Regulator for Constant Voltage, Dynamic Current? - Page 1
Forum section: electronics / beginners

author: SEnergy
author_rank: Contributor
timestamp: August 25, 2015, 08:38:20 pm
content: Hello,if I understood regulators correctly, they give constant Voltage & Current output, right?so my question is if there are any regulators, that have constant voltage (for example 12V), but have dynamic current based on the power input?e.g. input of 5V@2A would give ~12V@0.83A (10W)however the same regulator at 30V@5A would give ~12V@12.5A (150W)does something like that exist?Thanks!

author: Zero999
author_rank: Super Contributor
timestamp: August 25, 2015, 08:49:54 pm
content: No the current always depends on the load voltage and impedance. A 5V regulator will give 0A without anything connected to it and 1A with a 5 Ohm load connected to it.With a constant voltage regulator, the current rating is the maximum current which can be drawn before it overheats or limits the current to protect itself.I think you may be confusing some bench power supplies which specify a maximum power output, say 100W which can provide 10A at 10V or 5A at 20V, so long as V*A is less than 100.

author: retrolefty
author_rank: Super Contributor
timestamp: August 25, 2015, 08:56:18 pm
content: Quote from: SEnergy on August 25, 2015, 08:38:20 pmNo, a DC regulator can operate as a constant specified voltage or a constant specified current but not at the same time as that would violate Ohms Law with a constant specific load. Some 'lab' or 'bench' supplies can switch from one mode to the other but again not both. There is also a mode that has constant voltage output with a specific maximum current limit that will lower the voltage to maintain the current to the max or less.The main point to understand is that the power supply does not 'force' a given current into the attached load. The load's resistance determines the amount of current drawn from the supply by the Ohms Law calculation of current = voltage / resistance. Study basic Ohms law until you get a clear understanding of that law. A lot of understanding electronic circuits is to learn to instinctively think in Ohms Law.

author: SEnergy
author_rank: Contributor
timestamp: August 25, 2015, 09:05:08 pm
content: Quote from: Hero999 on August 25, 2015, 08:49:54 pmI'm asking this because I'm getting more into electronics every day and I've found some interesting videos on youtube and realized how easy it is to make a wind turbine, so I want to try making one, just for the sake of doing something usefulusing that turbine I want to charge 12V 10Ah battery, so I need about 12V under basically any wind conditions, transforming rest power to current... ppl were getting ~30V at 25km/h wind, which is pretty slow where I live, I have no idea about their current ratings thohow would something like that work? is something like that possible? what should I look for?

author: SEnergy
author_rank: Contributor
timestamp: August 25, 2015, 09:08:36 pm
content: Quote from: retrolefty on August 25, 2015, 08:56:18 pmI have no idea what you just said, it basically has nothing to do with what I asked and I understand Ohm's Law as good as I ever needed so far

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 25, 2015, 09:25:16 pm
content: Quote from: SEnergy on August 25, 2015, 08:38:20 pmYour earlier question begs to differ...Quote from: SEnergy on August 25, 2015, 08:38:20 pmAs SEnergy was saying, regulators do NOT put out constant voltage and current, because the current is decided by the load and it would be impossible for the regulator to control both voltage and current with a variable load, that would violate Ohm's Law. You also can't have a regulator that outputs a constant voltage and adjusts its output current based on power input, because again, the regulator cannot control both its output voltage and its output current...that would violate Ohm's Law.What you're looking for is an MPPT controller most likely. This is a more or less standard switching DC/DC converter with an adjustable output voltage that's controlled by a microcontroller. The microcontroller measures the input voltage and current, and adjusts the DC/DC converter's output voltage (which also changes the output current based on the load's properties via Ohm's Law) to maximize the power.You usually see MPPT controllers for solar installations, but AFAIK they have ones designed for wind as well.

author: Hideki
author_rank: Frequent Contributor
timestamp: August 25, 2015, 09:40:35 pm
content: Note that you need more than 12V to charge a "12V" lead acid battery. More like 14V plus/minus some depending on the type and usage etc.

