Joules's and Ohm's laws - Page 1
Forum section: electronics / beginners

author: Goran
author_rank: Contributor
timestamp: May 02, 2015, 11:46:46 pm
content: Can someone clarify some things to me about the Ohm's and Joule's laws? Here is a simple scenario:Power source: 12VDC, LED bulb: 12VDC 6W.By using the Joule's law, we calculate the current I = P/U = 0.42A. Using Ohm's law we calculate resistance R = U/I = 28.8Ohms.Lets say that at some point voltage starts dropping, and its current value is 11.5V.Q1: Is it safe to assume that the resistance is always the same, so if the voltage drops, then the current will drop also, and inevitably, the power will drop too?Q2: In case of above mentioned LED bulb, does it mean that if we connect it to power source of 12VDC, it will draw 1A current, as per manufacturer specification, or the actual value can be different? Q3: If the answer to the above two is yes, does this apply to any consumer, or it can behave differently for different consumers?Thanks.

author: IanB
author_rank: Super Contributor
timestamp: May 02, 2015, 11:54:09 pm
content: Firstly, there seem to be some issues with your arithmetic.If you had a 12 V, 6 W lamp, then the current would be 12/6 = 0.5 A. (How do you arrive at 0.42 A ?)Secondly, how do you then decide it will draw 1 A of current? If it did, this would make it a 12 W lamp, not a 6 W lamp.As to your main question, no, you cannot assume the resistance will remain constant. Electric lamps usually have a complex relationship between voltage and current. It is very hard to generalize about what will happen without testing a specific device and finding out.Normally when you decrease the voltage the current will also decrease, but it may not decrease in a proportional way.

author: flynwill
author_rank: Regular Contributor
timestamp: May 03, 2015, 12:08:16 am
content: Quote from: Goran on May 02, 2015, 11:46:46 pmWell 6W/12V = 0.5 A, not sure where you got 0.42A from.QuoteFor your example of an LED bulb the answer is "probably", You might find the current is constant with dropping voltage, (the power consumed will drop proportional to the voltage). For a simple resistive load the current will drop proportional to the voltage, so the powerwill fall in proportion to the voltage squared.QuoteWell there will be some tolerance in that, individual units may draw more or less, in the case of LED bulbs possibly by as much as 20%.QuoteI'm afraid I'm not sure what you mean by "consumer". If you're talking about electronic loads there are all sorts of possibilities. If, for example, you are powering a device which has a high efficiency power supply you will find that the current rises as you lower the voltage such that the power consumed is constant. Some loads might even draw more power at lower voltage (rare).

author: sleemanj
author_rank: Super Contributor
timestamp: May 03, 2015, 01:12:35 am
content: I get the feeling that your actual question might be along the lines of"I have a '6W 12V LED', can I connect it directly to a 12V power supply"the answer to that question is no that is not advisable if it is just a bare LED, if it's an LED with driver circuitry, then yes.LEDs should be provided with a constant current, not a constant voltage.

author: Goran
author_rank: Contributor
timestamp: May 03, 2015, 01:22:09 am
content: Thanks for the responses...Sorry for that calculation mistake, the numbers where from different scenarios, my mistake. When I wrote 1A, this is because in one real word scenario I will be having 2x 6W LEDs in parallel, so that means 1A total. As for the 0.42A, that was from another real world scenario, where the bulb is actually 5W. Nevertheless, there is no confusion about arithmetic.So, for anything than a "simple resistive load", it is not safe to apply Ohms law, since when the voltage changes, resistance could also change, and therefore we have two unknown variables? And by "simple resistive load" you mean - resistors, which have constant resistance? For everything else, we should not be assuming that resistance will be constant?

author: Goran
author_rank: Contributor
timestamp: May 03, 2015, 01:24:46 am
content: Quote from: sleemanj on May 03, 2015, 01:12:35 amMissed this one: well, the LEDs I will be using do have a driver. but your information is very welcomed. I have seen many examples where people replace halogen bulb with small led bulb (without a driver) and just add resistor for difference in load. So this approach is not correct?

author: sleemanj
author_rank: Super Contributor
timestamp: May 03, 2015, 01:29:58 am
content: Quote from: Goran on May 03, 2015, 01:24:46 amA resistor works to limit the current, which is fine, provided1. the supply voltage is relatively constant2. you don't mind that the extra power is just burnt off as heat in the resistor

author: Goran
author_rank: Contributor
timestamp: May 03, 2015, 01:42:07 am
content: Quote from: sleemanj2. And if I do mind, what would be the alternative?

author: flynwill
author_rank: Regular Contributor
timestamp: May 03, 2015, 04:09:03 am
content: If you've got a pre-packaged LED bulb that says it expects 12V then I would presume that it has some sort of current limiting built into it, probably just a series resistor.If you do mind the extra power being wasted in the resistor the solution is a switching circuit of some sort. The simplest is a "buck converter" (google that) and a series inductor setup to provide the proper operating current to the LEDs. There will still be some power "wasted" but this scheme can waste less over a wider range of supply voltages than just a resistor.

author: sleemanj
author_rank: Super Contributor
timestamp: May 03, 2015, 04:44:23 am
content: Quote from: Goran on May 03, 2015, 01:42:07 amA DC-DC converter designed to operate as a constant current supply.

author: fubar.gr
author_rank: Unknown
timestamp: May 04, 2015, 06:36:17 pm
content: Quote from: sleemanj on May 03, 2015, 01:29:58 amA series resistor also makes the LED's VI curve shallower. This allows for quite a lot of input voltage variation, even going twice as much, without damaging the LED (at least not instantly).