I've been searching for a way to add an extra pin to a driver board. The MCU board that i'm using only has 8 pins on the left side and i'm already using all of them. 2 are for powering the driver and the other 6 are analog inputs. I need only one more control which has to be a digital output. Is there a way to put some kind of multiplexer on the driver and use the MCU pins that i'm using as analog but as digital outputs to control the mux outputs? I know it is kinda strange what i'm asking, but i have to have a solution for the lack of MCU pins.
Electronic – Need an extra pin to control the driver board
analogdigital-logicinputmultiplexeroutput
Related Solutions
You are looking for an 8 to 1 MUX with a very fast propagation delay <10ns.
THe Analog Mux is a passive switch, is bi-directional (1:8 or 8:1) but meets your speed requirements if you can provide low impedance source and some terminator resistance (passive or active).
The Digital MUX is unidirectional 8 to 1 with complementary buffered outputs are slower. Both have OE for a similar tri-state output.
If you are considering digital MUX's and need optimal speed, it would seem the ANALOG Mux will satisfy your requirements. THere will be glitches on the output signal during address transitions from high output impedance and capacitance, but that's normal. There will be some source loading effects when output load is connected to each input, but Ok for say low 50 ohm source and say 500 ohm terminator.
If using this for an analog MUX/DEMUX then time constants will different from specs according to difference in RC loads and transmission line effects.
simulate this circuit – Schematic created using CircuitLab
If your microcontroller can output PWM, then set it to a high PWM frequency, and use a low-pass filter to generate a voltage. Something as simple as a resistor from PWM output to a capacitor, and a resistor across the capacitor to ground, as well as signal into the bar graph input, will probably work well (enough).
Try a 10k resistor in, and a 1 uF capacitor, and a 10k resistor in parallel with the capacitor. This is assuming that the bargraph has a high impedance input. If there is a significant load in the bargraph input, R2 is not needed.
The duty cycle of the PWM output will determine the voltage that gets sent to the bar graph. Just make sure the uC output voltage range is compatible with the bargraph input; else re-buffer with an opamp or source follower or whatever.
If I had this problem, though, I'd simply use 10 digital output pins of the uC to drive 10 LEDs arranged in a bar directly :-)
Best Answer
Rather than try to use the same pins as both analog inputs part of the time and digital output(s), I would do the following:
Add an 8-channel analog mux to the driver board, such as the HEF4051B. The output of the mux will then go into one of the analog input lines currently going to the MCU board. Tie the enable lead \$\mathsf{\small \overline{\text{E}}} \$ to ground (always enabled).
Reconfigure the remaining seven analog inputs as digital.
Use three of the seven digital lines to select the 1 of 8 channels for the mux.
You now have four digital lines available between the MCU board and driver board.