author: SEnergy
author_rank: Contributor
timestamp: August 25, 2015, 09:45:45 pm
content: Quotewhat load will be "decided" when I connect it to the battery?Quotehow would that violate ohm's law? you for example put in 10W (5V@2A), the regulator makes it 2.5V@5A (still 10W), the load will get destroyed or won't work properly or something, but there still is only 10WQuoteinput 24V@1A, this goes to regulator which is connected to a battery, the regulator makes it 12V@2A, 24W at both timeswe put 48V@1A to the same regulator (lets say it is capable of that), the output voltage of that regulator is 12V, so the current becomes 4A, 48W at both timeshow is that violating ohm's law?I'll check some info about those MPPT controllers... do you by any chance have a EU eshop where I could browse some?

author: retrolefty
author_rank: Super Contributor
timestamp: August 25, 2015, 09:54:35 pm
content: QuoteTrust me, your questions can all be answered by a better understanding of Ohms law and its derivative formulas (wattage, etc).Again your ratings are all trying to show what the maximum power/voltage/current source is but the actual load resistance is what tells the source supply how much current it demands, hopefully below the maximum power/current capacity of your source.QuoteDetermined by the difference of potential between the battery and your constant voltage source and the internal equivalent resistance of the battery. This will change as the battery draws current and with a constant voltage regulator the current will slowly drop as the battery charges until voltage source = battery terminal voltage. Again it all can be analyzed using Ohms law. Study it, use it, it will serve you well and long.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 25, 2015, 10:04:07 pm
content: Quote from: SEnergy on August 25, 2015, 09:45:45 pmThe battery IS the load (to be more specific, the battery's ESR is the load). The current that flows out of the regulator will be the output voltage of the regulator minus the battery voltage divided by the ESR of the battery. (Vr - Vbatt) / ESR. The regulator's ESR and the ESR of the wiring connecting them will also contribute, but usually the battery's ESR is dominant.Quote from: SEnergy on August 25, 2015, 09:45:45 pmConservation of energy != Ohm's LawOhm's Law is I=V/R. Current through the load is equal to the voltage dropped across the load divided by the resistance of the load. Neither the source nor the load can control two of these parameters, because then the equation would force the third parameter on the other device, which is impossible. If the regulator were to force both 12V and 1A output, then it would essentially be telling the load, "You are 12 ohms", but the load's resistance is a property of the load, the regulator can't tell the load what its own resistance is...One of the three values is controlled by the source, usually "V". One is controlled by the load, usually "R". The third one (usually "I") is the result of the equation.

author: SEnergy
author_rank: Contributor
timestamp: August 25, 2015, 10:31:01 pm
content: Quote from: suicidaleggroll on August 25, 2015, 10:04:07 pmlet's say I have 12V 10Ah battery and a 78S15 regulator (15V, 2A, I guess those are output LIMITS?)Regulator output resistance is 17 mOhmBattery internal resistance is 20 mOhminput limit for that regulator is 35V, so let's say my wind turbine is producing constant 30V, the output of the regulator will be 15Vwhat should I do to calculate the current going to the battery?2 scenarios that came to my mind, most likely both incorrect;1, the battery "eats" all 15V that it gets from the regulator, right? so Vr-Vbatt = 0, doesn't matter what resistance there is, I got no voltage... what now?2, let's say a battery somehow "eats" only 14V, so I have 1V left, the resistance of battery and regulator is 37 mOhm, I have no idea what resistance is from the wiring, but let's say they're all really close together, let's say the wiring resistance will be 63 mOhms, so total of 100 mOhm, since I have 1V left, 100 mOhm resistance would give me 10A, but since regulator has limit of 2A I'm getting 15V@2A and the rest is dissipated as a heat?I guess both points are absolutely wrong, so how do I correct them?in first case there's no voltage, kind of a dead point, but that seems to be illogical, as battery SHOULD "eat" all voltage - how do I get the current then?the second case would most likely destroy the regulator due to heat, also if there would be no limit for current, it would output 150W, since the voltage input was 30V, the current input logically would be 5A, which seems to be possible, or? therefore I'd be losing 120W as a heat, which is absolutely terrible

