Ohm's Law - Revision history

Jklingert: /* Ohm's Law */

2013-08-22T17:45:06Z

Ohm's Law

← Older revision		Revision as of 17:45, 22 August 2013
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	==Ohm's Law==		==Ohm's Law==
	[http://en.wikipedia.org/wiki/Ohm's_law Ohm's Law] is a formula that relates the basic electrical properties of [[Ohm%27s_Law#Voltage_.28V.29\|Voltage]], [[Ohm%27s_Law#Current_.28I.29\|Current]] and [[Ohm%27s_Law#Resistance_.28R.29\|Resistance]]. These properties can be measured with a [http://en.wikipedia.org/wiki/Multimeter multimeter].[[File:Multimeter.jpg\|100px]]<br/>		[http://en.wikipedia.org/wiki/Ohm's_law Ohm's Law] is a formula that relates the basic electrical properties of [[Ohm%27s_Law#Voltage_.28V.29\|Voltage]], [[Ohm%27s_Law#Current_.28I.29\|Current]] and [[Ohm%27s_Law#Resistance_.28R.29\|Resistance]]. These properties can be measured with a [http://en.wikipedia.org/wiki/Multimeter multimeter].<br/> [https://learn.sparkfun.com/tutorials/how-to-use-a-multimeter How to use a multimeter]<br/> [[File:Multimeter.jpg\|100px]]<br/>
	The common formula for Ohm's Law is:<br/>		The common formula for Ohm's Law is:<br/>
	:::'''<big>V=IR</big>'''<br/>		:::'''<big>V=IR</big>'''<br/>

Jklingert: /* Disclaimers */

2013-08-05T12:35:09Z

Disclaimers

← Older revision		Revision as of 12:35, 5 August 2013
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	==Disclaimers==		==Disclaimers==
	The standard disclaimers pertaining to the information contained on this wiki page are listed [[Disclaimers \| here.]]<br/>		The standard disclaimers pertaining to the information contained on this wiki page are listed [[Disclaimers \| here.]]<br/>
	~~'''THIS PAGE IS UNDER CONSTRUCTION AND IS NOT COMPLETE AND HAS NOT BEEN CHECKED FOR ERRORS YET!!'''<br/>~~
	~~<br/>~~

	==Ohm's Law==		==Ohm's Law==

Jklingert: /* LED Resistor Example */

2013-06-27T19:48:56Z

LED Resistor Example

← Older revision		Revision as of 19:48, 27 June 2013
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	===LED Resistor Example===		===LED Resistor Example===
	#What resistor would you use with a 5VDC power supply and a Blue 5mm LED with a V<sub>F</sub> of 3.0V?<br/> First you need to calculate the voltage that the resistor needs to drop. That is equal to <br/>V<sub>Drop</sub>=V<sub>PowerSupply</sub> - V<sub>F</sub>=5-3=2V<br/>The table above ~~recomends~~ a current I of 20 ma for a 5mm LED. The formula for resistance gives us:<br/>R=V/I=2/0.02=100Ω Generally you choose a resistor for that value or the next higher standard resistor value.<br/> To calculate the wattage resistor needed you use the formula for Power:<br/>P=VI=(2)(0.02)= 0.04W. You would use the next standard size up resistor, 1/8 W.<br/>		#What resistor would you use with a 5VDC power supply and a Blue 5mm LED with a V<sub>F</sub> of 3.0V?<br/> First you need to calculate the voltage that the resistor needs to drop. That is equal to <br/>V<sub>Drop</sub>=V<sub>PowerSupply</sub> - V<sub>F</sub>=5-3=2V<br/>The table above recommends a current I of 20 ma for a 5mm LED. The formula for resistance gives us:<br/>R=V/I=2/0.02=100Ω Generally you choose a resistor for that value or the next higher standard resistor value.<br/> To calculate the wattage resistor needed you use the formula for Power:<br/>P=VI=(2)(0.02)= 0.04W. You would use the next standard size up resistor, 1/8 W.<br/>
	#What resistor would you use with a 5VDC power supply and an optoisolator with a V<sub>F</sub> of 1.2V?<br/> First you need to calculate the voltage that the resistor needs to drop. That is equal to <br/>V<sub>Drop</sub>=V<sub>PowerSupply</sub> - V<sub>F</sub>=5-1.2=3.8V<br/>The table above ~~recomends~~ a current I of 6 ma for an optoisolator. The formula for resistance gives us:<br/>R=V/I=3.8/0.006=633Ω Generally you choose a resistor for that value or the next higher standard resistor value, 680Ω.<br/> To calculate the wattage resistor needed you use the formula for Power:<br/>P=VI=(3.8)(0.006)= 0.02W. You would use the next standard size up resistor, 1/8 W.<br/>		#What resistor would you use with a 5VDC power supply and an optoisolator with a V<sub>F</sub> of 1.2V?<br/> First you need to calculate the voltage that the resistor needs to drop. That is equal to <br/>V<sub>Drop</sub>=V<sub>PowerSupply</sub> - V<sub>F</sub>=5-1.2=3.8V<br/>The table above recommends a current I of 6 ma for an optoisolator. The formula for resistance gives us:<br/>R=V/I=3.8/0.006=633Ω Generally you choose a resistor for that value or the next higher standard resistor value, 680Ω.<br/> To calculate the wattage resistor needed you use the formula for Power:<br/>P=VI=(3.8)(0.006)= 0.02W. You would use the next standard size up resistor, 1/8 W.<br/>

