Calculate voltage drop to select the right resistor? - Page 1
Forum section: electronics / beginners

author: expertmax1
author_rank: Contributor
timestamp: April 11, 2013, 04:14:52 pm
content: Hi, I would love some help !I want a formula to calculate the R value of R1 and the W value of R1.Here is a picture :I think the formula to get the wattage is something like I squared times R... but I can't remember !Thank you so much, have a nice day.

author: c4757p
author_rank: Super Contributor
timestamp: April 11, 2013, 04:20:47 pm
content: Is this homework?

author: casper.bang
author_rank: Frequent Contributor
timestamp: April 11, 2013, 04:23:44 pm
content: Looks like very basic homework assignment you can easily Google. But Ok I'll bite: If there is 3.3V across the load and the reference is 5V, then there must be 1.7V (5V-3.3V) across the resistor. Since there is 0.5A current flowing though the circuit, the effect dissipated into the resistor must be 0.85W (1.7V * 0.5A).

author: Simon
author_rank: Unknown
timestamp: April 11, 2013, 04:24:21 pm
content: Quote from: c4757p on April 11, 2013, 04:20:47 pmGood point.To the original poster: go look up ohms law. V/I = R that is all you need

author: expertmax1
author_rank: Contributor
timestamp: April 11, 2013, 04:37:53 pm
content: My question is not only power dissipation but I want to know the R value of R1 (ohms) if I want a 3.3V output under a 0.5A load. This is not homework, just a little project for myself.

author: casper.bang
author_rank: Frequent Contributor
timestamp: April 11, 2013, 04:43:29 pm
content: Quote from: expertmax1 on April 11, 2013, 04:37:53 pmAs others have stated, this is ohms law - one of the most basic laws of electronics. In this case, R = U/I = 1.7V/0.5A = 3.4ohm.

author: mariush
author_rank: Super Contributor
timestamp: April 11, 2013, 04:59:33 pm
content: I like to write Ohm's law as V = IxR ... short for virus (which I remember because I work with computers daily.From that formula you can determine anything... I = V / R , R = V / I.In a series circuit, the current is the same, so if you want the first load to be limited to 0.5A, you want the resistor to allow 0.5 A, so now your formula is V = 0.5 x RThe voltage is the voltage of your supply, minus the voltage of the load, so it's 5v-3.3v = 1.7vTherefore, R = V / I becomes 1.7v / 0.5a = 3.4 ohm.The power dissipated by the resistor is calculated with the formula P = I x I x R = 0.85 watts, so you need to use a resistor that's rated for at least 1 watt, but ideally you should use a 2w or 3w resistor, at least.

author: MasterOfNone
author_rank: Regular Contributor
timestamp: April 11, 2013, 05:26:16 pm
content: Quote from: expertmax1 on April 11, 2013, 04:37:53 pmDon’t do it!If you are thinking of powering a device rated at 3.3V 0.5A and you are going use a Resistor to drop the voltage down from 5V then don’t do it.Depending on what you are powering, it is likely that it won’t always draw 0.5A, that value is probably the Maximum current.When the 3.3V device is drawing less current the voltage drop across the resistor will be less (V=IxR see above). However since the 5V supply/battery is fixed the voltage across your device will increase, because the total voltages across your device and the resister must still add up to 5V.What you need to do give us more info on what you are trying to do and/or Google “5V to 3.3v converter” “5V to 3.3v regulator” “5V to 3.3v psu” etc until you see something that will fit your needs.

author: mariush
author_rank: Super Contributor
timestamp: April 11, 2013, 05:39:39 pm
content: I agree. Even a low drop linear voltage regulator will be better than a simple resistor.Here's a very simple IC that's a linear regulator which outputs a fixed 3.3v as long as it receives at least 1-1.2v above that value ( so 3.3v + 1.2v = 4.6v or more) :http://uk.farnell.com/stmicroelectronics/ld1117v33c/ic-voltage-regulator/dp/1703357http://uk.farnell.com/stmicroelectronics/ld1117v33/ic-v-reg-ldo-3-3v-1117-to-220-3/dp/9755837http://uk.farnell.com/stmicroelectronics/ld1117av33/ic-v-reg-ldo-3-3v-1117-to-220-3/dp/1087165The 1117 chip is a very generic chip made by various companies, and it's made in several fixed output voltage versions (usually those have the voltage at the end, -33 for 3.3v, -50 for 5v and so on) or as an adjustable version, on which you can use 2 resistors to set the exact voltage you want.For 0.5A current, it wouldn't hurt to use a tiny heatsink with it, for example something like this:http://uk.farnell.com/aavid-thermalloy/577102b00000g/heat-sink-to-220-25-9c-w/dp/1703172http://uk.farnell.com/aavid-thermalloy/507302b00000g/heatsink-to-220/dp/1611415http://uk.farnell.com/wakefield-solutions/273-ab/heat-sink-to-218-to-220/dp/1700005Once you've learned more about circuits, you may want to search for dc-dc converters or to use a switching regulator which works a bit differently and may have better efficiency, therefore your device will have more battery life.But it really depends on how your device uses power, if it only occasionally peaks at 0.5A, switching regulators or other types of dc-dc converters may have so small efficiency improvements over these linear regulators that it wouldn't be worth using those more expensive solutions.

