Introduction: 3 Easy Steps to Make a Voltage Divider

For another project of mine, I needed a voltage divider from 12V to 5V. Looking on internet I found many fancy, complicated, powerful, etc schematics. So, I decided to build very simple one. Is not my original idea, you can find it on internet as well but was looking some time for it. I like it because is simple and I decided to share it with community.

Step 1: What You Need

- 15k resistor

- 10k resistor

- wires

- soldering

Step 2: Prepare

Strip three wires on both ends. Solder the two resistors at one end and one wire between them. Solder the other two wires to free ends of resistors. That's all folks.

Step 3: How to Use It

15k resistor free end is to be connected to 12V, 10k resistor free end is to be connected to GND and middle wire is the 5V output.

Comments

author
Omnivent made it! (author)2016-02-19

That's not really an instructable.

You probably know the expression "give a man a fish and he's full for a day, give a man a fishing rod and he's full for the rest of his life"

You don't even give a whole fish here, rather just the tail.

Or, in more tangible terms, you give slightly incorrect info, according to your own design parameters and nothing about calculating for other voltages.

Your voltage divider as it stands will give you 4.8V out (unloaded) for 12V in. I know that this voltage will have no ill effect on your inductive sensor, as it just needs the output to be above 1.9V (on your Atmel controller when driven from +5V), but for 5V out, the top resistor needs to be 14k (found in E-rows E-48 and up).

You didn't even give the formula for a voltage divider, so here it is:

U_out = U_in x R2 / (R1+R2)

Where U_in is your input voltage, U_out is your output value and R1 is the top resistor, like in your pic.

But that's not the whole story either, as you also need to look at input and output impedance. For the input voltage, the datasheet of your sensor says 200mA max. (at 10 to 30V).

Feeding the sensor with 12V, as you do, its minimum output impedance will be 12V/0.2A=60 Ohm. Your load should ideally be larger than 60 Ohm as the 200mA is the max. allowed current. Then you'll typically look at the other extreme - how little current is needed. The fact that you use the output voltage unbuffered makes it a bit involved to calculate precisely and I'll not go there - just say that you'll be best off, using a current draw of perhaps 50..100mA, to minimize the effect of loading.

All that aside, you would be just as well off with using a 4.7V zener diode for R2 and a 2.2kOhm resistor for R1. And if you had just bought the NPN version of the sensor, you'd only need a pull-up to +5V.

Hope that was more rod than fish :)

author
AnsoGroup made it! (author)AnsoGroup2016-02-19

Thanks for all that theory. I just wanted to keep it as simple as possibile. All that you explain, is on internet. I did not wanted that. Got it?

But anyway, thanks for all that. Will be helpfull for others.

author
John_Singleton_ made it! (author)2016-02-19

That's really cool. I didn't know how to do that until now.
Thx

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