	===Single LED Calculator===		===Single LED Calculator===

Jklingert: /* Voltage Drop Example */

2013-06-25T04:21:58Z

Voltage Drop Example

← Older revision		Revision as of 04:21, 25 June 2013
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	===Voltage Drop Example===		===Voltage Drop Example===
	# What is the voltage drop when using 5V and drawing 500ma over a 10M pair of Cat5 wire?<br/>Since Cat5 wire has a loop resistance of about 0.19 Ω/M, the 10 M length will have a resistance of (10)(0.19)=1.9 Ω<br/> By Ohm's Law we know that V=IR, then<br/>V=IR=(0.5)(1.9)=0.95 V drop, so only 4.05 V will be at the far end of the wire.<br/>		# What is the voltage drop when using 5V and drawing 500ma over a 10M pair of Cat5 wire?<br/>Since Cat5 wire has a loop resistance of about 0.19 Ω/M, the 10 M length will have a resistance of (10)(0.19)=1.9 Ω<br/> By Ohm's Law we know that V=IR, then<br/>V=IR=(0.5)(1.9)=0.95 V drop, so only 4.05 V will be at the far end of the wire.<br/>
			#What is the voltage drop when using 5V and drawing 30A over a 2ft length of 18ga wire?<br/> Since 18 ga wire has a loop resistance of about 0.013 Ω/ft, the 2ft wire will have a resistance of (2)(0.012) = 0.024 Ω<br/> By Ohm's Law we know that V=IR, then<br/>V=IR=(30)(0.024)=0.72 V drop, so only 4.28 V will be at the far end of the wire. The wire will get hot because it is dissipating P=VI=(0.72)(30)= 21W! You must use larger wire for high currents!<br/>

	This is a [http://doityourselfchristmas.com/forums/showthread.php?20242-New-tools-for-estimating-pixels-string-voltage-drops thread] that discusses voltage drop.<br/>		This is a [http://doityourselfchristmas.com/forums/showthread.php?20242-New-tools-for-estimating-pixels-string-voltage-drops thread] that discusses voltage drop.<br/>

Jklingert: /* Current Examples */

2013-06-25T04:04:36Z

Current Examples

← Older revision		Revision as of 04:04, 25 June 2013
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	#How much current is used by a 75W floodlight using 120V?<br/>I=P/V, I=75/120=0.625A		#How much current is used by a 75W floodlight using 120V?<br/>I=P/V, I=75/120=0.625A
	#What is the current thru a 5W lamp with a resistance of 2 Ω?<br/>V=√<span style="text-decoration: overline">P</span>/R=√<span style="text-decoration: overline">(5)</span>/2=2.24/2=1.12A		#What is the current thru a 5W lamp with a resistance of 2 Ω?<br/>V=√<span style="text-decoration: overline">P</span>/R=√<span style="text-decoration: overline">(5)</span>/2=2.24/2=1.12A
			#How much current is required by 1 RGB pixel that draws 20ma (0.02A) each per color?<br/>I<sub>Parallel</sub>= I<sub>1</sub>+I<sub>2</sub>+I<sub>3</sub>=0.02+0.02+0.02=0.06A, or 60mma
			#How much current is required by 4 RGB pixels that draw 60ma (0.06A) each when lit white?<br/>I<sub>Parallel</sub>= I<sub>1</sub>+I<sub>2</sub>+I<sub>3</sub>+I<sub>4</sub>=0.06+0.06+0.06+0.06=0.24A, or 240 ma

	==Resistance '''(R)'''==		==Resistance '''(R)'''==

Jklingert: /* Current Formulas */

2013-06-25T03:58:57Z

Current Formulas

← Older revision		Revision as of 03:58, 25 June 2013
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	*'''I=P/V'''		*'''I=P/V'''
	*'''I=√<span style="text-decoration: overline">P</span>/R'''		*'''I=√<span style="text-decoration: overline">P</span>/R'''
			*'''I<sub>Parallel</sub>= I<sub>1</sub>+I<sub>2</sub>+I<sub>3</sub>+...I<sub>n</sub>'''

