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Just add 100nF cap beween positive and negative leads of Your sensor.
So, a 100nF instead of the 10uF from datasheet is better?
That's all? Only 100nF cap between Vcc and Gnd and I solve all the false reading?
Does this make problem to GP2Y0A02YK0F reading speed?
I do hope You use regulated voltage and not 8x 1.2V = 9.6V as MAX sensor supply voltage is 7V and recommended one is 4.5V to 5.5V.
Could you please tell me how to give 5.5V to my sensor out of 9.6V I give to Arduino?
Quote from: GS88 on October 21, 2011, 02:19:45 AMCould you please tell me how to give 5.5V to my sensor out of 9.6V I give to Arduino?So You were running Your sensor on 9.6V You must use voltage regulator such as LM7805.
How do You know that Your IR Sharp sensor gives false readings if You did not plug it in?
What do you think about something similar to use instead of LM7805? http://en.wikipedia.org/wiki/Voltage_divider
Going to learn how to connect those LM7805 to my configuration...
Does this LM7805 have problem giving much than 150mA to the output?
The sharp ir sensor has peak around 220mA.
Quote from: GS88 on October 21, 2011, 07:13:27 AMGoing to learn how to connect those LM7805 to my configuration...Always analyze datasheet as it tells/shows You everything about the device. LM7805 datasheet shows You the following:What it means is:Battery pack +terminal goes to INPUT pin of LM7805Battery pack -terminal goes to COMMON pin of LM7805IR Sharp positive wire goes to OUTPUT pin of LM7805IR Sharp negative wire goes to COMMON pin of LM7805IR Sharp signal wire goes to ADC pin of Arduino
Again, datasheet is Your best friend. As it states at the top - regulator can handle 1A.
How do You know? Datasheet says: "Consumption current : Typ. 33 mA [...] Max. 50mA".
Searching in google the 33mA is the average value, but it has a peak power consumption of about 200 mA:http://www.emartee.com/product/42073/Arduino%20Sharp%20IR%20Sensor%20%20%20GP2Y0A21YK0FAren't they right?
You want the pulses to look like this: ______ | | | | | | | |_| |_Not like this: _ |\ | \ | \ | \_| |_
I'm going out to buy some capacitors and a LM7805 to give it a try
Note 2: The maximum allowable power dissipation at any ambient temperature is a function of the maximum junction temperature for operation (TJMAX = 125°C or 150°C), the junction-to-ambient thermal resistance (θJA), and the ambient temperature (TA). PDMAX = (TJMAX − TA)/θJA. If this dissipation is exceeded, the dietemperature will rise above TJMAX and the electrical specifications do not apply. If the die temperature rises above 150°C, the device will go into thermal shutdown. For the TO-220 package (T), θJA is 54°C/W and θJC is 4°C/W.
I'm trying the circuit and it seem work fine. In input line I have 8 1.2V rechargeable battery and 5V on output. I'm using 10uF capacitor for input line and a 0.1uF capacitor for output line.
I don't have a fuse didn't buy it this morning.
Could you please explain while I have to swap the two capacitors? Or maybe could you let me know how to calculate right capacitors value base don input and output needed voltage?
Unfortunatly no luck 5V output but spikes remain as you can see from serial output. Any hints?236236228 <--238237[Snip lots of numbers]238237243 <--