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 25, 2015, 10:49:16 pm
content: Quote from: SEnergy on August 25, 2015, 10:31:01 pm15V is its output voltage, 2A is its current limit. It will put out 15V and you can draw up to 2A from it.Quote from: SEnergy on August 25, 2015, 10:31:01 pmYes, the output of the regulator will be 15V as long as the input is anywhere between ~18-35V and the output current is <2A.Quote from: SEnergy on August 25, 2015, 10:31:01 pmI don't know what you mean by "eats"...what is the battery voltage? Batteries put out voltage, so what is it in each case?.In #1 you said "Vr-Vbatt=0", since Vr=15, then you must be saying Vbatt=15? In that case, the regulator is putting out 15V, but the battery is also floating at 15V, so you get 0V differential and no current. In other words, the battery is already charged, at least as much as that regulator is capable of charging it.In #2, you're saying the battery is at 14V, so since the regulator is putting out 15V, you have 1V being used to push current into the battery. Yes if you assume a total ESR of 100m then you'll get a current of 10A. That regulator is only capable of 2A though, so realistically once you passed 2A the short circuit protection in the regulator would kick in and it would drop its output voltage to whatever is required to maintain 500mA (table 9 in the datasheet). Given those numbers, that would be about 14.05V.Also, the 78S15 is a linear voltage regulator, which means its input current is equal to its output current, and any residual power ((Vi-Vo)*I) is burned off internally as heat. So if your input voltage was 30V and the output was 15, it would have an efficiency of 50% and you would fry that regulator with any more than say a half an amp of charging current (7.5W), and even with that you'd need a big heatsink.If you want to EFFICIENTLY change voltages, then you need a switching converter, not a linear.

author: SEnergy
author_rank: Contributor
timestamp: August 26, 2015, 08:24:25 am
content: Quote from: suicidaleggroll on August 25, 2015, 10:49:16 pmwell, I have no idea what a correct term is, but for example when you connect a 3V device to a 5V source, it will "use"/"eat" etc all 5V unless there's something else in the load, like the resistor, right?how does it work with a battery? If I have 12V battery and put 15V in it, what happens? the battery needs more input voltage for charging than it outputs, so will it "use" (as described above) all 15V or will it use the voltage it needs and rest can be used to calculate current?again, if I'd have 15V regulator and battery would use <= 14V from that, then the current would be above the limit, so what should I do in this case?also what would happen if I'd put several regulators in parallel and then connect their outputs together? would that divide the current going thru regulators?

author: Hideki
author_rank: Frequent Contributor
timestamp: August 26, 2015, 01:34:41 pm
content: Quote from: SEnergy on August 26, 2015, 08:24:25 amIt's not that it "eats" or "uses" anything. If we assume that the battery is at exactly 12V when you connect the 15V there will be a voltage difference of 3V.Assuming a perfect voltage source and no other regulation, the current will depend on the resistance of the wires, connectors, internal battey resistance etc.If that's 0.1 ohms the current will be 3V / 0.1ohms = 30A.This current will slowly charge the battery, so some time later the battery voltage will no longer be 12V. Let's say it has increased to 13V. The voltage difference is now 2V so the current will drop from 30A to 20A and charge the battery a bit slower. After even more time the battery will be at 14V and the current will have dropped to 10A. At 14.9V the current is only 1A. So you see that as the battery charges, the voltage rises towards 15V and the current slowly drops. It is very similar to charging up a capacitor.If for example you want a constant current, you have to change the charge voltage as the battery charges. If you make it 0.2V higher than whatever the battery is at, then you will maintain a 0.2V / 0.1ohm = 2A current.Since the actual resistance will be unknown and changing, that's not a good solution. It's better to measure the current and adjust the charge voltage based on that.QuoteIf we're still talking about 78S15 -- yes, it would divide the current, but not very well. There will be small differences in output voltage between the regulators, so they will be fighting each other.A somewhat inefficient solution is to put small resistors in series with each regulator output before you connect them all together, but is this case it's better to use a switching regulator instead. Rated for the maximum charge current you want to use.Edit: At 14.9V the current will be 1A, not 0.1A.