	===Current Examples===		===Current Examples===
	#How much current flows thru a 680 Ω resistor that has 5V applied to it?<br/>I=V/R, I=5/680=0.007A (7 ma)		#How much current flows thru a 680 Ω resistor that has 5V applied to it?<br/>I=V/R, I=5/680=0.007A (7 ma)

Jklingert: /* Voltage Drop Example */

2013-01-26T00:09:54Z

Voltage Drop Example

← Older revision		Revision as of 00:09, 26 January 2013
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	<br/>		<br/>

			[[Category:Beginners Info]]
	[[Category:DIYC Index]]		[[Category:DIYC Index]]
	[[Category:General Info]]		[[Category:General Info]]
	[[Category:Pixel]]		[[Category:Pixel]]

Jklingert: /* Voltage Drop Example */

2012-11-26T16:11:05Z

Voltage Drop Example

← Older revision		Revision as of 16:11, 26 November 2012
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	[[Category:DIYC Index]]		[[Category:DIYC Index]]
	[[Category:General Info]]		[[Category:General Info]]
			[[Category:Pixel]]

Jklingert: /* LED Resistor Example */

2012-11-20T02:35:15Z

LED Resistor Example

← Older revision		Revision as of 02:35, 20 November 2012
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	===LED Resistor Example===		===LED Resistor Example===
	#What resistor would you use with a 5VDC power supply and a Blue 5mm LED with a V<sub>F</sub> of 3.0V?<br/> First you need to calculate the voltage that the resistor needs to drop. That is equal to <br/>V<sub>Drop</sub>=V<sub>PowerSupply</sub> - V<sub>F</sub>=5-3=2V<br/>The table above recomends a current I of 20 ma for a 5mm LED. The formula for resistance gives us:<br/>R=V/I=2/0.02=100Ω Generally you choose a resistor for that value or the next higher standard resistor value.<br/> To calculate the wattage resistor needed you use the formula for Power:<br/>P=VI=(2)(0.02)= 0.04W. You would use the next standard size up resistor, 1/8 W.<br/>		#What resistor would you use with a 5VDC power supply and a Blue 5mm LED with a V<sub>F</sub> of 3.0V?<br/> First you need to calculate the voltage that the resistor needs to drop. That is equal to <br/>V<sub>Drop</sub>=V<sub>PowerSupply</sub> - V<sub>F</sub>=5-3=2V<br/>The table above recomends a current I of 20 ma for a 5mm LED. The formula for resistance gives us:<br/>R=V/I=2/0.02=100Ω Generally you choose a resistor for that value or the next higher standard resistor value.<br/> To calculate the wattage resistor needed you use the formula for Power:<br/>P=VI=(2)(0.02)= 0.04W. You would use the next standard size up resistor, 1/8 W.<br/>
			#What resistor would you use with a 5VDC power supply and an optoisolator with a V<sub>F</sub> of 1.2V?<br/> First you need to calculate the voltage that the resistor needs to drop. That is equal to <br/>V<sub>Drop</sub>=V<sub>PowerSupply</sub> - V<sub>F</sub>=5-1.2=3.8V<br/>The table above recomends a current I of 6 ma for an optoisolator. The formula for resistance gives us:<br/>R=V/I=3.8/0.006=633Ω Generally you choose a resistor for that value or the next higher standard resistor value, 680Ω.<br/> To calculate the wattage resistor needed you use the formula for Power:<br/>P=VI=(3.8)(0.006)= 0.02W. You would use the next standard size up resistor, 1/8 W.<br/>

	===Single LED Calculator===		===Single LED Calculator===

Jklingert: /* LED Resistor Calculator */

2012-11-19T03:51:33Z

LED Resistor Calculator

← Older revision		Revision as of 03:51, 19 November 2012
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	<table border="1">		<table border="1">
	<tr><td>LED Color</td><td> Typical V<sub>F</sub></td></tr>		<tr><td>LED Color</td><td> Typical V<sub>F</sub></td></tr>
	<tr><td>Optoisolator</td><td>1.25 V</td></tr>		<tr><td>Optoisolator</td><td>1.2-1.6 V</td></tr>
	<tr><td>Red</td><td>1.9-2.3 V</td></tr>		<tr><td>Red</td><td>1.9-2.3 V</td></tr>
	<tr><td>Green</td><td>2.9-3.5 V</td></tr>		<tr><td>Green</td><td>2.9-3.5 V</td></tr>