Print Page | Close Window

resistor 5v to 2.8v and 2.0v

Printed From: the12volt.com
Forum Name: Motorcycle Electronics
Forum Discription: Installing Stereos, Alarms, Remote Starters, Lights, Garage Door Openers and other electronics on motorcycles.
URL: https://www.the12volt.com/installbay/forum_posts.asp?tid=126612
Printed Date: April 29, 2024 at 12:47 AM


Topic: resistor 5v to 2.8v and 2.0v

Posted By: katman1260
Subject: resistor 5v to 2.8v and 2.0v
Date Posted: March 17, 2011 at 10:19 PM

Can someone explain what this statment means? I don't understand where both ends of the resisters are being connected? are all resisters connected at one end to 5v source and the other leg to pin (23) and (16) There are 4 resistors and only two connection points?

33k resistor to +5v

(23) and 22k resistor to gnd(16) to obtain 2v and 33k to +5v and

47k to gnd to obtain 2.8v




Replies:

Posted By: i am an idiot
Date Posted: March 18, 2011 at 5:57 AM
What are you trying to do?



Posted By: KPierson
Date Posted: February 18, 2012 at 2:19 PM
It sounds like a voltage divider.

For the first voltage, connect one side of the 33K resistor to 5vdc and the other side to (23) and one leg of a 22K resistor. Connect the other side of the 22K resistor to ground. Where the 33K and 22K resistors connect there will be 2vdc on that connection (according to the voltage divider formula).

For the next voltage connect one leg of a 33K resistor to 5vdc and connect the other leg to the 47K resistor. However, according to the formula this will give you 2.938vdc where the 33K and 47K connect. To get 2.8vdc you would need a 42K resistor.

There is a great voltage divider calculator here: https://www.raltron.com/cust/tools/voltage_divider.asp

Voltage dividers are great for voltage references, but they can not be used to drive anything. This is because the resistors restrict current to virtually nothing. Even trying to pull 1mA you'll pull your voltage way down!

-------------
Kevin Pierson


Print Page | Close Window