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### Author Topic: \$50 robot question  (Read 1245 times)

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#### I Am Blashyrkh

• Jr. Member
• Posts: 18
##### \$50 robot question
« on: April 14, 2009, 04:01:56 PM »
So I was going to make the photoresistors for my project but I have a couple questions.
1. My multimeter sucks and I don't really understand the part of measuring for the resistance but in the tutorial he uses a 1.5 kohm resistor. If I were to use a 1 kohm resistor what would it change? would it be more or less sensitive?

2. He doesn't mention in the instructions but when I was looking at resistors there are 1/8 watt, 1/4 watt, 1/2 watt etc. resistors. Which one was used in the tutorial? or does it even matter?

I apologize for the stupid questions.

#### SmAsH

• Supreme Robot
• Posts: 3,959
• SoR's Locale Electronics Nut.
##### Re: \$50 robot question
« Reply #1 on: April 14, 2009, 04:31:45 PM »
the resistors are 1/4 watt, if you change the resistance values it will vary your data outputted but you can always compensate by changing something in the code. hence why some people like putting pots with their sensors to tweak certain elements. i mean hell i even used 1 or 2 kohms off when i made my \$50 robot sensors
« Last Edit: April 14, 2009, 04:39:51 PM by SmAsH »
Howdy

#### GearMotion

• Supreme Robot
• Posts: 489
• Two decades+ of Embedded Design
##### Re: \$50 robot question
« Reply #2 on: April 15, 2009, 09:28:45 AM »
Photoresistors pass a low current, so a 1/4 W resistor is fine. Even 1/10 W is OK.

Photoresistors have a resistance specified when completely dark and when brightly illuminated. If a cell has 10k ohm when brightly lit, and you have a 10k ohm series resistor, then at the connection between the two components your voltage will be half of the voltage you applied to the circuit. As it gets darker and the photoresistor resistance increases, that voltage will change.

It is helpful to pick components that provide as large of a readable change (analog input) as possible.