$$
you finally obtain formula for missing resistance $R_{2}$:
$$R_{2} = \frac{R_{1}V_{out}}{V_{in}-V_{out}}$$

Based on this formula and notation from the following schema you can calculate desired resistances $R_{21}$, $R_{22}$ and $R_{23}$.

• Calculations for $R_{21}$ -- in this case $V_{out}$ should be equal to 0.5V:
  $$
  R_{2} = R_{21} = \frac{R_{1}\cdot V_{out}}{V_{in}-V_{out}} = \frac{1.0 \cdot 0.5}{5.0 - 0.5} = \frac{0.5}{4.5} = 0.11
  $$
  To get resistance close to theoretical value of 0.11kO which is 110 Ohm, you will use one 100 Ohm resistor and one 10 Ohm so $R_{21}$ would be equal to 110 Ohm.
• Calculations for $R_{22}$ -- in this case $V_{out}$ should be equal to 1.0V:
  $$
  R_{2} = R_{21} + R_{22}
  $$
  On the other hand you know that:
  $$R_{2} = \frac{R_{1}V_{out}}{V_{in}-V_{out}}$$
  so in consequence:
  $$R_{21} + R_{22} = \frac{R_{1}V_{out}}{V_{in}-V_{out}}$$
  and then:
  $$R_{22} = \frac{R_{1}V_{out}}{V_{in}-V_{out}} - R_{21}$$
  $$R_{22} = \frac{1.0 \cdot 1.0}{5.0-1.0} - 0.11 = \frac{1}{4} - 0.11 = 0.25 - 0.11 = 0.14$$
  To get resistance close to theoretical value of 0.14kO which is 140 Ohm, you will use one 150 Ohm resistor so $R_{21}$ would be equal to 150 Ohm.
• Calculations for $R_{23}$ -- in this case $V_{out}$ should be equal to 1.5V:
  $$
  R_{2} = R_{21} + R_{22} + R_{23}
  $$
  On the other hand you know that:
  $$R_{2} = \frac{R_{1}V_{out}}{V_{in}-V_{out}}$$
  so in consequence:
  $$R_{21} + R_{22} + R_{23} = \frac{R_{1}V_{out}}{V_{in}-V_{out}}$$
  and then:
  $$R_{23} = \frac{R_{1}V_{out}}{V_{in}-V_{out}} - R_{21} - R_{22}$$
  $$R_{23} = \frac{1.0 \cdot 1.5}{5.0-1.5} - 0.11 - 0.15 = \frac{1.5}{3.5} - 0.26 = 0.428 - 0.26 = 0.168$$
  To get resistance close to theoretical value of 0.168kO which is 168 Ohm, you will use one 120 Ohm resistor and one 51 Ohm so $R_{23}$ would be equal to 171 Ohm.

The above procedure is quite mechanical and can be easily automated with simple code. Below you can find Python script implementing this procedure (even if you have no Python installed on your system, you can run it online, for example on OnlineGDB):

The variable resistanceToUse defines resistances that must be used; for example you may require that third button should use 150 Ohms resistor:

When executed, above code should return the following output:

Finally you can implement more universal method for detecting the pin which was pressed:

and measure real voltage and ADC values as it is presented in the table below:

Switch pressed	Resistance value of variable divider part $R_{2}$ (for constant $R_{1}=1kO$)		Resistance distance		Voltage
	used	measured	calculated	used	as integer from ADC	converted to volts from ADC value	measured by multimeter	theoretical value
0	---	---	---	---	1023	5.00	4.95	5.00
1	0k$\Omega$	0	0	0	0	0.0	0.02	0.0
2	$R_{21}$ = 110 $\Omega$	113 $\Omega$	$R_{21}=110 \Omega$	$R_{21}=110 \Omega$	100	0.5	0.51	0.5
3	$R_{21}+R_{22}$ = 260 $\Omega$	260 $\Omega$	$R_{22}=140 \Omega$	$R_{22}=150 \Omega$	208	1.0	1.04	1.0
4	$R_{21}+R_{22}+R_{23}$ = 431 $\Omega$	430 $\Omega$	$R_{23}=168 \Omega$	$R_{23}=171 \Omega$	306	1.5	1.51	1.5

Below you have a real life example of resistor ladder application:

You may ask: What if I press more than one button at once? For example: two or three? Nothing. Nothing in the sense that pressing multiple buttons at once is equivalent to pressing only one of them. Let's make a test in Tinkercad.

