# Arduino Mega 2560 - Stepping down voltage as input

Maddy from Little Bird Electronics suggested I go direct to the experts so here goes..

My requirement is:- 20 Rooms temperature to monitor

In Place:- We currently monitor these rooms back to a patch panel that displays this info. Using TP2000 Probes. These use resistance which from the output of the dislpays creates a output voltage which I wish to use as input.

New Requirement:- If the Temp gets above or below n degrees C a SMS alert is sent to a tech. Data to be logged on a pc. I can code any program for the data logging / sending alerts etc but electronics is not my strong point. Intermediate understanding though. (Plus google).

It has been suggested to use the Arduino Mega 2560 along with a ArduinoGSM Shield (for connectivity / sms).

Issue.. As this Arduino board only accepts max 5v as input (able to have 50 inputs) how can I step the voltage down. Ie. If the shed is 30 degrees c then the input will be 5v If the shed is 15 degrees c then the input will be 2.5 etc etc..

Is the Arduino the answer to use along with sonething to step the voltage down for input.. or is there something better that may meet my needs.

Regards Mark

• What is the maximum voltage input? – Leon Heller Apr 5 '13 at 1:53
• Notice that ATmega2560 only has 16 analog inputs. – jippie Apr 5 '13 at 19:57

You are saying you need to step down the voltage, but then you are describing the voltage levels as if they range from 0 to 5 volts. Is the voltage range of your signals 0 to 5V? If yes, then you can used the A/D converters that are built into the AtMega2560. If you have more inputs than supported by AtMega2560, you can use an analog switch, such as 74HC4851. I don't know how the analog inputs are configured on the Arduino board, you might have to use Op Amp configured as unit gain as a front end. Also, you might want to use zener diode to limit the voltage to 5 volts.

• Hi Suirder... The input voltage is up to 30 volts dc but my understanding of the Mega2560 is it only acepts up to 5 volts. Therefore that is why I need to have some way of converting the power before it gets to the board.. like the example above.. – Mark Fellowes Apr 5 '13 at 5:53
• Mark, You can use the voltage divider as recommended by Raaymaan. Be sure to use resistors that rated 1% or better. If you end up using external A/D, you can also look at AD7880; This A/D has very easy to use paralle interface. – Suirnder Apr 5 '13 at 14:28
• Hi Team.. Really appreciated your advice and help.... I have gone and purchased the 2560 R3 and will use the resistor method to limit the voltages coming into the unit.. Once again thanks heaps.. Mark – Mark Fellowes May 1 '13 at 3:29

The Arduino Mega 2560 has a recommended input voltage between 7-12v, with absolute limits of 6-20v. It has a 5v linear regulator on board, which supplies the 5v.

Usually a sensing device is powered from the Arduino's 5v (or at least shares a common ground), and then the sensor's output pin (spits out an analogue voltage representing it's measurement) is fed into the one of the Arduino's analog pins, which contain the ADC input. These are 10bit ADCs. If you need greater resolution, I would personally recommend NOT using a Mega 2560, but instead use a Duemilanove/Diecemilia/Leonardo and use an SPI connected ADC such as Microchip's MCP3208.

It is a great 12bit ADC, tried and tested, with an existing example code base for Arduino. By using this SPI ADC, you don't need all of the pins that the larger Arduino's have, saving precious ££/.

Note the Mega has 8x the flash memory than those mentioned.

Can you be a little more specific about the temperature probes probes? Who is the manufacturer? Do you already have a link to their datasheet?

If you want to step down 30v to a safe voltage for the Arduino's supply, you can use something like this Integrated Switching Regulator, SLTS059A.

If you want to 'step down' the probe's output voltage of 30v to the Arduino's maximim input voltage of 5v, you can use a 5:1 voltage divider, such as this 50k : 10k

• Thank you for the fantastic answer. It's helping me right now. +1 – Chris K Aug 1 '13 at 5:52