Electronic – arduino – Lower Voltage With A Zener Diode

arduinovoltagezener

I am trying to lower the voltage from a 9V battery to 5V for use on an Arduino pin. I bought a 5.1V zener diode from RadioShack. Will this do what I want? Do I have to lower the amps as well? It looks like the battery delivers 9V and 1A. Should I first use a resistor to lower the amps, then use the zener diode to lower the voltage? I tried using a 330 ohm resister directly with the battery and a multimeter just to see how much it lowered the amps, and it got very hot so I am not sure if I should be using it.

Zener Diode:

IN4733A

Voltage: 5.1V

Current: 49mA

Maximum power dissipation: 1.0W

I looked at some example equations here: http://www.electronics-tutorials.ws/diode/diode_7.html but am still not comprehending it. I am not even sure which side the black band on the diode should point. So my question is: Will the diode work to drop from 9V to 5V and do I need to do anything extra?

I wanted to use this: http://www.amazon.com/Voltage-Sensor-Detector-Divider-Arduino/dp/B00S4PCCG8/ref=sr_1_1?ie=UTF8&qid=1424025730&sr=8-1&keywords=voltage+divider+arduino, but don't want to wait so thought I would try something else until I can get this ordered.

Best Answer

The zener diode you reference has a zener-voltage of 5.1V and a power rating of 1W. You want to place a resistor in series with the 9V source and the zener diode to limit current draw.

The datasheet for the IN4733 states that at a zener current of 49mA the zener voltage is at the rated 5.1V. The only calculation you need to do is to size the resistor such that you have at least 49mA, although lower current would also likely give you around the 5.1V zener rating. Normally you would look at the datasheet for a chart to give you this information, but the one I found left this information out.

$$ R = \frac{9-5.1}{10mA} = 390 \Omega$$

But, again, larger sized resistors are likely to give you the voltage you want with lower power dissipation. Test to verify.