author: SEnergy
author_rank: Contributor
timestamp: August 26, 2015, 02:17:17 pm
content: Quote from: Hideki on August 26, 2015, 01:34:41 pmbut that specific regulator is not capable of delivering 10A, not to mention 30A, so I have to go with a different onethis might be a repeating question, but for total understanding I need a specific example, solet's assume I have a voltage source (wind turbine) that varies from 20-50V, 12V battery, and a L4970A switching adjustable regulator (5.1-40V, 10A) - or more of themdatasheet;http://www.gme.sk/img/cache/doc/332/008/l4970-datasheet-1.pdfI can't find what internal resistance that regulator has, so let's assume we have 0.1ohm resistance excluding the regulatorusing these parts as base for the circuit, what should I do and what else should I buy to make a fully functional circuit that will safely charge the battery, assuming I already have a circuit that will break the connection to the battery when it's fully charged ?

author: kezat
author_rank: Contributor
timestamp: August 26, 2015, 03:16:58 pm
content: I think this is getting overly complicated, what you need is a simple constant current. Set the current to what you need and the voltage will automatically be adjusted to give you that current. Supply or battery voltage will not matter other then then making sure your not dissipating to much power in the constant current transistor/Mosfet.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 26, 2015, 03:18:12 pm
content: Quote from: SEnergy on August 26, 2015, 02:17:17 pmA simple DC/DC converter is not capable of operating correctly off of a dynamic power source like a wind turbine or solar panel. This is why you need a special charge controller, such as an MPPT, to adjust the output voltage (and therefore the output current, and therefore the output power) based on the power available from the source.

author: SEnergy
author_rank: Contributor
timestamp: August 26, 2015, 03:35:02 pm
content: Quote from: suicidaleggroll on August 26, 2015, 03:18:12 pmI've checked MPPTs yesterday, they are big compared to what I want to build, expensive and for solar panels, there were none for wind turbines, also they have special ports for the wiring, making that another obstacle in what was supposed to be a simple wind turbinethis guy uses a cigarette lighter adapter & charger to get stable 5V@1A, I was hoping to do something similar, except I wanted to charge 12V battery with itwould it work if I'd connect the wind turbine to the regulator, which would then output power to the digital voltmeter, which would feed my MC, so I could control the output voltage using transistors and PWM?

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 26, 2015, 03:52:32 pm
content: First off if you're charging a battery with a simple DC/DC converter you need constant current with a voltage limit, not constant voltage with a current limit.The problem with a simple DC/DC converter and no intelligent controller is the set point. You have to pick something...so what do you pick? If you pick a current that's too high you'll get -nothing- from the system, if you pick one that's too low you'll be throwing away power.So say you design it to output 1A up to 14V, that means charging a nominal 12V battery you'll get a nominal 12W out of it. If the wind generator can't provide 12W, you'll get zero power, nada. If the wind generator is capable of providing 30W, you'll only get 12. It's a waste of your wind generator's capabilities. An MPPT would deliver to the battery whatever the generator is capable of providing. If it can only do 2W, you'll get 2W to the battery (minus some small loss due to efficiencies). If it can do 40W, you'll get 40W to the battery, as long as the MPPT controller can handle at least 40W.Many solar MPPTs work with wind as well, but I don't see any at such a low power. You can always design your own with an adjustable output voltage DC/DC converter, microcontroller, current shunt, and some ADCs (usually built into the MCU). There are many papers online that evaluate and compare MPPT algorithms for wind energy. If you aren't up to designing your own, well...many people aren't, that's why they cost money.It doesn't have to be MPPT either, but you do want SOME kind of controller. Unless you enjoy buying/installing a wind generator that's significantly larger than necessary so after going through your simple voltage converter you can get the power you need. You shouldn't be comparing size X wind generator with an MPPT vs size X wind generator with a simple DC/DC converter...yes of course the MPPT will make things bigger and more expensive. You should be comparing size X wind generator with an MPPT to size X*3 wind generator with a simple DC/DC converter, since now you're comparing apples to apples with regards to the amount of power you're actually going to get out of each system. Which is actually bigger and more expensive, the normal size wind generator and MPPT controller, or the wind generator that's 3x more powerful with a simple DC/DC converter that won't work half the time and won't give you as much power as the generator is capable of the other half?