author: bitwelder
author_rank: Super Contributor
timestamp: April 11, 2013, 09:16:18 pm
content: Quote from: expertmax1 on April 11, 2013, 04:14:52 pmAnd if you don't remember a formula (and you can't ask to EEVblog forum), how would you find out which one is it?

author: expertmax1
author_rank: Contributor
timestamp: April 12, 2013, 04:40:23 am
content: It's not to power anything, I just wanted the formula. The real use is to power a LED (0.08A 5V) from a 12V power supply. So R1 would be 87.5 ohm and 0.56 watts. Thanks.

author: smashedProton
author_rank: Frequent Contributor
timestamp: April 12, 2013, 02:31:32 pm
content: This is just ohms law. I sugguest that you read your textbook. Or you should get a humanities major

author: Excavatoree
author_rank: Frequent Contributor
timestamp: April 12, 2013, 03:11:20 pm
content: Quote from: expertmax1 on April 12, 2013, 04:40:23 amI'll embarrass myself by saying that doesn't look correct. If you "ballpark" .6 volts for the diode, that's 11.4 volts divided by 87.5, which gives .137 amperes, not .08. (If you use .7, that's still 11.3/87.5 = .129 amperes.)Shouldn't it be:R = 11.4/.08 which gives 142.5, then select the closest standard value?

author: Excavatoree
author_rank: Frequent Contributor
timestamp: April 12, 2013, 03:14:11 pm
content: The resistor is dropping 11.4 volts (approximately) so 11.4 * .08 amperes is .9 watts?I shouldn't have done this. If I made a silly mistake I've embarrassed myself in front of everyone on a High school electrical class problem.EDIT: D'oh!!! Thanks to the poster below for seeing my mistake.

author: hlavac
author_rank: Frequent Contributor
timestamp: April 12, 2013, 03:46:27 pm
content: 0.6V looks more like a schottky diode than LED - LEDs have much higher voltage drop (~1.8V for red, ~2.1 for orange and green, 3V or more for blue and white)

author: Excavatoree
author_rank: Frequent Contributor
timestamp: April 12, 2013, 03:56:56 pm
content: Ah, I KNEW I was messing up somewhere. I just couldn't realize it. It must be too many meetings this morning. Thanks.

author: mariush
author_rank: Super Contributor
timestamp: April 12, 2013, 04:01:31 pm
content: Quote from: Excavatoree on April 12, 2013, 03:11:20 pmA LED's forward voltage is between 1.8v and around 3.4v and the basic LEDs are usually rated for up to 10mA - but you'll find out that at 10mA some LEDs are very bright. You should aim for about 3-5mA at best.But anyway, this below is assuming the LED needs 2.5v and 10mAV = I x R5v (supply) - 2.5v (led voltage) = 0.01 (10mA) x R => R = V / I = 2.5v / 0.01 = 250 ohm (this is not a standard E12 or E24 value, so you go for 240ohm or 270 ohm which would give you 0.0104A (for 240ohm) or 0.009A for 270 ohm).Power consumption is P = IxIxR = 0.01 x 0.01 x 240 ohm = 0.024 watts, so you can use a 1/8w resistor (0.125w).For a 5v 0.08A LED, you have 12v-5v = 0.08 x R => R = 87.5 ohm , so you'd probably want to go with 82 or 91 ohm (E24 values). Power then is 0.08 x0.08 x 91 = 0.58w so you use a 1w resistor.

author: expertmax1
author_rank: Contributor
timestamp: April 13, 2013, 02:34:04 am
content: Thanks for the help ! But I just can't understand why everyone thinks that new people who posts on this forum are all high school students who just are too lazy to read a textbook. I work as a programmer and I enjoy electronics but as a hobby.