Start with pressing one button at a time (instead of buttons I use wire to make "hard" connection; this is due to inability to press more than one button in simulation which will be needed soon -- instead of pressing I make a connection):

No button is pressed:

First button is pressed:

Second button is pressed:

Third button is pressed:

Fourth button is pressed:

Now, press two buttons and more:

Fourth and third button is pressed:

In this case result is the same as you would press third button. If you continue this experiment you will not get different results in a sense that what you get agrees with corresponding one button case:

Fourth, third and second button is pressed:

All buttons are pressed:

Multiplexing is the generic term used to describe the operation of sending one or more analogue or digital signals over a common transmission line at different times or speeds. The device you use to do that is called a multiplexer and demultiplexer. An electronic multiplexer (also mux or data selector) makes it possible for several signals to share one device or resource, for example, one A/D converter or one communication line, instead of having one device per input signal. It selects between several analog or digital input signals and forwards it to a single output line:

You should be familiar with this concept, as every day you use it (or rather it is used by some devices) to transmit your data through an network:

The selection of each input line in a multiplexer is controlled by an additional set of inputs called control lines and according to the binary condition of these control inputs, one selected data input is connected directly to the output. Normally, a multiplexer has an even number of $2^{n}$ input lines and a corresponding number of $n$ control inputs. An electronic multiplexer can be considered as a multiple-input, single-output switch. The schematic symbol for a multiplexer is an isosceles trapezoid with the longer parallel side containing the input pins and the short parallel side containing the output pin. The schematic below shows a 2-to-1 multiplexer (also denoted as 2:1 multiplexer or 2:1 mux) on the left and an equivalent switch on the right:

Multiplexers can be either digital circuits made from high speed logic gates used to switch digital or binary data or they can be analogue types using transistors, MOSFET’s or relays to switch one of the voltage or current inputs through to a single output.

To start out easy, you’ll create a multiplexer taking two inputs and a single selector line as it was depicted above. With inputs $X_{0}$ and $X_{1}$ and select line $S$, if $S$ is 0, the $X_{0}$ input will be the output $Y$. If S is 1, the $X_{1}$ will be the output $Y$.

To find a digital circuit implementing this functionality, you will apply approach described in my Logic gates tutorial. If you are not familiar with logic gates, it may be a good idea to read this material first.

You begin with creating corresponding truth table defining the dependence of the output signal on the value of the input signals:

Inputs			Output
$S$	$X_{0}$	$X_{1}$	$Y$
0	0	0	0
0	0	1	0
0	1	0	1
0	1	1	1
1	0	0	0
1	0	1	1
1	1	0	0
1	1	1	1

Having this table, you can derive a corresponding Boolean formula:
$$
\begin{align}
Y =
& \overline{S} \cdot X_0 \cdot \overline{X_1} + \overline{S} \cdot X_0 \cdot X_1 + S \cdot \overline{X_0} \cdot X_1 + S \cdot X_0 \cdot X_1 = \\
& \overline{S} \cdot X_0 \cdot (\overline{X_1} + X_1) + S \cdot X_1 \cdot (\overline{X_0} + X_0) = \\
& \overline{S} \cdot X_0 + S \cdot X_1\\
\end{align}
$$

The boolean formula for the 2-to-1 multiplexer looks like this:
$$Y= \overline{S} \cdot X_{0} + S \cdot X_{1}$$
or the same as programming logic expression:

or in the pseudocode style:

Based on the formulas given above, you can make a corresponding logic circuit:

Of course it can be completed also with other set of gates, for example NAND gates. From the previous material, Completeness of NAND gate set, you know that using only NAND gates you can have equivalents of all three basic gates: AND, OR and NOT as it is showed again below:

Using these "replacements" to the schema with basic set of gates you obtain:

This schema could be simplified. Let's introduce some additional symbols:

Now you can write:
$$
Y = \overline{E \cdot F} = \overline{\overline{C \cdot C} \cdot \overline{D \cdot D}}
$$
Because in Boolean algebra the following is true:
$$
X \cdot X = X
$$
you can write:
$$
Y = \overline{\overline{C} \cdot \overline{D}} = \overline{\overline{\overline{A \cdot A}} \cdot \overline{\overline{B \cdot B}}}
$$
Again from the law given above you obtain:
$$
Y = \overline{\overline{\overline{A}} \cdot \overline{\overline{B}}}
$$
and applying the following rule:
$$
\overline{\overline{X}} = X
$$
you finally get:
$$
Y = \overline{A \cdot B}
$$
Other words, none of the gate with output $C$, $D$, $E$ or $F$ is needed, and final schema of multiplexer build only with NAND gates looks like below:

You can combine two lower order multiplexers like 2:1 or 4:1 to get higher order multiplexer like 8:1. Now, for example you will implement a 4:1 multiplexer using a 2:1 multiplexer. To construct it, you have to combine three 2:1 multiplexer together, as I shown on the image below:

In result you have a circuit with:

• 4 input pins ($X_{0}$, $X_{1}$, $X_{2}$ and $X_{3}$),
• 2 control/select pins ($S_{0}$ and $S_{1}$)
• and one output pin ($Y$).

Dependence of the output signal on the combination of control signals $S_{0}$ and $S_{1}$ along with flow path of the signal is depicted below:

Conversely to multiplexer, a demultiplexer (or demux) takes only one input and pass through to one of multiple output lines, using a select line to choose which output the input goes to.

To find a digital circuit implementing this functionality, first you have to create corresponding truth table:

Inputs		Outputs
$S$	$X$	$Y_{0}$	$Y_{1}$
0	0	0	0
0	1	1	0
1	0	0	0
1	1	0	1

Having this table, you can derive a corresponding Boolean formula which in case of the 1-to-2 demultiplexer is really simple and looks like this:

$$Y_{0}= \overline{S} \cdot X$$
$$Y_{1}= S \cdot X$$
or the same as programming logic expression:

or in the pseudocode style:

Based on the given above formulas, you can make a corresponding logic circuit:

Of course it can be completed also with other set of gates, for example only with NAND gates:

There are a lot of different ready to use (de)multiplexer.

• The basic trio of analog multiplexers stems from the venerable 4000-series CMOS logic chips: the 4051 is a single eight-way multiplexer, the 4052 has two four-way multiplexers, and the 4053 has three two-way switches.
• The 74HC405x series chips with higher switching speed or low-voltage analog multiplexers meant for battery-powered use: 74LV405x.
• The high-speed and low-resistance 3251, 3252, and 3253 series chips: CBT3251, FST3252, SN74CBT3253, etc.

In most cases multiplexer can work also as a demultiplexer, so further I will use the term multiplexer in both meanings.

In this part I will show how to work with CD4051B and CD4052B chip (CD4051B, CD4052B, CD4053B data sheet ).

CD4051 is an eight (8) channel multiplexer and has three control input named as A, B and C. These control inputs selects one of 8 channels to the output in order to obtain the desired connection. Channel I/O terminals became outputs and common O/I become input terminals when CD 4051 is used as a demultiplexer. Pinouts for all chips from this family are depicted on the image below and for CD4051 chip is summarized in the following table:

Pin	Meaning
A	selector line coding first bit of binary channel number
B	selector line coding second bit of binary channel number
C	selector line coding third bit of binary channel number
0-7	independent inputs/outputs channels
Vee	negative supply voltage (connected to ground (GND) in our case)
Vss	ground (GND, 0V)
Vdd	positive supply voltage (from 5 to 20 volts)
COM AUT/IN	common input/output
INH	enable input active on LOW (connected to ground (GND) in our case)

Static hardware channel selection

I start with very basic example and show how multiplexer can be used as a static hardware switch:

Left: Basic schema to test CD4051 chip