author: SEnergy
author_rank: Contributor
timestamp: August 26, 2015, 04:25:31 pm
content: Quotewhy? if I deliver high current to the battery isn't that just supposed to charge it faster?the current delivered to the regulator doesn't matter, does it? only the output is what matters, or?see, I can't even ask a question about the other statements you made because you just don't explain them at all, especially the "nada" part, like what is that? did your teachers as well just said to you "no, you can't do that, nada" and then sat in silence? why wouldn't I get more than 12W if my wind turbine is capable of delivering 30W? I have a switching regulator that can provide 5.1 - 40V and 10A, that's up to 400W, more than 13times more than 30W, with 12V battery that would require for example 14V at the worst case it still would be 140W as it can output 10AQuotehow? what is so special about MPPT that it can deliver 2W to the battery? it's supposed to push the elctricity to the battery at something above 12V, let's say 13V, so with 2W that would be about 0.15A -> how do I make a 13V@0.15A out of 2V for example?Quoteyou seem to ignore the question I placed above, as I asked about how to do exactly that... I don't need a high efficient algorithm, just a simple one that will keep my system up and running, I actually don't need an algorithm at all, I can do that, I just need to know what parts I need to build this and the understanding of those parts based on EXAMPLES (thus the specific regulator, power source and battery mentioned above) - that's why I came here, I can always read books, papers, online threads about how someone made it, but a static information as that can't answer my questions if I have any nor can teach me anything comparable to what any person on this forum can, THAT is the reason I'm here and THAT is the reason I'm not googling 10 different online papers, as they wouldn't teach me much and my project my end up like an exact copy of something someone else made - not my styleQuote

author: kezat
author_rank: Contributor
timestamp: August 26, 2015, 04:51:14 pm
content: I guess the most simple solution would be a simple diode in between the generator and battery as long as the generator current output is not exceeding the max charge rate of the battery I see no reason why that would not work. However if the goal is learning something new that might not be the most helpful:)

