| Lecture: | 33 |
| Objective: | To understand how to build a circuit and program firmware to have
multiple buttons connected to a single pin and identify which button was pressed.
|
Today we will construct an interesting circuit that allows you
to do a lot with a little. A general problem when working with
MCUs is the limited number of I/O pins.
All is not lost however. If you are willing to trade
off some time, you can arrange several push buttons on a single
MCU pin. This seemingly miraculous accomplishment is realized
using the following schematic.
To understand how this circuit works lets look at it in
three different ways.
High Level
At its highest level this is a circuit with three inputs
(button) which are connected to a single MCU pin. The MCU
will run a program which will allow it to know which of the
three buttons has been pushed. If more than one button is
pressed, then the button with the lowest index (closest to
the MCU pin) will be identified.
Mid Level
This circuit operates in two modes; charging and discharging.
While charging the node labeled "MCU pin" must be configured
as an output. While configured as an output, this pin supplied
3.3v to the positive side of capacitor C1 charging it. After
the cap is charged "MCU pin" is configured as an input.
At this time the charge on C1 will move to ground through
whatever resistor stands in its way. The more resistance
that stands in its way, the longer it will take to discharge
C1. Measuring the amount of time that it takes "MCU pin"
to go from logic 1 to 0 indirectly tells you the total
resistance and consequently the button which is pressed.
Detailed
The circuit operates by successively charging and discharging
the node labeled "MCU pin". When in charging mode MCU pin
will be an output at logic 1 causing cap C1 to carry a
3.3v charge. When the node labeled "MCU pin" is switched
to input mode the positive end of the cap will discharge
to ground through any intervening resistance. Lets look
at the case of what happen when the cap discharges and the
button S2 is being held.
Button S2 is pressed. The charge on the cap will go
to ground through the series combination of the 2, 1000
ohm resistors. This series pair has an effective resistance
of 2000 ohms. Hence the time constant of the voltage decay is
RC = (2000)(47nF) = 94uS
Give the discharge voltage of an RC circuit is V(t) = Vo*e^(-t/RC)
What is the time required for the voltage to drop from 3.3v to
1.0v when the first, second, or third button is pressed?
1.0 = 3.3e^(-t/x)
1/3.3 = e^(-tx)
3.3 = e^(t/x)
1.19 = t/x
t=1.19 x
First Second Third
RC = 47uS RC=94uS RC=141uS
t=1.19(47uS) t=1.19(94uS) t=1.19(141uS)
t=56uS t=112uS t=168uS
MCU considerations
At this point its appropriate to consider which of the MCU pins
to utilize. Using a general purpose I/O pin would be be straight
forward, but the deviation in the threshold voltages (the voltage
at which a pin goes from a 1 to a 0), would require some calibration
for each Microcontroller which was flashed. Using RA0-RA3 has the
important advantage of allowing the comparator module (on page 234)
and comparator voltage reference module (page 239) to set an
absolute voltage threshold.
Simultaneous button presses
If more than one button is pressed only the resistance of the button
closest to the MCU pin effects the discharge time, because the two
button forms a parallel resistor network where one branch (the button
closer to the MCU) has a 0 resistance and the other branch (the
button further away from the MCU) has a 1k resistance. The effective
resistance of this pair is 0, negating the effect of the button further
away from the MCU. Thus, when more than one button is pressed, the
discharge time is governed by the button closest to the MCU pin.
No button press
When no button is pressed, the cap discharges thought the (very high)
resistance of the MCU pin. See the advanced topics below. The resistance
in this case is about 1M ohm, hence the time constant is on order to tenths
of a second.
Firmware
Do you think that you can write some firmware for this problem?
Advanced topics
If we take into consideration the input impedance of the PIC pin
the analysis above is effectively unchanged. By Thevenin, the
input of the PIC can be viewed as an impedance (resistance)
connected to ground. The input impedance of the PIC is really
large, on the order of mega ohms plus the 100 ohms sitting in
front of the input. The resistance of a series circuit is the
sum of the individual resistors, hence the PIC input is on the
order of mega ohms. Now the charge on C1 has two paths to
ground, one through the R1, R2, R3 resistor network and the
other through the PIC input. These two resistors are in parallel.
The effective resistance of a pair of parallel resistors is the
reciprocal of the sum of the reciprocal of the resistances.
Working this out you can see that the equivalent resistance is
virtually the same as that of the R1, R2, R3 resistor network.
Hence, taking into consideration the input resistance, only
slightly changes the calculations shown above.