As the title implies, I’m interested in utilizing the Artemis for a project which involves taking analog readings (temperature readings using thermistor probes). With my current MCU this is easy, as I can just use a high output from a digital pin, but that isn’t really possible here.
I am concerned with power consumption, so I need to be able to turn the probe circuit on and off. Again, easy enough using the digital pin.
I’m fairly novice at this, and have a couple of general ideas, but would really like to hear from more experienced folks on best practices and theory.
The possibilities I’ve considered are:
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Using a voltage divider, but the probe circuit is itself a voltage divider circuit so I’m not sure what the effect will be of chaining them, nor what the power considerations look like here
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Using a BJT transistor as a switch from an external power source and then using an LDO or zener diode(s) to further adjust the voltage down as needed.
Thanks
If your thermistor is already a voltage divider, then you can reduce it’s output range by adding resistors in series with it (two resistors in series of Ra and Rb are the same as one big resistor of Ra+Rb). So add a resistor between it and 3.3V to lower the peak of the range (you’ll need to do the maths).
You can use a digital pin output pin to power the whole lot, and an analog in pin to read the result of thermistor.
If your thermistor is already a voltage divider, then you can reduce it’s output range by adding resistors in series with it (two resistors in series of Ra and Rb are the same as one big resistor of Ra+Rb). So add a resistor between it and 3.3V to lower the peak of the range (you’ll need to do the maths).
You can use a digital pin output pin to power the whole lot, and an analog in pin to read the result of thermistor.
Had the same question, thanks this worked for me!
stephenf’s answer is spot-on. I just want to add the tidbit of information that the Artemis (Apollo3) analog pins are in fact 3.3V tolerant - that is to say it is OK to apply a full 3.3V signal to them. The ADC does have a 2V reference however (as noted of course). So any analog signal over 2.0V will be reported as 2.0V.
The benefits of the additional voltage divider recommended above is for the purpose of signal conditioning so that the optimal combination of range and resolution can be obtained.
Try using 1N914 switching diodes for voltage drops. Each diode has about a 0.6v drop. They’re dirt-cheap on Amazon, Mouser, . . .
Any switching diode will work – I just happen to have a pile of the 1N914s.
I have a quick questions: can I safely power a LED with 3.3v without a resistor? I have some LEDs that I typically power with 5v and a 220Ω resistor, but I can only supply 3.3v at the moment.
Additionally I have push-buttons that I typically pull-down with 10kΩ resistors when using 5v. What resistor should I use in this case? Is there a general rule on what resistor should be used given the voltage?
Thank you in advanced for your help!
Generally it is safest to ensure that you have a current limiting resistor for LEDs. Diodes, once connected to voltages above their forward voltage, begin to draw lots of current. Red LEDs have lower forward voltages than blue LEDs. Know the danger - if your LED draws a lot of current it has the possibility of damaging the microcontroller pin that is driving it. You can decide whether that risk is acceptable for you. That being said I often do power LEDs directly.
My “rule of thumb” is that I always start with a 10k pull up/down resistor. It’s more of a habit really. So when do I change that? It’s all about timing (and sometimes impedance and maybe current limits). When restoring a line to the pull state the resistor will interact with capacitance on the line to form an RC circuit. A lower resistance will get you there faster (more current can flow through a smaller value resistor).
Too “weak” of a pull: Large value pull resistors may respond too slowly for you, and also don’t handle any stray charges very well (they tend to stick around longer, imparting voltage in your line).
Too “strong” of a pull: Small value pull resistors respond quickly but can allow lots of current through (which could possibly damage supplying components). They also require the active portion (when the line is in the non-pull state) to provide more current so they can be a power waste. Finally, depending on the effective impedance of the current source they can begin interacting with the supply to form a voltage divider - in the non-pull state you might be fairly far off from the desired voltage.
https://learn.sparkfun.com/tutorials/di … cteristics