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 26, 2015, 05:42:00 pm
content: Quote from: SEnergy on August 26, 2015, 04:25:31 pmWhen you "pick" a current with a constant current regulator, you're picking a set point. The regulator WILL OUTPUT THAT CURRENT as long as the output voltage required to do so doesn't exceed its limit. If you set the regulator up to output 1A, and the battery is at 12V, that will force the regulator to output somewhere around 12.1V. 12.1V @ 1A is 12.1W. The regulator will use however much current from the source as is necessary to supply that 12.1W plus its own inefficiencies, given the input voltage. If there is more available from the generatorit won't be used. It won't be burned off as heat, it won't go anywhere else, it just won't exist, because the DC/DC converter won't be driving the generator at a V/I combination that provides that much power. If there is LESS THAN 12.1 watts available from the generator, the input voltage to the regulator will drop, the regulator will go into a brownout shut down, and it will deliver nothing to the load.This is the problem with standard voltage regulators, they operate with a fixed set point under the assumption that there is infinite power available from the source. If all of a sudden the source can't provide enough power to supply the regulator and its output, the regulator freaks out and shuts down. THIS IS WHAT THE MPPT CONTROLLER IS FOR. This is what it does, this is the point of it, this is why you use them with wind/solar power. When it detects that the available power from the generator starts to drop, it adjusts the DC/DC controller to provide less power to the load/battery, so that everything stays up and running. When it detects that the available power from the generator increases, it adjusts the DC/DC controller to provide more power to the load/battery. A standalone regulator cannot do either of these. It has a set point. It operates at that set point. If the power source (wind generator) can't provide enough power to keep up, the whole system shuts down. If the power source can provide more power than necessary, the regulator doesn't care, it doesn't use it, because that extra power isn't necessary given its set point and load.QuoteIt's a DC/DC converter with an adjustable output voltage and an microcontroller running an algorithm that tailors that output voltage to optimize/maximize the input power, that's what so special about an MPPT.QuoteExamples aren't for copying, they're for learning from. Look them up, once you understand how they work you'll know how to design your own. That's the kicker, you need tolearn how they work, and to do that you REALLY need to do more reading about Ohm's Law, regulators, etc.We're just going in circles here.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 26, 2015, 05:43:25 pm
content: Quote from: kezat on August 26, 2015, 04:51:14 pmIf the battery's nominal voltage is a relatively close match for the generator's peak power voltage given the current wind conditions, and you add a high power zener or similar to shunt power away from the battery once it's full, yes that will work.But as you can see here, the peak power voltage changes dramatically with wind speed, so a battery voltage that works well at 5 m/s wind speed might be horrible at 10 m/s or vice versa:SEnergy - The point of the MPPT is to continuously "search" this V/W curve to find the peak, and dwell on it. If the wind changes and the peak moves to a different voltage, the MPPT adjusts its output voltage and output power to move with it.

author: SEnergy
author_rank: Contributor
timestamp: August 26, 2015, 06:01:59 pm
content: Quoteyes, we go in the circles here, mainly because when I ask "HOW" a MPPT works, you tell me that it makes 12V out of xV to maintain the system up and running, but that is NOT how it works, that's what it does, I need to know how to make my own MPPT, so I need to know how I can transform 1V@0.5A (for example) to 12V@0.04A, I know how a transformer works, but that works kinda differently and steps up/steps down a voltage Xtimes all the time, it doesn't output SPECIFIC voltageI understand that MPPT takes whatever power there is from the source and transforms it to the whatever voltage is currently needed by the battery based on the internal resistances of battery, MPPT and wires, BUT I do not know HOW that magic happens, THAT is what I want to know, the few articles I found were repeating the same thing as you do -> it makes the power output of power source stable so it charges the battery, but I haven't found how that happens

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 26, 2015, 06:39:21 pm
content: An MPPT is the combination of a more-or-less standard DC/DC switching regulator and an intelligent controller for that regulator that sets its output voltage to whatever is necessary to optimize the power drawn from the source.Are you asking how a DC/DC switching regulator works?https://en.wikipedia.org/wiki/DC-to-DC_converter#Switched-mode_conversionIf you're asking how the controller does its intelligent management, that's what all of those articles you're finding on MPPT control algorithms are about. It measures the input voltage and current, and then iteratively adjusts the DC/DC output voltage to maximize the input I*V.

author: SEnergy
author_rank: Contributor
timestamp: August 26, 2015, 07:00:02 pm
content: Quote from: suicidaleggroll on August 26, 2015, 06:39:21 pmlike this one?http://www.gme.sk/img/cache/doc/332/008/l4970-datasheet-1.pdfor should I look for something different?

author: Hideki
author_rank: Frequent Contributor
timestamp: August 26, 2015, 08:06:58 pm
content: L4970 is better than nothing. There are so many DC/DC converters on the market that it can be hard to make a sensible choice.I think the problem here is this: For a given wind speed you assume that the windmill outputs a certain voltage and current, which when multiplied together gives you a certain power that you need to transform.It doesn't work that way!There is no current without a load, and the load determines how fast the windmill spins and what voltage you get out. The load in this case is the DC/DC converter, which in turn feeds the battery.- If the load is small, current is low, the voltage is high and it spins fast. This gives you a certain power value.- If the load is large, current is high, but it will now spin slower and so the voltage will be lower. This gives you a certain power value too. It could even be the same as before. Load it way too much and might practically stop. You won't get much power out by doing that.If you plot a curve for different loads you will find that the maximum power peaks somewhere between the two extremes. You load it just enough so that you don't slow it down too much, thereby maximizing the combination of current AND voltage. The windmill is not a "constant power" system.What an MPPT controller does is to continually measure the voltage and current and multiply them together. It then uses some algorithm to adjust the output voltage of the DC/DC converter, which in turn affects how much current is going to the battery. If the calculated power drops compared to the previous measurement it will have to make some semi-intelligent decision about increasing or decreasing the load. It is a continuous trial and error to make the calculated power as high as possible.

author: SEnergy
author_rank: Contributor
timestamp: August 26, 2015, 09:33:25 pm
content: QuoteBUT the propeller is connected to the DC motor, which, when spinning is producing voltage AND current even without load connected to it

author: Hideki
author_rank: Frequent Contributor
timestamp: August 26, 2015, 10:06:34 pm
content: Quote from: SEnergy on August 26, 2015, 09:33:25 pmOh really? Where exactly does that current go when nothing is connected?

author: sdg
author_rank: Regular Contributor
timestamp: August 26, 2015, 10:40:35 pm
content: Quote from: SEnergy on August 26, 2015, 09:33:25 pmThere. Is. No. Current. Without. A. Load.None.Never.

author: retrolefty
author_rank: Super Contributor
timestamp: August 27, 2015, 01:14:16 am
content: Quote from: Hideki on August 26, 2015, 10:06:34 pmOP is just refusing to get a handle on Ohms law before trying to design his system. Many have tried, many have failed. Perhaps your comment will turn on a light bulb for him.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 27, 2015, 01:33:32 am
content: Quote from: SEnergy on August 26, 2015, 09:33:25 pmOnce again, Ohm's Law...I=V/RIf there's no load, R=infinity and I=0.

author: SEnergy
author_rank: Contributor
timestamp: August 27, 2015, 07:41:48 am
content: Quote from: sdg on August 26, 2015, 10:40:35 pmso you are saying my multimeter is broken?

author: SEnergy
author_rank: Contributor
timestamp: August 27, 2015, 07:44:19 am
content: Quote from: retrolefty on August 27, 2015, 01:14:16 amI'm not refusing anything, Im just stating the factdoesn't DC motor have copper wires in it? oh wait, it does...don't copper wires have resistance? oh wait, they do...so, who's the one ignoring the ohm's law and ignoring the resistance within the motor? oh wait, that's you

author: Hideki
author_rank: Frequent Contributor
timestamp: August 27, 2015, 08:03:35 am
content: Quote from: SEnergy on August 27, 2015, 07:44:19 amIf you measure the current by connecting your multimeter in amps range... guess what the load is... --> it's the multimeter!We are saying: WITHOUT a loadDid you measure the voltage at the same time, WITH the load connected? If not, the measurement is mostly meaningless if you want to calculate the power.

author: SEnergy
author_rank: Contributor
timestamp: August 27, 2015, 08:13:30 am
content: Quote from: Hideki on August 27, 2015, 08:03:35 amQuotebut wait, if there's load it actually produces voltage AND current! and there's load ALL the time, otherwise the wind turbine would just spin around without any effect, which could be achieved just by placing the propeller on a stick faced towards the wind -> that is not a wind turbine, that is a propeller on a stick, isn't it?also what about the copper that's DC motor full of? isn't THAT considered a LOAD? it has a resistance, motor is producing voltage, how come suddenly in this case an ohm's law doesn't take place?

author: SEnergy
author_rank: Contributor
timestamp: August 27, 2015, 08:27:18 am
content: Quotesince you said the multimeter IS the load in this case, I did the same thing but measured the voltage with second multimeter, guess what? blowing slightly on the fan was producing ~100mV and ~5mA, at least those are the numbers I was able to catch at the same time

author: Hideki
author_rank: Frequent Contributor
timestamp: August 27, 2015, 08:40:16 am
content: If the two terminals of the moter is open - nothing connected to them - the circuit is broken. The current can not jump through the air unless there's several kilovolts between them. This is the "no load" situation.If you short the two terminals together, it is the resistance of the copper that limits the current through the motor. In addition you're using the magnetic field produced to actively brake it and slow it down.You have also determined that slightly blowing can generate 0.5 mW in your multimeter shunt resistor.

author: SEnergy
author_rank: Contributor
timestamp: August 27, 2015, 08:53:13 am
content: Quote from: Hideki on August 27, 2015, 08:40:16 amso this statementQuotein fact is not true, because a wind turbine without a load is absolutely useless and is just a "tower" with propeller attached to it, NOT a wind turbine, correct?

author: Hideki
author_rank: Frequent Contributor
timestamp: August 27, 2015, 09:02:18 am
content: Do you REALLY want to use all your energy on attacking people who try to help your understanding? It looks like the answer is yes. Well, it's your choice.It is no more useless than a lightswitch that's turned off because you don't want the light to shine at the moment. There will still be voltage, but no current.Once you turn the switch on, hopefully some kind of light is connected (a load) and current starts flowing through it.

author: SEnergy
author_rank: Contributor
timestamp: August 27, 2015, 09:18:54 am
content: Quote from: Hideki on August 27, 2015, 09:02:18 ami'm not using all my energy on attacking people who try to help me here, as the posts that were actually helpful could be squished to one page, instead of 3, while the rest are just replies from ignorant people that live and enjoy the fact that they know something more (e.g. retrolefty), and instead of actually providing an information with explanation why I'm wrong and why they are correct so I can understand it they are just arrogantthe ppl who actually helped me here and taught me something are suicidaleggroll and you, while some of the rest were here just to boost their ego...hopefully last question; I still don't quite understand how the regulator (at least the one I posted above) handles current -> linear regulator outputs the same current as it inputs, how about the switching one? how is the output current managed?

author: sdg
author_rank: Regular Contributor
timestamp: August 27, 2015, 09:58:47 am
content: Quote from: SEnergy on August 27, 2015, 07:41:48 amNo.Your multimeter is a load.As soon as you connect it, the "even without load connected to it" part isn't true anymore.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: August 27, 2015, 01:11:43 pm
content: Quote from: SEnergy on August 27, 2015, 09:18:54 amYou're working in the wrong direction. The output current is not "managed" by the regulator in either case, it's managed by the load. The regulator controls its input current.A linear regulator, wired normally, outputs a voltage (V). When you connect a load (R), the load draws a current equal to I=V/R. In order to supply this current to the load, the regulator draws the same amount of current from its input. The efficiency is calculated as Pout/Pin, but since Iout=Iin, you can get the efficiency just by Vout/Vin. Linear regulators can only drop the voltage down, they can't raise it up.A switching regulator also outputs a voltage (V). When you connect a load (R), the load draws a current equal to I=V/R. In order to supply this current to the load, the regulator must draw AT LEAST as much current from its input to satisfy Pout <= Pin. It does this by pulsing current from the source and storing it in a magnetic or electric field before delivering it to the load, which allows it to step the voltage up or down. So the current drawn from the source isn't continuous, it's pulsed, but stick a big cap on there and you can smooth most of that out. The best way to estimate the average current drawn from the source is to use the manufacturer's provided efficiency numbers. You should find some chart or table in the datasheet that gives you the approximate efficiency as a function of input/output voltage and output current. Once you know the efficiency, then the power input will be Pout/eff, and the current input will be Pin/Vin. So for example with a voltage input of 30V, output voltage of 15V, current output of 2A, and efficiency of 90%, you get 30W out, 33.3W in, and 1.11A in.

author: SEnergy
author_rank: Contributor
timestamp: August 27, 2015, 04:04:25 pm
content: Quote from: suicidaleggroll on August 27, 2015, 01:11:43 pmgreat explanation, thanks!